expansion of gas

Fundamental Physics-I: by Dr. Abhijit Kar Gupta (email: [email protected]) 1

Expansion of Gas

Basic facts: • Thermal expansion or contraction occurs in a gas similar to that in a liquid. As in

the case of a liquid, the expansion of volume is only important. The discussion of the expansion of length or surface of gas is useless as a gas has no definite shape.

• The amount of volume expansion in a gas is much larger compared to that in a solid or a liquid. Thus the thermal expansion of the solid vessel or container, which contains the gas, is usually neglected in comparison to the expansion of the gas.

• Experiments show that all gases at very low densities exhibit the same expansion behaviour. A thermometer that uses a gas therefore shows the same readings irrespective of which gas is used.

• The change in volume of a gas also occurs due to pressure applied on it. The role of pressure on the volume of liquid or solid is not important, in general. The volume of a gas increases when the pressure over it is decreased and it decreases when the pressure is increased.

Ideal Gas and Gas Laws: The volume expansion in a gas occurs due to both temperature and pressure. To know the behaviour or the state of a certain quantity (mass) of a gas, we have to know its pressure ( P ), volume (V ), and temperature (T ). These three variables are related to one other. If one of the variables is kept constant and the other two are allowed to vary, we find three different relations. These are called gas laws. Within a certain range of temperatures and pressures, many gases have been found to follow three simple laws, namely Boyle’s law, Charles’ law and Gay-Lussac’s law. A gas that follows these laws completely is an idealization called an ideal gas.

Boyle’s law: When the temperature is held constant, the volume of a certain mass of a gas is inversely proportional to its pressure. The relation was discovered in 1660 by the British Chemist Robert Boyle (1627-1691).

Note: 1. The temperature referred to in the gas laws is always the absolute

temperature. 2. The pressure that is referred to is the pressure exerted due to the gas

molecules on the inner walls of the vessel. In equilibrium, the pressure due to gas is equal to the external pressure on the gas.


If V is the volume of a certain mass of a gas and P is the pressure, we can write

according to Boyle’s law: ∝V P1 at constant temperature.

Thus we have, =PV K (constant); the value of the proportionality constant K depends on the nature, mass and temperature of the gas. If the volume of the gas is 1V at pressure 1P , 2V at pressure 2P and so on, we can write,

2211 VPVP = =….= K . Therefore, Boyle’s law can also be stated as follows: When temperature is held constant, the product of pressure and volume of a certain amount of gas is constant: =PV constant. The relation between density and pressure from Boyle’s law: Let us suppose, the mass of a certain amount of gas is m . The gas is held at constant temperature. The volume and density of the gas are 1V and 1ρ at pressure 1P and that are

2V and 2ρ at pressure 2P . According to Boyle’s law,

2211 VPVP = Or, mVP

mVP 2211 = [Dividing both sides by m ]

Or, 2

2

1

1

ρρPP

= [Q 1

1 Vm

=ρ and 2

2 Vm

=ρ ]

∴ =ρP const. Or, ρ∝P

Therefore, we can write: If the temperature is held constant, the density of a certain mass of a gas is proportional to its pressure. Graphical representation of Boyle’s law: Boyle’s law can be expressed by a number of ways through graphical plots. The plots involve pressure ( P ) and volume (V ) at constant temperature. Thus they are called isotherms.

• VP − graph: If we plot a graph considering the volume,V as abscissa and the pressure, P as ordinate keeping the temperature of a gas constant, we obtain a rectangular hyperbola. This is an isotherm. We get different isotherms (different rectangular hyperbolas) for different constant temperatures as shown in the following plot [fig.**].

• V

P 1− grap

Fig. to be included


When the pressure P is plotted as abscissa and the inverse of volume V1 as

ordinate keeping the temperature of the gas constant, we get a straight line passing through origin. Different straight lines (with different slopes) are obtained for different constant temperatures [see fig.**].

The P

V 1− graphs will also be similar straight lines.

• PPV − graph: When the pressure P is plotted as abscissa and the product PV as ordinate for a gas whose temperature is kept constant, we get a horizontal straight line (parallel to P -axis). Different parallel straight lines are obtained for different constant temperatures [fig.**]. The VPV − graphs will also be similar straight lines.

Charles’ law: When the pressure is held constant, the volume of a quantity (mass) of gas is directly proportional to the absolute temperature of it. If the volume of a gas is V and its temperature in absolute scale isT , we can write according to Charles’ law: TV ∝ .

We can then write, =TV const. Or,

2

2

1

1

TV

TV

= …and so on (at constant pressure).

If the volume of the gas is tV at Ct 0 and 0V at 0 C0 , then we can write,

2732730V

tVt =+

Or,

+

=273

2730

tVVt [Q Ct 0 = )273( +t K ]

∴

+=

27310

tVVt ………………..(1)

Similarly, if the volume of the gas is tV at Ct 0− , then we can write,

=tV

−

27310

tV .

Therefore, Charles’s law can also be written in the following form:

Fig. to be included

Fig. to be included


When the pressure is held constant, the volume of a certain mass of gas increases or

decreases by 2731 fraction of its volume at 0 C0 for one degree Celsius increase or

decrease of temperature. In 1787, French physicist Jacques Charles (1747-1823), who made early hot air balloon, observed that the volume of a gas under constant pressure increases or decreases with temperature. This behaviour was quantified around 1808 by another French scientist, Joseph Gay-Lussac, who measured the thermal expansion of a gas as 1/267 of its original volume per degree Celsius. In 1847, Henri Regnault refined this value to 1/273, and also discovered that many gases violate this rule, which in principle holds only for so-called ideal gases. Gay-Lussac’s law: When the volume is held constant, the pressure exerted by a gas is directly proportional to the absolute temperature of the gas. This law is named after Joseph Louis Gay-Lussac as he made the observation in 1802. The above law is sometimes called pressure law. If P is the pressure exerted by the gas at temperature T measured in absolute scale, then according to Gay-Lussac’s law: TP ∝ .

We can then write, TP = const. Or,

2

2

1

1

TP

TP

= …and so on (at constant volume).

If the pressure of the gas at Ct 0 is tP and at 0 C0 it is 0P , then we can write as before,

2732730P

tPt =+

Or,

+=

27310

tPPt ……………………(2)

Remember:

Examples with Solutions

Example#1 The volume of a gas of certain mass is 500 cc at standard temperature and pressure. What will be the volume at a pressure of 700 mm if the temperature is kept constant?

Boyle’s law: =PV const., when T is constant

Charles’ law: =TV const., when P is constant

Gay-Lussac’s law: =TP const., when V is constant.


Solution: Since the temperature is held constant, we get from Boyle’s law, 2211 VPVP = . Here, 1P = 76 cm Hg. (pressure of 76 cm long mercury column), =1V 500 cc, 2P = 700 mm Hg = 70 cm Hg, =2V ?

∴ 27050076 V×=× Or, 85.54270

500762 =

×=V cc.

Example#2 During recording of pressure against volume of a certain mass of gas at a fixed temperature, a student forgets to write some reading. Fill in the blanks in the following recording table.

Pressure (mm) 100 125 200 ----Volume (cc) 80 ---- 40 32

Solution: As the temperature is fixed, we can apply Boyle’s law: PV = constant. Here, in the 1st observation, 80008010011 =×=VP and in the 3rd observation,

80004020033 =×=VP .

Thus in the 2nd observation, 800022 =VP Or, 8000125 2 =×V Or, 641258000

2 ==V cc.

In the 4th observation, 800044 =VP Or, 8000324 =×P Or, 25032

80004 ==P cc.

Example#3 The volume of a gas is reduced to 61 th of its initial volume by applying

pressure at constant temperature. If the initial pressure is equal to atmospheric pressure, what will be the final pressure? Solution:

As the temperature remains constant, we get from Boyle’s law: 2211 VPVP = .

Here, 1P = 1 atm. (atmospheric pressure), =1V x cc (say,), =2P ?, xV61

2 = cc.

∴ xPx611 2 ×=× Or, 62 =P atm.

Example#4 A bubble of 1 mm diameter is formed at the depth of 238 ft in a lake. What will be the diameter of the bubble when it reaches the surface of the lake? The temperature is same everywhere and the height of the water barometer is = 34 ft. [J.E.E. ‘00] Solution:

The volume of the bubble at the bottom of the lake, 31 )05.0.(

34π=V cc.

The pressure on the bubble at the bottom, 342381 +=P = 272 ft water column’s pressure. If the diameter of the bubble is d cm at the surface of the lake, the volume of the bubble

at that place, 3

2 234

=

dV π cc.


The pressure on the bubble at that place is =2P 34 ft water column’s pressure. The pressure inside the bubble is outside pressure on the bubble at equilibrium. Keeping in mind the temperature to be same everywhere, we can apply Boyle’s law:

2211 VPVP = Or, ×272 3)05.0.(34π = ×34

3

234

dπ

Or, 008.034

8)05.0(272 33 =

××=d Or, 2.0=d cm.

Example#5 An air bubble of volume 20 cc is formed at the 40 m depth of a lake. This bubble rises up the water surface. What will be its volume just before it comes up the water surface? (The standard atmospheric pressure = 76 cm Hg.) [H.S. ‘01] Solution.

The volume of air bubble, 201 =V cc at the bottom of the lake. The pressure on the bubble at the bottom of the lake,

980)140006.1376(1 ××+×=P dyne/cm 2 Let the volume of the air bubble just before it comes out of water surface is 2V . The pressure on the air bubble at this time, 9806.13762 ××=P dyne/cm 2 . As the temperature of the bubble does not change, we apply Boyle’s law,

2211 VPVP = Or, 29806.137620980)40006.1376( V×××=××+×

Or, 4.976.1376

20)40006.1376(2 =

××+×

=V cc.

Example#6 A movable piston is fitted at the upper end of a 100 cm long vertical cylinder. The cylinder contains an ideal gas. At the initial stage, when the system is at equilibrium with the piston between the confined gas and the atmosphere, the length of the column of gas in the cylinder is 90 cm. Now mercury is being poured on the piston from top. When the mercury is about to overflow, the piston goes down by 32 cm. Find the atmospheric pressure. Assume that the temperature of the gas remains constant. The thickness and the mass of the piston are negligible. [I.I.T.] Solution: Let us say, the atmospheric pressure = P cm Hg. At the initial stage at equilibrium, the pressure of the confined gas = the atmospheric pressure, P cm Hg. The initial volume of the gas, α90=V cc [ =α area of cross-section of the cylinder, in sq. cm] When the mercury is about to overflow, the length of mercury column on the piston = 42 cm (see fig.**). The pressure of gas, )42(1 += PP cm Hg, the volume of gas, α581 =V cc. As the temperature remains constant, we have from Boyle’s law:

11VPPV = Or, αα 58).42(90. += PP Or, 4258.58.90 ×+= PP

Fig. to be included


Or, 425832 ×=P Or, 125.7632

4258=

×=P cm Hg.

Example#7 The volume of a gas is made double at constant pressure by increasing temperature. What is the final temperature of the gas if the initial temperature is 13 C0 ? [H.S.(T)] Ans.

As the pressure is kept constant, we apply Charles’ law: 2

2

1

1

TV

TV

= .

According to question the final volume, 12 .2 VV = . We have, 286132731 =+=T K , =2T ?

∴ 2

11 .2286 T

VV= Or, 5722 =T K

∴In Celsius scale, the final temperature is = 572 – 273 = 299 C0 . Example#8 The volume of a certain mass of gas at 47 C0 is 640 cc and the pressure is 75 cm Hg. When will the pressure be double if the gas is heated keeping the volume unchanged? [H.S.] Ans.

Here, 751 =P cm Hg, 472731 +=T = 320 K , and 1507522 =×=P cm Hg, ?2 =T Applying Gay-Lussac’s law (pressure law),

2

15032075

T= Or, 6402 =T K . In Celsius scale, 3672736402 =−=t C0 .

Example#9 The temperature of a gas having volume 5 litre is changed from 0 C0 to 35 C0 and its volume is increased by 640 cc. Find the absolute zero temperature for the gas in Celsius scale. [H.S.] Ans.

Let the absolute temperature for the gas is = CT 0− . ∴The initial temperature, 1T = 0 C0 = T K and the final temperature is 352 =T C0 =

)35( +T K . The initial and final volumes are 50001 =V cc and 564064050002 =+=V cc. Following Charles’ law,

2

2

1

1

TV

TV

= Or, 35

56405000+

=TT

Or, TT .564)35.(500 =+

Or, 17500.64 =T Or, 43.27364

17500==T

∴ The absolute temperature at Celsius scale is C043.273− . Example#10 Someone measures the air pressure of his car tyre to be 30 lb/sq. inches. At this time, the temperature of air is 27 C0 and the air pressure is 15 lb/sq. inches. Later in


the afternoon, the person goes to a city where the atmospheric temperature is 12 C0 and the atmospheric pressure is 10 lb/sq. inches. How much reading will the pressure gauge show for the pressure of a car tyre at that time? Consider the volume of a car tyre to be the same in two cases. [J.E.E. ‘89] Ans. The pressure gauge measures a pressure that is something above the atmospheric pressure. Therefore, the actual air pressure inside the tyre in the first case is 4515301 =+=P lb/sq. inches. The temperature in the first case, 300272731 =+=T K . Let the reading in the pressure gauge in the second case = x lb/sq. inches. ∴The actual pressure of air in the tyre is 2P = )10( +x lb/sq. inches. The temperature in this case is 285122732 =+=T K . As the volume remains constant,

2

2

1

1

TP

TP

= Or, 285

1030045 +

=x Or, 75.42

3002854510 =

×=+x Or, 75.32=x lb/sq. inches.

Example#11 A hydrogen cylinder can withstand an internal pressure of 1000 lb/sq. inches. When the temperature is 15 C0 , the pressure of hydrogen gas in the cylinder is 240 lb/sq. inches. At what temperature the cylinder will about to explode? [J.E.E.] Ans.

The initial pressure, 2401 =P lb/sq. inches and temperature, 288152731 =+=T K . When the cylinder is about to explode, the pressure inside will be 10002 =P lb/sq. inches; =2T ? As the volume of the gas remains constant in the cylinder,

2

2

1

1

TP

TP

= Or, 2

1000288240

T= Or, 1200

2401000288

2 =×

=T K .

∴The required temperature in Celsius scale = 1200 – 273 = 927 C0 . Volume and Pressure Coefficients of Gas: A certain mass of gas can be heated in two ways:

• When the pressure is kept constant, the volume changes according to Charles’ law

as given by equation (1):

+= tVVt .

273110 …………….(3)

• When the volume is kept constant, the pressure changes according to Gay-

Lussac’s law as given by equation (2):

+= tPPt .

273110 …………(4)

Thus the equations (3) and (4) can be viewed as the expressions for thermal expansion of volume and pressure of the gas:


).1(0 tVV pt γ+= ……………….(5) and ).1(0 tPP vt γ+= ………………...(6) Here, pγ is called the volume coefficient at constant pressure as the expansion of volume is considered when the pressure is held constant. Similarly, vγ is the pressure coefficient at constant volume as the increase of pressure is considered when the volume is held constant. Comparing equations (5) and (6) with equations (3) and (4),

2731

== vp γγ = 0.00366 C0/ .

Note that t is the increase in temperature measured from C00 . The definitions of the volume and pressure coefficients can be obtained from equations (5) and (6):

tV

VtVVVt

p .. 00

0 ∆=

−=γ ……….(7) and

tP

PtPPPt

v .. 00

0 ∆=

−=γ . ………..(8)

If we put 10 =V and 1=t in (7), we have, Vp ∆=γ . For a value of 1=t , we have to

consider the increase in temperature from 0 C0 to 1 C0 . Thus we have the following definition of volume coefficient. Definition of

Volume coefficient ( pγ ): When the temperature of a certain mass of gas is increased

from C00 to 1 C0 at constant pressure, the increase in volume of the gas for each unit volume is called the volume coefficient. Similarly, if we put 10 =P and 1=t in (8), we have, Pv ∆=γ . Thus we have the following definition of pressure coefficient. Definition of

Pressure coefficient ( vγ ): When the temperature of a certain mass of gas is increased from C00 to 1 C0 at constant volume, the increase in pressure of the gas for each unit pressure is called the pressure coefficient. Examples with Solutions Example#1 The pressure of a gas becomes 900 mm Hg when the temperature is raised to 50 C0 at constant volume. Find the pressure coefficient of the gas. Solution.


The pressure of a gas at 0 C0 is 760 mm Hg (the standard temperature and pressure). We know, ( )tPP vt γ+= 10 , =vγ the pressure coefficient at constant volume. Here, 900=tP mm Hg, 7600 =P mm Hg, 50050 =−=t C0 .

∴ CtPPPt

v0

0

0 /00368.050760

14050760760900

.=

×=

×−

=−

=γ .

Example#2 The volume of a certain amount of gas is 500 cc at 80 C0 at 600 cc at 150 C0 . What is the volume expansion coefficient of the gas? Solution.

We know, ( )tVV pt γ+= 10 , where pγ is the volume expansion coefficient at constant pressure and 0V is the volume at 0 C0 . ∴ )801(500 0 ×+×= pV γ ……….(1) and )1501(600 0 ×+×= pV γ ………..(2)

Dividing (2) by (1), p

p

γγ

.801.1501

56

+

+= Or, pp γγ .7505.4806 +=+

Or, 1.270 =pγ Or, Cp0/

2701

=γ .

Example#3 A glass vessel contains air at 30 C0 . At what temperature has the vessel to

be heated, keeping pressure constant, so that 31

fraction of air comes out? Cp0/

2731

=γ .

[H.S. ‘02] Ans. Method-1

Suppose, the volume of air at 0 C0 is 0V and the volume is 30V at 30 C0 where the pressure is kept constant. So, we can write, )301(030 ×+= pVV γ ………………….(1) Let the required temperature is t C0 . If the volume is tV at this temperature, we can write, )1(0 tVV pt ×+= γ ……………..(2) Dividing (2) by (1) we get,

p

pt

tVV

γγ

.1.301

30 +

+= ………………..(3)

According to question, 31 fraction of the volume of gas comes out when heated up to a

temperature, t C0 .

Hence, 31 fraction of gas remains and this volume is equal to the volume of the vessel

which in turn is equal to the volume of gas at 30 C0 .


Therefore, we can write, 3032 VVt = Or,

23

30

=VVt ……….(4)

Comparing (3) and (4) we get,

p

p

t γγ

.1.301

23

+

+= Or, pp t γγ ..22.903 +=+

Or, 5.1812

902732901

=+

=×

×+=

p

ptγγ

C0 .

Method-2

(In this method, we do not need to use the pressure coefficient, pγ .) Suppose, the initial volume of air is 1V cc.; the temperature, 303302731 =+=T K .

If the volume of air is 2V cc at a temperature of 2T K , the volume of emergent air = 231V

cc; the volume of air that remains in the vessel should be 1V cc.

∴ 212 31VVV += Or, 12 2

3VV = cc.

As the pressure remains constant,

2

2

1

1

TV

TV

= Or, 2

11 2/3303 T

VV= Or, 5.454

2909

2 ==T K .

∴The required temperature in Celsius scale is = 454.5 – 273 = 181.5 C0 . Equation of State of an Ideal Gas (Ideal Gas Law): Let us consider Boyle’s law: =PV constant, when the temperature T is constant and

Charles’ law: =TV constant, when the pressure P is constant.

When P, V and T all vary at the same time, we can combine Boyle’s law and Charles’ law to get,

=T

PV const.

∴We can write, TkPV .= , where the value of the constant k depends on the mass of the gas and on the units of P , V and T . If the volume of a gas is 1V at a pressure 1P and at temperature 1T and the volume is 2V at pressure 2P and at temperature 2T , we can write,

2

22

1

11

TVP

TVP

= .



Example#1 Some amount of gas is taken at normal temperature and pressure and then it is heated to make its pressure and volume double. Find the final temperature of the gas in Celsius scale. [H.S. ‘93] Solution. From the equation of state of an ideal gas we can write,

2

22

1

11

TVP

TVP

=

Here, 761 =P cm, xV =1 cc (say,) 2731 =T K , 7622 ×=P cm, xV 22 = cc, ?2 =T

∴ 109276

2732276

11

1222 =

××××

==xx

VPTVP

T K .

The final temperature in Celsius scale = 1092 – 273 = 819 C0 . Example#2 At a temperature of 27 C0 and a pressure of 76 cm Hg, 100 cc gas is collected over water. The volume occupied by the gas is saturated by water vapour. What is the volume of dry gas at Normal temperature and pressure (N.T.P.)? The highest pressure of water vapour at 27 C0 is = 17.4 mm Hg. Solution.

Initially, the volume of the gas is =1V 100 cc, the temperature, 300272731 =+=T K and the pressure of dry gas is 26.7474.1761 =−=P cm Hg. The temperature and pressure at N.T.P. are 762 =P cm Hg, and 02 =T C0 =273 K . Suppose, the volume of dry gas at N.T.P. is = 2V cc. We apply gas law,

2

22

1

11

TVP

TVP

= Or, 273

76300

10026.74 2V×=

×

Or, 91.88763

27326.742 =

××

=V cc.

Avogadro’s hypothesis: Avogadro’s hypothesis, formulated in 1811, states that equal volumes of gas at the same pressure and temperature contain equal numbers of molecules. One mole or one gram-mole: One mole of a gas is the quantity that contains 2310023.6 ×=AN molecules (Avogadro’s number) and one gram-mole is the molecular weight of one mole gas expressed in grammes. Example: The mass of one mole of oxygen is 32.0 gm.


Universal Gas Constant: When one gram-mole or one mole gas is taken, the constant k is denoted by R . According to Avogadro’s hypothesis, one gram-mole of all gas at same temperature and pressure has same volume. Thus the constant R for any one gram-mole ideal gas will be the same. For this reason, R is called universal gas constant (or molar gas constant). Therefore, we can write for one gram-mole of any gas: RTPV = . If we take n -gram mole a gas, the ideal gas equation becomes: nRTPV = .

If we take a gas of m -gram whose molecular weight is M , we have Mmn = .

∴ RTMmPV = .

If we now compare the above with the equation of state kTPV = , we find MRmk .= ; the

constant, MRr = is called specific gas constant. This is different for different gases as the

molecular weights of different gases are different. Value of Universal Gas Constant:

For one gram-mole gas, 0

00

TVP

TPVR == , where 0P = Normal pressure, 0T = Normal

temperature and 0V = the volume of one gram-mole gas at normal temperature and pressure (N.T.P.). We know, 0P = 76 cm Hg = 9806.1376 ×× dyne/cm 2 , 27300

0 == CT Kelvin and following Avogadro’s hypothesis, 4.220 =V litre/mole = 22400 cm 3 /mole.

∴273

224009806.1376 ×××=R = 71031.8 × dyne-cm/(mole Kelvin)

= 71031.8 × erg/(mole K) = 31.8 Joule/(mole K). The value of specific gas constant for different gases:

Hydrogen: Molecular mass = 2

∴ 16.4231.8

2===

Rr J/(mole K)

Nitrogen: Molecular mass = 28

∴ 297.02831.8

28===

Rr J/(mole K)

Oxygen: Molecular mass = 32


∴ 26.03231.8

32===

Rr J/(mole K)

Carbon di-oxide: Molecular mass = 44

∴ 189.04431.8

44===

Rr J/(mole K)

Boltzmann Constant: The total number of molecules in a gas, N , is equal to the product of the number of moles of the gas and the number of molecules per mole (i.e., Avogadro’s number, AN ): AnNN = . Therefore, the ideal gas law for n -gram mole a gas can also be written in the following form:

TNkTRNNnRTPV B

A

=== . ,

where 231038.1 −×==A

B NRk J/K is known as Boltzmann constant.

Examples with Solutions Example#1 If the mass of 1 litre hydrogen at N.T.P. is 0.0896 gm, find the value of the universal gas constant .R Solution. At N.T.P., the volume of 0.0896 gm hydrogen is = 1000 cc. ∴At N.T.P., the volume of 1 gram-mole or 2 gm hydrogen ,

V = 20896.0

1000× = 22321.4 cc.

Normal (standard) pressure, P = 76 cm Hg = 9806.1376 ×× dyne/cm 2 and normal temperature, 27300 == CT K .

∴ 71028.8273

4.223219806.1376×=

×××==

TPVR erg/(mole Kelvin).

Example#2 If the volume of 10 gm oxygen at 20 C0 and at 2 atmospheric pressure is 3.76 litre, find the universal gas constant R . Solution. We know, nRTPV = . Here, =P 2 atm. = 9806.13762 ××× dyne/cm 2 , 76.3=V litre = 3760 cc,

3210

=n gm-mole, and 29320273 =+=T K .

∴ 7103.8293

3210

37609806.13762×=

×

××××==

nTPVR erg/(mole K).


Example#3 Find the gas constant k of 1 gm air. The density of air at NTP is = 1.293 gm/litre and the density of mercury = 13.6 gm/cc. [J.E.E. ‘96] Solution.

If the gas constant for 1 gm air is k , we can write, T

PVk = .

Here, the volume of 1.293 gm air = 1000 cc.

∴The volume of 1 gm air = 293.1

1000 cc., the pressure 9806.1376 ××=P dyne/cm 2 ,

273=T K .

∴ 710287.0293.1273

10009806.1376×=

××××

=k erg/(gm K).

Variation of Density with Pressure and Temperature: Let us take an ideal gas of mass, m . If the volume of the gas is 1V and density, 1ρ at a temperature, 1T and the volume and density are 2V and 2ρ respectively, at a temperature,

2T , we can write, 1

1 ρmV = and

22 ρ

mV = .

If 1P be the pressure of the gas at temperature, 1T and 2P be the pressure at temperature,

2T , we have,

2

22

1

11

TVP

TVP

= Or, 22

2

11

1

TmP

TmP

ρρ= Or,

22

2

11

1

TP

TP

ρρ=

∴ (i) When the temperature of the gas is changed, keeping the pressure constant,

=Tρ const. Or, T1

∝ρ

(ii) When the pressure of the gas is changed, keeping the temperature constant,

=ρP const. Or, P∝ρ .


Example#1 The temperature and atmospheric pressure on a hill are C07 and 70 cm Hg, respectively; the temperature and pressure at the foot of the hill are 27 C0 and 76 cm Hg respectively. Compare the densities of air at the top and bottom of the hill.

=TPρ

const.


[H.S. ‘92] Solution.

We know, 22

2

11

1

TP

TP

ρρ=

Here, =1P 70 cm Hg, 28072731 =+=T K ; =2P 76 cm Hg, 300272731 =+=T K .

∴300

76280

70

21 ×=

× ρρ Or,

7675

2807630070

2

1 =××

=ρρ

∴ 1ρ : 2ρ = 75 : 76. Example#2 The density of air at N.T.P. is 1.29 gm/litre. Find the mass of 10 litre air at pressure of 5 times the normal pressure and at a temperature of 127 C0 . Solution.

We know, 22

2

11

1

TP

TP

ρρ=

Here, =1P standard pressure = 76 cm Hg, 2731 =T K , =1ρ 1.29 gm/litre; =2P 765× cm Hg, 4001272732 =+=T K

∴ 4.440076

76527329.1

21

2112 =

××××

==TP

PTρρ gm/litre

∴The mass of 10 litre air = 104.4 × = 44 gm. Example#3 The density of Argon is 1.6 gm/litre at a temperature of 27 C0 and a pressure of 76 cm Hg. A 200 cc glass bulb is filled with Argon. If the pressure of Argon gas in the bulb is 75 cm Hg and the average temperature is 127 C0 , find the mass of Argon. Solution.

Suppose, the density of argon at 75 cm Hg pressure and at 127 C0 temperature is 2ρ gm/litre.

We know, 22

2

11

1

TP

TP

ρρ= .

Here, =1P 76 cm Hg, =1ρ 1.6 gm/litre, 300272731 =+=T K ; =2P 75 cm Hg, 4001272732 =+=T K .

∴400

753006.1

76

2 ×=

× ρ Or, 184.1

40076753006.1

2 =×

××=ρ gm/litre

∴The mass of 200 cc Argon is = 2001000

184.1× = 0.2368 gm.

Example#4 The atmospheric pressure is 75 cm and the temperature is 27 C0 at some place. The atmospheric pressure is 70 cm and the temperature is 17 C0 at some other place. Compare the density of air at this two places. [J.E.E. ‘99] Solution.


We know, 22

2

11

1

TP

TP

ρρ= .

Here, =1P 75 cm Hg, 300272731 =+=T K ; =2P 70 cm Hg, 290172732 =+=T K .

∴290

70300

75

21 ×=

× ρρ Or,

2829

3007029075

2

1 =××

=ρρ

∴ 1ρ : 2ρ = 29 : 28 More Examples with Solutions Example#1 A bubble of diameter 1 mm is created at the bottom of a lake. The diameter of the bubble becomes 2 mm when it comes to the water surface. If the air pressure is 76 cm Hg, what is the depth of the lake? The density of mercury is 13.6 gm/cc. [ H.S. ‘90] Solution. Let the depth of the lake = h cm. At the bottom of the lake, the pressure, 1P = atmospheric pressure + water pressure = 98119816.1376 ××+×× h = 981)6.1376( ×+× h dyne/cm 2 and

the volume, 1V = 3

201

34

π cc.

At the water surface, the pressure, =2P the atmospheric pressure = 9816.1376 ×× dyne/cm 2 and

the volume, 2V = 3

101

34

π cc.

According to Boyle’s law, 2211 VPVP =

Or, ××+× 981)6.1376( h3

201

34

π = ××× 9816.1376

3

101

34

π

Or, 6.137681)6.1376( ×=×+× h Or, 6.13766.13768 ×−××=h

Or, 76.1376)18(6.1376 ××=−××=h = 7235.2 cm = 72.352 m. Example#2 During making an electric vacuum tube, it is sealed at 27 C0 and at a pressure of 61012 −× cm. The volume of the tube is 100 cc. Find the number of gas molecules that are left in the tube. The Avogadro number, i.e., the number of molecules of any gas in a volume, 22.4 litre at N.T.P. is 231002.6 × . [J.E.E.] Solution. Let us assume, there are n - molecules that are left in the sealed tube.

The density of gas at N.T.P. = 3

23

104.221002.6×× molecules/cc.


We know, 22

2

11

1

TP

TP

ρρ=

Here, 61 106.12 −×=P cm Hg,

1001n

=ρ molecules/cc, 300272731 =+=T K ;

762 =P cm Hg, 3

23

2 104.221002.6××

=ρ molecules/cc, 2731 =T K .

∴2731002.6

104.2276300

100102.123

36

××××

=×

×× −

n

Or, 133

236

1086.3104.2276300

2731002.6100102.1×=

××××××××

=−

n molecules.

Example#3 During making an electric bulb of volume, 250 cc has been sealed at a temperature of 27 C0 and at a pressure of 310− mm. Find the number of molecules in this bulb. (Avogadro number = 23100.6 × ). [H.S. ’99; I.I.T.] Solution.

The volume of air in the bulb, 2501 =V cc, the air pressure inside the bulb, 31 10−=P mm

Hg = 410− cm Hg, the temperature, 300273271 =+=T K . Let there are n -molecules inside the bulb.

∴The density of air in the bulb is 2501n

=ρ molecules/cc.

At normal temperature and pressure (N.T.P.), that is at a pressure, 762 =P cm Hg, and at a temperature, 2732 =T K the volume, =2V 22.4 litre = 22400 cc. contains 23100.6 × number of molecules.

∴The density of air at NTP is 22400

100.6 23

2×

=ρ molecules/cc.

We know, 22

2

11

1

TP

TP

ρρ=

∴273100.6

2240076300

2501023

4

×××

=××−

n

Or, 1516234

10018.810224763273625

2240076300273100.625010

×=×××××

=××

××××=

−

n .

Example#4 If the temperature of a gas is increased from 15 C0 to 25 C0 at fixed pressure, the volume increases in the ratio 1:1.035. Determine the absolute zero temperature in Celsius scale. Solution. Let the absolute zero temperature is x C0 below the freezing point (0 C0 ). Then the temperature of C015 in absolute scale will be )15(1 xT += K and the temperature of 25 C0 in absolute scale will be )25(2 xT += K . According to Charles’ law,


2

2

1

1

TV

TV

= Or, xx

TT

VV

++

==2515

2

1

2

1

Or, xx

++

=2515

035.11 Or, )15(035.125 xx +×=+ Or, 7.270=x

∴The absolute temperature is = -270.7 C0 . Example#5 A 5 litre and a 3 litre container contain air at 3 times and 7 times atmospheric pressure, respectively. What will be the common air pressure in two containers when they are joined by a narrow pipe? [I.I.T.] Solution.

Let us assume, there are 1n and 2n gm-moles of air in the two containers. Therefore, using nRTPV = formula we can write, RTn135 =× ; RTn173 =× . Adding the above two expressions, RTnn )(2115 21 +=+ ∴ RTnn )(36 21 += …………….(1) When the two containers are connected by pipe, there will be total )( 21 nn + gm-moles of air in the system of total volume, (5+3) = 8 litre, at the same temperature. Let the common pressure be P atm ( P times the atmospheric pressure). Then we can write,

RTnnP )(8 21 +=× …………….(2) Dividing (1) and (2),

368 =×P Or, 5.46

36==P atm.

Example#6 Two bulbs of equal volume are filled with a gas at NTP after they are joined by a narrow pipe. Now, one bulb is kept in melting ice and the other is immersed in water of 62 C0 . What will be the pressure of gas? Ignore the volume of t he pipe? [I.I.T ‘85] Solution. Let the volume of each bulb is V cc and there are n gram-mole gas in each of them. Since, the bulbs are filled with the gas at NTP, we can write for each of the bulb,

27376 ×=× nRV [Q ]nRTPV =

∴Total number of gm-moles in the two bulbs is 2273

762 ×××

=R

Vn .

If the common pressure in the system in second case is P ,

the number of gram-moles in one bulb, 2731 ×

=R

PVn and

the number of gram-moles in the other bulb, )27362(2 +×

=R

PVn .

We can write, nnn 221 =+ .

∴273×R

PV + )27362( +×R

PV 2273

76×

××

=R

V


Or, 273

2763351

2731 ×

=

+P Or, 36.84

)335273(273335273276

=+××××

=P cm Hg.

Example#7 An air bubble is rising from the bottom of a lake. The diameter of the bubble at the bottom is 3.6 mm and at the top it is 4 mm; the depth of the lake is 2.5 m and the temperature above is 40 C0 . What is the temperature at the bottom of the lake? Neglect the variation of density of water with depth. Atmospheric pressure = 76 cm Hg and =g 980 cm/s 2 . [I.I.T.] Solution. The pressure at the bottom of the lake,

980)3506.1376(98013509806.13761 ×+×=××+××=P dyne/cm 2 . The volume of the bubble at the bottom of the lake,

31 )18.0(

34π=V cc; the temperature, =1T ?

At above, the pressure, 9806.13762 ××=P dyne/cm 2 , the volume, 32 )2.0(

34π=V cc;

the temperature, 313402732 =+=T K .

We know, 2

22

1

11

TVP

TVP

=

Or, 313

)2.0(349806.1376)18.0(

34980)2506.1376( 3

1

3 ππ ×××=

××+×

T

Or, 313

)2.0(6.1033)18.0(6.1283 3

1

3 ×=

×T

Or, 3

3

1 )2.0(6.1033313)18.0(6.1283

×××

=T

∴ 37.2831 =T K ; the temperature in Celsius scale is = 283.37 – 273 = 10.37 C0 . Example#8 A balloon can lift a total 175 kg of weight at NTP. During the rise, when the barometer reads 50 cm and the temperature is -10 C0 , how much weight the balloon may lift? Assume the volume of the balloon to be constant. Solution. Suppose the volume of the balloon is = V .cc According to Archimedes’ principle, the volume of the displaced air is V cc. If the density of air is 1ρ kg/cc at NTP, we can write,

1751 =ρV Or, V

1751 =ρ

Let us assume that at the given height, the balloon can carry a mass of M kg. If the

density of air at that height be 2ρ kg/cc, we can write, MV =2ρ Or, VM

=2ρ

We know, 22

2

11

1

TP

TP

ρρ= Or,

2

2

1

1

175 TMP

TP

×=

×


Or, 263

50273175

76×

=× M

[Q 01 =T C0 = 273 K ; 102 −=T C0 = 263 K ]

Q 26376

50273175×

××=M = 119.5 kg.

Example#9 If a tyre is filled with air at 27 C0 and then the temperature is raised to 57 C0 , what will be the percentage increase of pressure inside the tyre? Solution.

In the first case, the pressure = 1P , the temperature, 300272731 =+=T K . In the second case, the pressure = 2P , the temperature, 330572732 =+=T K . As the volume remains constant, we can write,

2

2

1

1

TP

TP

= = 12

12

TTPP

−− Or,

1

12

1

12

TTT

PPP −

=− .

Here the quantity, )( 12 PP − is the increase of pressure.

∴The percentage increase in pressure = %1001

12 ×−P

PP

= %10%100300

300330%1001

12 =×−

=×−T

TT .

Example#10 A compartment contains a gas of mass 1m at pressure, 1P and another compartment contains the gas of mass 2m at pressure, 2P . If a passage is created between the two compartments, find the pressure of the gas mixture. Solution.

Let the volumes of the 1st and 2nd compartments are 1V and 2V , respectively. In the 1st compartment, there is gas of mass 1m at a pressure, 1P and in the 2nd compartment, there is gas of mass 2m at a pressure, 2P . Therefore, we can write,

RTMmVP 1

11 = and RTMmVP 2

22 = at a constant temperature, T ; where =R universal

gas constant and =M molecular weight of the gas.

∴ RTM

mmVPVP )( 212211

+=+ ……………….(1)

After mixture, the total volume = )( 21 VV + and the total mass = )( 21 mm + . If the pressure of the gas mixture is now P , we can write,

RTM

mmVVP )()( 2121

+=+ …………………(2)

Comparing (1) and (2),

21

2211

VVVPVPP

++

= ………………….(3)

Again we have, RTMPmV

1

11 = and T

MPmV

2

22 = .


∴The expression (3) becomes

RTMPmRT

MPm

RTMmRT

Mm

P

2

2

1

1

21

+

+= =

2

2

1

1

21 )(

Pm

Pm

mm

+

+ =1221

2121 )(PmPmPPmm

++ .

Example#11 The length of a capillary glass tube, having two sides sealed, is 100 cm. In horizontal position, a mercury column of 10 cm is in the middle of it. On the two sides of this mercury column, there are air columns (of same length) at a pressure of 76 cm Hg and at a temperature of 27 C0 . If now the tube is kept horizontal in such a way that the temperature of the air column on one side becomes 0 C0 and the temperature of the air column on the other side becomes 127 C0 . Find the length and pressure of the air column at 0 C0 . Neglect the thermal expansion of glass and mercury. [I.I.T.] Solution. Let the area of cross-section of the tube = α cm. ∴The volume of mercury column = α10 cc, the volume of air column, in the first case, on either side of mercury column, α451 =V cc. [see fig**]

In the first case, the temperature, 300272731 =+=T K ; the air pressure, 761 =P cm Hg. In the second case, the air temperature at the left end is 27300

2 == CT K . If the length of the air column is l cm, the volume of air on the left, αlV =2 cc. Suppose the air pressure in the second case is P (Since the system is in equilibrium, the air pressures on two sides of mercury column are same.). Applying the following formula,

2

22

1

11

TVP

TVP

= , we get,

2733004576 αα lP×

=× ………………..(1)

Now the air temperature at the right end is 127 C0 = 273 + 127 = 400 K . The length of this air column is )90( l− cm. Thus we can write,

400)90(

3004576 αα lP −×

=× ………………(2)

Comparing (1) and (2) we get,

400)90(

273αα lPlP −×

=× Or, 5.36

67327390

=×

=l cm.

From equation (1),

Fig. to be included


2735.36

3004576 αα ×

=× P Or, 3.85=P cm Hg.

∴The length of air column at 0 C0 is = 36.5 cm and the pressure is = 85.3 cm Hg. Example#12 Two thermally insulated containers of volume 1 litre and 3 litre are connected by a pipe. The first container is filled with Nitrogen at a temperature, 0 C0 and at a pressure, 0.5 atm and the second container is filled with Argon at a temperature, 100 C0 and a pressure, 1.5 atm while keeping the tap on the connected pipe closed. If now the tap is made open, the temperature of the gas mixture becomes 79 C0 . What will be the pressure of the gas mixture? Solution.

In the first container, the pressure of gas, =1P 0.5 atm, the volume, =1V 1 litre, the temperature, =1T 273 K . In the second container, the pressure, 5.12 =P atm, the volume, =2V 3 litre, the temperature, =2T 100 + 273 = 373 K . The total volume of the gas mixture, 431 =+=V litre and the temperature,

35279273 =+=T K . Suppose, the pressure of the gas mixture after the tap is open is = P . Applying ideal gas law we can write,

TPV

TVP

TVP

=+2

22

1

11 Or, 352

4373

35.1273

15.0 ×=

×+

× P

Or, 0138.00120.00018.088

=+=P

∴ 2144.1880138.0 =×=P atm. Example#13 A 3 litre and a 1 litre glass bulb are joined by a narrow pipe. This system is filled with air at a temperature of 30 C0 and a pressure of 76 cm. The 3 litre bulb is immersed in vapour of 100 C0 and the other one is kept at 30 C0 . What will be the air pressure in the two bulbs? Neglect the expansion of 3 litre bulb. [J.E.E.] Solution.

In the first case, the total volume of the two bulbs, 4131 =+=V litre, the air pressure, 761 =P cm Hg, air temperature, 303302731 =+=T K .

In the second case, the air pressure = 2P cm Hg (say,). The temperature of air in the 3 litre bulb = 100 + 273 = 373 K , the temperature of air in the 1 litre bulb = 30 + 273 = 303 K .

We know the ideal gas law is nRT

PV= , where n is number of moles in the gas and R is

the universal gas constant. As the total number of moles in the gas remains constant, we have the total number of moles of gas in the system in the first case should be equal to the sum of moles of gas in the two bulbs kept at two different temperatures: 21 nnn += We have here,


1

11

TVPnR = ,

37332

1×

=PRn and

30312

2×

=PRn

∴303

1373

3 22

1

11 ×+

×=

PPTVP Or,

+=

×3031

3733

303476

2P

Or, 4.883733033373476

2 =+×××

=P cm Hg.

Example#14 One end of a narrow and uniform tube is sealed. Some air is trapped inside the tube with the help of a mercury column of length h cm. If the pipe is kept vertical with the closed end up, the length of air column is 1l cm. When the tube is held vertical keeping the open end up, the length of air column is 2l cm. What is the air pressure? Solution. Suppose, the air pressure = P cm Hg; the area of cross-section = α sq. cm. In the case when the closed end of the tube is on the upper side (see fig.**): the volume of enclosed air, α11 lV = cc; the pressure of this enclosed air pressure,

)(1 hPP −= cm Hg. In the other case, when the open end is up, the volume of enclosed air, α22 lV = cc; the pressure of this enclosed air pressure,

)(2 hPP += cm Hg. According to Boyle’s law, 2211 VPVP = Or, αα 21 )()( lhPlhP +=−

∴ hllllP .

21

21

−+

= cm Hg.

Example#15 Some air is trapped inside a one end open tube by a 20 cm long mercury column. The length of the air column becomes 3 cm when the tube is held vertical, keeping the open end up. When the tube is held upside down, the air column becomes 6 cm. Find the atmospheric pressure. Solution. Let the atmospheric pressure = P cm Hg; the area of cross-section = α sq. cm. In the first case, the pressure of enclosed air = )20( +P cm Hg; the volume = α3 cc. In the second case, the pressure of enclosed air = )20( −P cm Hg; the volume = α6 cc. ∴According to Boyle’s law: αα 6)20(3)20( ×−=×+ PP , solving we get, 60=P cm Hg. Example#16 A one side open glass tube of uniform cross-section contains some air at 27 C0 which is confined by 4 cm long mercury column. If the tube is held vertical, the length of air column becomes 9 cm when the open side is up and the air column is 10 cm

Fig. to be included


when it is held upside down; find (i) the atmospheric pressure, (ii) the temperature when the air column is 9 cm in upside down position. [I.I.T.] Solution.

(i) Let the atmospheric pressure is P cm Hg, the area of cross-section is α sq. cm. When the open side of the tube is up in a vertical position (see fig.**), the volume of enclosed air, α91 =V cc, the pressure of enclosed air, )4(1 += PP cm Hg. When the open side is below, the volume of enclosed air, α102 =V cc, the pressure of enclosed air, )4(2 −= PP cm Hg. According to Boyle’s law, 2211 VPVP = Or, αα 10)4(9)4( ×−=×+ PP Or, 4010369 −=+ PP Or, 76=P cm Hg. (ii) Suppose, the temperature in this case is = 2T K . We have, the pressure,

724762 =−=P cm Hg, the volume, α92 =V cc.

∴2

22

1

11

TVP

TVP

= Or, 2

97227273

9)476(T

αα ×=

+×+ Or,

2

7230080

T= Or, 2702 =T K .

∴The required temperature in Celsius scale is = 270 – 273 = 3− C0 . Example#17 A glass tube of uniform cross-section has its both sides sealed. In the horizontal position, there is a 5 cm long mercury column in the middle of the tube and the air columns are on two sides having same length. The pressure of this air is P . The tube is now made inclined by 60 0 with respect to the vertical. In this position, the lengths of air column above the mercury column and that below the mercury column are 46 cm and 44.5 cm, respectively. Find the value of P . The temperature of the system remains fixed at 30 C0 . [I.I.T. ‘86] Solution.

Let in the horizontal position, the length of each air column on two sides is = l cm. This implies, lll == 21 according to the picture (see fig.**). The length of the tube = 44.5 + 46 + 5 = 95.5 cm.

∴ 25.452

55.95=

−=l cm.

In the inclined position, the air pressure, 02 60cos5+= PP = )5.2(

21.5 +=+ PP cm Hg.

If the volume of air above is 1V and the volume of air below is 2V , we can write according to Boyle’s law, 2211 VPVP = . ∴ αα 5.44)5.2(46. 11 ×+= PP [ =α area of cross-section of the tube] Solving, 17.741 =P cm Again we can write, 11VPPV = Or, αα 4617.7425.45 ×=×P

Fig. to be included

Fig. to be included


Or, 4.7525.45

4617.74=

×=P cm Hg.

Example#18 When the barometer reading is 75 cm, an amount of 10 cc air is inserted in the empty space of the barometer in atmospheric pressure. The mercury column of the barometer goes down by 25 cm due to this. How much space is occupied by the air inside the barometer tube? [H.S.] Solution.

Initially, the pressure of the confined air, 1P = 75 cm Hg; the initial volume of air, 101 =V cc. In the second case, the volume of air = 2V (say,) and the pressure,

5025751 =−=P cm Hg.

According to Boyle’s law, 2211 VPVP = Or, 1550

1075

2

112 =

×==

PVPV cc.

∴ The air will occupy a volume of 15 cc in the barometer tube. Example#19 The reading of a faulty barometer comes down from 75 cm to 65 cm due to air gets into the empty space of it. The length of the empty column, up in the barometer tube, was 6 cm at the initial stage. If the cross-section of the tube is 1 sq. cm, how much space will be occupied by enclosed air at normal atmospheric pressure? [H.S.] Solution.

Normal atmosphere, 761 =P cm Hg Let us say, the volume of enclosed air at that pressure is = 1V cc Pressure of enclosed air, 1065752 =−=P cm Hg; Volume of enclosed air,

161)}6575(6{2 =×−+=V cc.

Applying Boyle’s law: 2211 VPVP = Or, 105.276

1610

1

221 =

×==

PVPV cc.

Thus the enclosed air occupies 2.105 cc volume at normal pressure. Example#20 When the readings of an errorless barometer are 28.5 in and 31 in, the corresponding readings of the faulty barometer are then 28 in and 30 in. What will be the reading of the errorless barometer when the reading of the faulty barometer is 29 in? [H.S. ’03; J.E.E.] Solution. Suppose, the length of the tube of barometer = l in and the area of cross-section = α in. In the first case, we can write, the volume of enclosed air in the faulty barometer tube,

α×−= )28(1 lV cubic in; the pressure of this enclosed air, 5.0)285.28(1 =−=P in Hg. In the second case, the volume of enclosed air in the faulty barometer tube,

α×−= )30(2 lV cubic in; the pressure of this enclosed air, 1)3031(2 =−=P in Hg. Applying Boyle’s law: 2211 VPVP = , we get,

1)30(5.0)28( ×−=×− αα ll Or, 32=l in. Let the reading in errorless barometer be h in when the faulty barometer reading is 29 in. ∴The volume of enclosed air = α×− )29(l cubic in; the pressure = )29( −h in Hg.


Again applying Boyle’s law, )29()29(1)30( −××−=××− hll αα

Or, )29()2932(1)3032( −××−=××− hαα Or, )29(32 −×= h

Or, 67.293229 =+=h in.

∴The required reading in the errorless barometer is 29.67 in. Example#21 A small thread of mercury separates some air from surroundings in a tube. The tube can be rotated in the vertical plane. The length of air column in tube is 1l when the tube is horizontal and it is 2l in the vertical position (the air column is up) of the tube. If the tube makes an angle α with the vertical, find the length of air column in it. [J.E.E. ‘88] Solution. Suppose, the atmospheric pressure = P , the length of the mercury thread = h , the area of cross-section of the tube = A . In the horizontal position of the tube, the air pressure in the tube = P ; volume = Al1 . In the vertical position, the air pressure in the tube = hP − ; volume = Al2 . Let the length of air column is 3l when the tube makes an angle α with the vertical. According to Boyle’s law,

AlhPAlP 21 )( ×−=× ………..(1) AlhPAlP 31 )cos( ×−=× α ………..(2)

From (1), 21 )( lhPPl −= Or, Ph

ll

−= 12

1 Or, 2

11ll

Ph

−= ……(3)

From (2), 31 )cos( lhPPl α−= Or, αα cos1

1cos1

3

PhhP

Pll

−=

−= …….(4)

Now putting (3) in (4) we get,

αcos112

1

13

−−

=

llll =

αcos)( 122

21

lllll−−

.

Note that in a faulty barometer, air introduced in the empty space, above the mercury column, forces the mercury column to go down. So, the reading in a faulty barometer will be somewhat less than the correct reading. The difference in readings is equal to the pressure (in terms of length of mercury column) exerted by air column.


Example#22 When a barometer reads 75 cm, the length of Torricelli’s vacuum is 20 cm. When 10 cc air is introduced into the barometer tube at atmospheric pressure, the barometer reading is 45 cm. Find the diameter of the barometer tube. [J.E.E. ‘95] Solution. Let the area of cross-section of the barometer tube = A sq. cm, the diameter of the barometer tube = d cm. We have the initial volume of air, 101 =V cc and the pressure of air, 751 =P cm Hg; the final volume, [ ] AAV 50)4575(202 =×−+= cc and the final pressure, 45752 −=P = 30 cm Hg. According to Boyle’s law, 2211 VPVP = Or, A50301075 ×=×

Or, 21

50301075

=××

=A sq. cm.

∴21

4

2

=dπ Or, 798.02

==π

d cm.

Discussions of a few Questions

Q.1 For expansions of solid and liquid, only temperature is referred whereas in the case of volume expansion of a gas, temperature and pressure both are mentioned. Why? Ans. When temperature is changed, volume of all kinds of matter -solid, liquid, gas changes. Thus the temperature is mentioned in case of volume expansion. For solid and liquid, the influence of pressure in the expansion of volume is negligible. However, for a gas, volume is dependent on pressure. The volume of a gas decreases as the pressure is increased and the volume increases as the pressure is decreased. Hence, the volume of a gas is dependent on both pressure and temperature. For this reason, temperature and pressure both are mentioned in the case of expansion of volume of a gas. Q.2 Liquid has a coefficient of apparent expansion whereas for a gas such a coefficient is not mentioned. Why? Ans. Liquid is heated in a container. Along with the expansion of liquid, there is an expansion of the solid container too. The coefficient of expansion of solid is less but not negligible in comparison to the coefficient of expansion of liquid. Thus the expansion of the container is not ignored. If we do not consider the expansion of the container, the coefficient of expansion of the liquid is called the coefficient of apparent expansion. When the coefficient of expansion of the container is added with the coefficient of expansion of liquid, we get the coefficient of real expansion of liquid. Like a liquid, a gas is also heated keeping it in a container. But the expansion coefficient of a gas is approximately 100 times the expansion coefficient of a solid container. Thus the expansion of container is ignored with respect to the expansion of gas. Hence, it is assumed that the coefficients of real and apparent expansions of gas are essentially the same unless a very accurate measurement is required. For a gas, only one coefficient of expansion is mentioned instead of using ‘real’ and ‘apparent’ words.


Q.3 Two similar spherical glass bulbs , filled with air, are joined by a horizontal glass tube which contains a small thread of mercury. The air temperatures in the two bulbs are 0 C0 and 20 C0 . If now the temperatures of both the bulbs are increased by 10 C0 , will there be any change of position of the mercury thread? If so, then in which direction will it shift? [J.E.E. ‘88] Ans. The mercury thread will shift towards the bulb which has higher temperature. We can assume that the mercury thread was in equilibrium in the horizontal tube at the initial stage. Let the common pressure at this stage is P . When the temperatures of the bulbs are increased by 10 C0 , the temperatures of the bulbs are now C010100 =+ and C0301020 =+ . Considering the volume of the bulbs to remain constant, we can write for the first bulb,

TP

TP

=1

1 Or, 037.1273283

27302731011 ==

++

==TT

PP

Similarly, for the second bulb,

034.1293303

273202733022 ==

++

==TT

PP

∴ We can write, 21 PP > . Hence, the mercury thread shifts towards the bulb of higher temperature. Q.4 In the definition of the coefficient of expansion of gas, the initial volume or pressure is always referred to at 0 C0 . But for a solid or liquid, this is not done – why? Ans. The value of the coefficient of expansion for a solid or liquid is small. Thus the value of pressure or volume at any temperature can be considered as the initial value. The error involved due to this is negligible. But the value of the coefficient of expansion of a gas is large. Thus the calculations done by assuming the initial value of volume or pressure of gas at different temperatures give significantly different results. Hence, for a gas, the initial value of volume or pressure is always referred to be that at 0 C0 . Q.5 What is universal gas constant? Is it same for all gases? What is the value of this constant? Ans.

For one gram mole of a gas, the value of T

PV obtained from the ideal gas law is a

constant. This constant

=

TPVR is called the universal gas constant.

The value of R is same for all ideal gases. 71031.8 ×=R erg/(mole Kelvin). Q.6 Determine the value of the universal gas constant R and the gas constant k for 1 gm mole air. [At N.T.P., the density of air = 1.293 gm/litre and the density of mercury = 13.6 gm/cc.] [J.E.E. ‘96]


Ans.

Pressure, 9806.1376 ××=P dyne/cm 2 , Temperature, 27300 == CT K . At NTP, the volume of 1 gm-mole of any gas, 4.22=V litre = 22400 cc.

71031.8273

224009806.1376×=

×××==∴

TPVR erg/(mole K).

The volume of 1 gm-mole air, 293.11

=′V litre = 293.1

1000 cc.

∴ 710287.0273293.1

10009806.1376×=

××××

=′

=TVPk erg/(gram K).

Q.7 Draw CtP 0− and TP − K graphs for a certain mass of gas at constant volume. Can one find the value of absolute zero from the first graph? [H.S. ’04, ‘90] Ans. If for a certain mass of gas at constant volume, the temperature, Ct 0 is considered as abscissa and the pressure, P is considered as ordinate, we get a straight line as shown in the figure**. The tP − relationship is the following:

+=

27310

tPP .

The pressure, 0P is the pressure of gas at Ct 00= which is the intercept on the vertical

P -axis by the straight line; 273

0P is the slope of the line. For different constant volumes

of gas, we get different values of 0P . Thus we obtain a set of straight lines having different slopes and different intercepts. If the straight lines are extrapolated towards the negative temperature, they all cut the −t axis at a certain point )0( =P . The value of this point is C0273− . The pressure of

gas is zero at this point as can be seen from the graph. Therefore, the temperature Ct 0273−= is the absolute zero temperature. The straight lines can not be extrapolated

further as that will indicate negative pressure which is unphysical.

The TP − relationship is =TP constant.

If we plot temperature T (in absolute scale) as abscissa and pressure P as ordinate of a certain mass of gas at constant volume, we get a straight line passing through the origin. For different constant volumes, we get different TP − straight lines and all of them pass through the origin (see fig.**). Q.8 A certain mass of gas is heated first in a small vessel and then in large vessel. Assume that the volumes of the vessels remain unchanged during heating. How will be the pressure-temperature ( )TP − graphs in two cases? Ans.

Fig. to be included


As the gas is heated, keeping the volume constant, we can write =TP constant.

Thus the ( )TP − graphs will be straight lines passing through origin in both the cases. The equation for ideal gas is kTPV = , k is a constant. Suppose, the volumes of the two vessels are 1V and 2V , where 21 VV < . If the pressure of the gas is 1P in the first vessel and 2P in the second vessel at a certain temperature T , we can write, 2211 VPVP = . ∴We have, 21 PP > . This is true for any temperature. Thus the pressure of the gas is always higher in small volume than the pressure of gas in larger volume. Therefore, the ( )TP − straight line corresponding to smaller volume ( 1V ) is above the one corresponding to larger volume ( 2V ) at all temperatures. The slope in the first case will be higher than the slope in the second case. Q.9 At first m gm and then m2 gm of a gas are heated in a container of fixed volume. Draw pressure-temperature ( TP − ) graph in each case. Ans.

If some amount of gas is heated in a fixed volume, we can write =TP constant.

So, the ( TP − ) graphs for different masses of gas will be straight lines passing through origin. Now, the ideal gas law is nRTPV = , where n is the number of gram moles of the gas and R is the universal gas constant. If M is the molecular weight, we can write for m gm gas,

RTMmPV = Or, T

MVmRP .= ……….(1) and for m2 gm gas,

RTMmPV 2

= Or, TMVmRP .2

= ……….(2)

Therefore, in the ( TP − ) graph, MVmR is the slope in the first case and

MVmR2 is the slope

in the second case. Hence, the slope in the second case is higher than the first one. Thus the )( TP − straight line graph corresponding to m2 gm gas will be above the graph corresponding to m gm gas. The above can also be analyzed in a different way. Suppose, 1P and 2P are the pressures of m gm gas and m2 gm gas, respectively at any temperature, T ′ . As the volume,V is constant, we can write,

MVTR

mP ′

=1 = constant; MV

TRm

P ′=

21 = constant.

Fig. to be included

Fig. to be included


∴2

21

PP = ; the value of pressure in the ordinate is higher in the first case than the second

case. So, the ( TP − ) straight line for m gm gas will be above the one for m2 gm gas. Q.10 In the fig.**, TV − graphs for a certain amount of gas have been shown at two pressures 1P and 2P . It can be said from the graphs that the pressure 1P is greater than the pressure 2P . Is it correct? [I.I.T.] Ans. It is seen from the graphs, at some temperature T ′ (say), the volume of the gas 1V at pressure 1P is greater than the volume 2V at pressure 2P .

According to Boyle’s law, 2211 VPVP = Or, 1

2

2

1

VV

PP

= .

Q 2V > 1V ∴ 1P > 2P . Hence, we can say from the given graphs, the pressure 1P is greater than the pressure 2P . Q.11 When a gas expands, it obeys the law: 2PV = constant. Show that there will be cooling due to this kind of expansion. [J.E.E. ‘98] Ans.

Let the initial volume, pressure and temperature of the gas are 1V , 1P and 1T , respectively. If the corresponding quantities are 2V , 2P and 2T after expansion, we can write from ideal gas law:

2

22

1

11

TVP

TVP

= Or, 2

1

22

11

TT

VPVP

=

Or, 22

112

22

211

TVTV

VPVP

= [Multiplying by 2

1

VV on both sides]

Since 2PV = constant, we can write 222

211 VPVP = .

∴ 122

11 =TVTV Or,

1

2

2

1

VV

TT

= .

As the gas expands, the final volume will be greater than the initial volume: 2V > 1V . ∴ 1T > 2T which implies that the final temperature is lower than the initial temperature. Hence, in this kind of expansion, the gas cools down. Q.12 A faulty barometer tube contains some air above mercury. How can the correct value of atmospheric pressure be determined by this? Ans. Let the correct barometer reading = H cm Hg and the area of cross-section of the barometer tube = α sq. cm.

Fig. to be included


Suppose, in the faulty barometer, we find the height of mercury column = 1h cm and the length of air column above mercury = 1l cm. ∴The volume of the air column, 1V = α1l cc and the pressure, )( 11 hHP −= cm Hg. Next, the mercury column is altered by slightly lifting the barometer tube while keeping its open end still immersed in mercury. Suppose this time, the height of mercury column = 2h cm and the length of air column above mercury = 2l cm. ∴In this case, the volume of the air column, 2V = α2l cc and the pressure,

)( 22 hHP −= cm Hg. Applying Boyle’s law:

αα 2211 )()( lhHlhH −=− or, 21

2211

lllhlhH

−−

= .

∴The correct barometer reading or the actual atmospheric pressure H can be determined if we know 1h , 2h , 1l , 2l .

Questionnaire

Very Short Questions: Mark: 1 (Answer in one or two words)

1. What is the volume coefficient of gas? [ C0/2731 ]

2. What is the pressure coefficient of gas? [ C0/2731 ]

3. What will be the temperature in degree Celsius at which the volume of gas becomes zero according to Charles’ law? [-273 C0 ] 4. What is the volume of one mole of any ideal gas at NTP? [22.4 litre]

(Fill in the blanks)

1. At constant temperature, the volume of some amount of gas is _________ its pressure. [Inversely proportional to] 2. At constant pressure, the volume of some amount of gas is ____________ its absolute temperature. [Proportional to] 3. At constant volume, the pressure of some amount of gas is ____________ its absolute temperature. [Proportional to]

(Multiple choice type)

1. The coefficient of volume expansion of gas, when compared to solid and liquid, is (a) same (b) relatively large (c) relatively small. [(b)]


2. The volume and pressure of some amount of gas are seen to increase. This is possible when the temperature of the gas (a) remains same (b) decreases (c) increases. [(c)] 3. If the volume expansion coefficients of solid, liquid and gas are Sγ , lγ and gγ , respectively, (a) The values of Sγ , lγ and gγ for different solids, liquids and gases are different (b) The values of Sγ , lγ for different solids and liquids are different, but gγ is same for all gases (c) The value of Sγ is different for different solids, but lγ is same for all liquids and gγ is same for all gases (d) The values of Sγ , lγ and gγ for different solids, liquids and gases, respectively are same. [(b)] 4. The volume expansion coefficients of solid, liquid and gas are Sγ , lγ and gγ , respectively. In general, (a) Sγ < lγ < gγ (b) Sγ > lγ > gγ (c) lγ < Sγ < gγ (d) lγ > Sγ > gγ [(a)] 5. The volume of a gas at N.T.P. is 150 cc. If the temperature at constant volume is 25 C0 , the pressure becomes 850 mm. What is the pressure coefficient of the gas? (a) C03 /1073.4 −× (b) C03 /1073.5 −× (c) C03 /1073.6 −× (d) C0/1 [(a)] 6. The value of relative gas constant for hydrogen is (a) 71016.4 × erg/gm K (b) 71026.0 × erg/gm K (c) 71080.4 × erg/gm K (d)

71016.5 × erg/gm K [(a)] 7. When the temperature is increased keeping the pressure constant, the density of a gas (a) remains same (b) decreases (c) increases (d) may increase or decrease, depends on gas. [(b)] 8. For determining the volume expansion coefficient of gas, the volume at some temperature is always taken as initial value. The temperature is (a) 273 C0 (b) 0 C0 (c) 100 C0 (d) -273 C0 [(b)] 9. The isotherm of a gas is (a) VP − graph (b) TP − graph (c) TV − graph (d) TPV − graph. [(a)] 10. At constant pressure, the graph of volume of a certain amount of gas with temperature, ( TV − ) graph is (a) horizontal (b) straight line with a positive slope (c) straight line with a negative slope (d) rectangular [(b)] 11. The PPV − graph is


(a) parallel to P -axis (b) parallel to PV -axis (c) straight line passing through the origin (d) rectangular hyperbola [(a)] 12. The volume coefficient and the pressure coefficient are same for (a) ideal gas (b) real gas (c) hydrogen gas (d) inert gas [(a)] 13. There is 1 mole 2O gas (relative molecular weight 32) in a vessel at a temperatureT . The pressure of this gas is P . There is 1 mole He gas (relative molecular weight 4) at a temperature T2 in a similar vessel. The pressure of this gas will be

(a) 8P (b) P (c) P2 (d) P8 [I.I.T. ‘97]

[(c)] 14. Two containers A and B having frictionless pistons attached, contain same ideal gas. In both the containers, the volume of gas (V ) and temperature are same. The mass of gas in the containers A and B are Am and Bm , respectively. Now the gas is made to expand up to a volume V2 at fixed temperature in both the containers. The change in pressure in A and B containers are P∆ and P∆5.1 , respectively. Which of the following statements is correct? (a) BA mm 94 = (b) BA mm 32 = (c) BA mm 23 = (d) BA mm 49 = [I.I.T. ‘98] [(c)] 15. A closed horizontal cylinder of length 50 cm, has a freely moving piston separating two chambers in it. The left chamber contains 25 mg and the right chamber contains 40 mg of He gas. When equilibrium is established, what will be the ratio of the lengths of left and right chambers? (a) 1:2 (b) 3:2 (c) 5:8 (d) 8:5 [(c)] Short Questions: Marks: 2 1. How is the expansion of gas different with respect to solid and liquid? 2. State Boyle’s law and explain. [H.S. ’95, ‘92] 3. State Charles’ law and explain. [H.S. ’02, ’97, ’95, ’93, ‘92] 4. Name the law of variation of volume of a gas with temperature at constant pressure and explain. 5. How does the concept of absolute temperature come from Charles’ law? [H.S. ’05, ‘02] 6. Establish the equation for ideal gas from Boyle’s law and Charles’ law. [H.S. ’01, ’99, ’95, ‘92] 7. Determine the combined form of Boyle’s law and Charles’ law. [H.S. ‘93] 8. What is universal gas constant? [H.S. ‘02] 9. Establish the mathematical form of Charles’ law using absolute temperature. 10. Determine the relation among pressure, temperature and density of gas. 11. ‘To determine the state of a gas, we need to know three variables- volume, temperature and pressure’ – Explain the meaning of this statement. 12. What are the values of absolute zero in Celsius and Fahrenheit scale?


13. Why the absolute zero temperature rather than 0 C0 in Celsius scale is considered fundamental? 14. What do you mean by ideal gas law? Is this law applicable for real gas? 15. What is the value of universal gas constant in C.G.S. unit? 16. Why the expansion coefficient of gas is of two types? 17. Why do we consider 0 C0 as the initial temperature when we calculate expansion coefficient of a gas? Why is this difference with respect to solid and liquid? 18. Is the value of expansion coefficient same for all gases?

19. What do you mean by the pressure coefficient of gas to be C0/2731 ?

20. What is Normal temperature and pressure (N.T.P.)? 21. What is the relation between the volume coefficient and pressure coefficient of a gas? What are the values? 22. Is it possible to reach a temperature below the absolute temperature? Why? 23. A container is filled with oxygen gas and it is taken to Moon. What will happen if the container is a (i) steel cylinder and (ii) rubber balloon? 24. When is the density of a gas proportional to the pressure? 25. When is the density of a gas inversely proportional to its absolute temperature? 26. For the expansions of solid and liquid, only temperature is referred whereas in the case of volume expansion of a gas, temperature and pressure both are mentioned. Why? 27. There is a coefficient of apparent expansion for liquid, but no such coefficient is mentioned for a gas. Why? 28. What is the relative gas constant? Is this constant same for all gases. 29. Equal numbers of hydrogen and helium molecules are kept in two similar containers. What will be the ratio of their pressures? 30. What is an ideal gas? 31. What is a mole? What is Avogadro’s number? 32. Draw PPV − isotherm for an ideal gas of a certain mass and explain. [H.S. ‘03] 33. Draw CtV 0− and TV − K graphs for a certain mass of a gas at constant pressure. Is it possible to get the value of absolute temperature from the first graph? 34. A certain amount of gas is first heated in a container of small volume and then it is heated in a container of larger volume. Assume that the volume of the containers remain unchanged during heating. What will be the pressure temperature ( TP − ) graphs in the two cases? 35. During blowing up, the volume of a balloon increases as well as the pressure also increases. Is Boyle’s law violated here? [H.S. ‘04] Medium level Questions: Marks: 4 1. Define the volume coefficient and pressure coefficient of a gas. Show that for an ideal gas, the values of the two coefficients are same. [1+1+2] [H.S.(XI) ’06; H.S. ’05, ’03, ’00, ’98, ’96, ‘94] 2. What do you mean by absolute scale of temperature? How does the concept come from Charles’ law? Why is it called absolute scale? What will be the volume of an ideal gas at zero degree of the absolute scale? [1+1+1+1] [H.S. ’00, ’97, ‘93]


3. What is universal gas constant? Why is it called universal? Calculate the value of it for one gram-mole of any gas. [1+1+2] [H.S. ‘01] 4. State the law of ideal gas when volume is constant (Pressure law). Write the expression when absolute temperature is used and draw a graph. [2+1+1]

5. Draw VP − , V

P 1− , PPV − , and VPV − graphs following Boyle’s law.

[1+1+1+1] 6. How does the melting point of ice depend on external pressure? [H.S. ‘05] Short Problems: Marks: 2 1. At 27 C0 and 70 cm pressure, the volume of some oxygen gas is 400 cc. What will be its volume at normal temperature and pressure? What will be the change in product of the pressure and volume of the gas due to this variation? [Ans. 335.3 cc, 2517 C.G.S. unit] 2. The volume of some amount of gas is 2 litre at normal pressure and temperature. What is the volume of this gas at temperature 91 C0 and at pressure 570 mm? [Ans. 3.5 litre] 3. The volume of some amount of nitrogen is 50 cc at 50 C0 . What will be the volume of this gas at -50 C0 if the pressure does not change? [Ans. 34.5 cc] 4. The pressure of a gas is 60 cm Hg at a temperature -73 C0 . What will be the pressure at 27 C0 if volume remains constant? [Ans. 90 cm Hg] 5. When the temperature of 5 litre of a gas is raised from 0 C0 to 35 C0 at constant pressure, the volume is increased by 640 cc. Determine the value of absolute zero in Celsius scale from this data. [Ans. C04.273− ] 6. The atmospheric pressure is 74 cm Hg and temperature is 27 C0 at some place. The atmospheric pressure and temperature are 70 cm Hg and 23 C0 , respectively at some other place. Compare the densities of air at the two places. [Ans. 2738 : 2625] 7. One litre helium gas of temperature 27 C0 and of pressure twice the atmospheric pressure is heated such that the volume and pressure of the gas both become double. Find the final temperature. [Ans. 927 C0 ] 8. A glass vessel is filled with air at 50 C0 . Up to what temperature is the vessel to be

heated so that 41 fraction of air comes out? Assume the pressure to remain constant in the

vessel. [Ans. 157.67 C0 ]


9. What will be the volume of 1 mole oxygen at 27 C0 temperature and two atmospheric pressure? 71031.8 ×=R erg/(mole Kelvin). [Ans. 12.31 litre] 10. The volume of a dry gas is 1520 cc at 27 C0 temperature and at 700 cc Hg. What is the volume of that gas at NTP? [Ans. 1274 cc] 11. The weight of 1 litre gas is 1562 gm at NTP. What will be the weight of 1 litre gas at 25 C0 temperature and 78 cm Hg pressure? [Ans. 1460 gm] 12. The volume of a bulb is 1 litre. What percentage of air will come out if the bulb is heated from 0 C0 to 27 C0 ? Assume that the pressure inside remains constant. [Ans. 9.89%] 13. The density of oxygen is 1.429 gm/litre at normal temperature and pressure. Find the mass of 2.5 ltre oxygen gas at 27 C0 temperature and 780 mm pressure contained in a cylinder. [Ans. 3.336 gm] 14. If the mercury pressure in a barometer is 70 cm, the volume of a gas is 650 cc. What is the volume of that gas at standard temperature and pressure? [Ans. 598.68 cc] 15. When the mercury pressure in a barometer is 75 cm, the volume of some amount of hydrogen gas is 150 cc. If the volume of this quantity of hydrogen gas becomes 160 cc in the next day, what will be the barometer reading? [Ans. 70.31 cm] Medium level Problems: Marks: 4 1. An air bubble comes up from the bottom of 34 ft deep river. The temperature of water at the depth of river us 7 C0 and the volume of the bubble is 14 cc. The temperature of water above is 27 C0 and the pressure, 75 cm Hg. If the density of mercury is 13.6 gm/cc, what will be the volume of the bubble just above the water surface? [Ans. 65 cc] 2. The volume of an air bubble becomes 4 times as it rises above from the depth of a sea. If the atmospheric pressure is 76 cm Hg and the temperature below and under and above the sea is same what is the depth of the sea? Density of mercury = 13.6 gm/cc. [Ans. 3101 cm] 3. The perimeter of the tyre of a car is 1 m and diameter 10 cm. How much air has to be introduced in the tyre at atmospheric pressure so that the pressure inside it will be 10 times the atmospheric pressure? [Ans. 78.5 litre] 4. The density of air at standard temperature and pressure is 0.00129 gm/cc. If the barometer height comes down from 76 cm to 74 cm, what will be the difference in the weight of 15 litre air? [Ans. 0.51 gm] 5. The volume of a both side sealed cylinder is 22.4 litre and it contains 4 gm hydrogen at 0 C0 . What will be the pressure if the temperature is 60 C0 ? If now 14 gm nitrogen is


inserted in the cylinder at 0 C0 instead of hydrogen what will be the pressure of this gas at 100 C0 ? [Ans. 2.44 atm; 0.683 atm] 6. The volume of a vessel is 10 litre. This is filled with 2O gas at NTP. The vessel is heated to 27 C0 and opened at 75 cm Hg. How much gas will come out? At NTP, 1 litre

2O gas weighs 1.43 gm. [Ans. 1.452 gm] 7. There is a 10 cm long thread of mercury resting in the middle of a both side sealed narrow horizontal tube. Air is trapped at 76 cm Hg on the two sides of the mercury thread. If now the tube is kept vertical, what will be the displacement of the mercury column? The length of the tube = 100 cm. [Ans. 2.95 cm] 8. A car tyre is filled with 15 litre air at a temperature of 17 C0 and at a pressure of 2.5 times the atmospheric pressure. If the temperature is increased to 37 C0 and the volume is increased to 15.5 litre, find the pressure inside the tyre. The atmospheric pressure = 15 pound/in 2 . [Ans. 38.8 pound/in 2 ] 9. A container is filled with 4 gm gas at 12 C0 and then heated to 50 C0 . Some gas comes out and the inside pressure remains the same. What is the mass of outgoing gas? [Ans. 0.471 gm] 10. Some air, at 20 C0 and at atmospheric pressure, is confined in a flask with the help of a cork. Due to rise in temperature, the pressure increases by 1.7 times which pushes the cork out. What is the value of increased temperature? [Ans. 225.1 C0 ] 11. When a air bubble comes up to the surface of water from the bottom of the lake, its volume increases 10 times. Find the depth of water in the lake if the barometer height is 75 cm. [Ans. 91.8 m] Harder Problems: 1. A weightless and freely movable piston is fitted inside a cylindrical tube which contains some ideal gas. The tube is held vertical. The piston is in equilibrium at a temperature 27 C0 when it is at 10 cm above the bottom. The whole system is now immersed in a water bath. The temperature of the bath is then gradually increased to 100 C0 . Calculate the present height of the piston. [Ans. 12.43 cm] 2. The combined volume of some amount of gas and a piece of glass in it is 100 cc at 27 C0 . If the pressure and temperature are increased by two times, the volume becomes 60 cc. What is the volume of the glass piece? [Ans. 12.1 cc] 3. The diameter of a glass sphere is 20 cm. This is filled with 1 litre air at 30 in Hg pressure and at a temperature of 0 C0 . What is the pressure inside the sphere when it is heated to 27 C0 ? Neglect the expansion of the sphere. [Ans. 7.86 in Hg]


4. The volume of an oxygen tank is 50 litre. The pressure of this is measured by a pressure gauge (Pressure gauge measures the amount in excess of atmospheric pressure). When oxygen is taken out from the tank, the reading of the pressure gauge drops from 300 lb/in 2 to 100 lb/in 2 , and the temperature of the gas that remains in the tank, drops from 30 C0 to 10 C0 . How much oxygen was there in the tank initially? How much oxygen have been withdrawn? [Ans. 1.376 kg; 0.838 kg] 5. The volume of some air with saturated water vapour is 80 cc at a pressure of 74 cm. If the pressure is made 146 cm keeping the temperature constant, the volume becomes half the initial volume. What will be the pressure due to water vapour at this time? [Ans. 2 cm Hg] 6. The area of cross-section of a barometer tube is 1 sq. cm and the height of this is 85 cm from the surface of mercury contained in the cup. The height of mercury column is 75 cm. What will be the height of mercury column when 1 cc air is introduced into the Torricelli’s empty space at atmospheric pressure? [Ans. 70 cm] 7. A helicopter is flying at the height of 400 m from the ground. If the average density of air is 3102.1 −× gm/cc below this and the air pressure is 1010 m bar on the ground, what is the air pressure inside the helicopter? [Ans. 962.96 m bar] 8. There is some air above the mercury column in an erroneous barometer. When the barometer readings are 760 mm and 742 mm, the correct readings are 770 mm and 750 mm. What is the length of the confined air column in the first case? What is the correct barometer reading if the erroneous barometer reading is 752 mm? Assume the temperature is unchanged. [Ans. 7.2 cm; 761 mm] 9. A mercury barometer gives erroneous reading because an air bubble enters in the empty space above the mercury column. When the air pressure is equal to the pressure of 760 mm mercury column, the barometer reads 740 mm. Again, the reading is 710 mm at a pressure of 727.5 mm Hg. Find the length of barometer tube (above the mercury surface in the cup). [Ans. 95 cm] 10. A bottle of volume 500 cc is being immersed in a water tank keeping the open face down. To what depth the bottle has to be taken so that 100 cc of water will enter in it? The height of mercury column in barometer is 75.6 cm and the density of mercury is 13.6 gm/cc. [Ans. 2.57 m] 11. The temperatures at the bottom and at the surface of a lake are 7 C0 and 27 C0 , respectively. At what ratio the diameter of the bubble will change when it reaches the surface from the bottom of the lake? The atmospheric pressure is 75 cm Hg and the density of mercury is 13.6 gm/cc. [Ans. 1:1.61] 12. The space between the lower sealed end of a vertical tube and a mercury column in the tube is filled with some ideal gas. The open end of the tube is exposed to the atmosphere (atmospheric pressure = 76 cm Hg). The lengths of the mercury column and that of the column of gas are 20 cm and 43 cm, respectively. When the tube is slowly


inclined with respect to the vertical by 60 0 , what will be the length of the column of gas? Assume the temperature of the gas to remain constant. [Ans. 48 cm] 13. A 80 cm long both side open narrow tube is immersed halfway in mercury. Now, as the tube is taken out from mercury by closing its upper end, a 23 cm long mercury column remains in the tube. What is the atmospheric pressure? [Ans. 77.12 cm Hg] 14. A one end sealed glass tube contains some air in it with the help of a 8 cm long mercury thread. If the tube is held vertical keeping the sealed end up, the length of air column becomes 40 cm at 27 C0 . What will be the length of the air column if the temperature is increased to 60 C0 and the tube is inclined by an angle of 60 0 with the vertical? The area of cross-section of the tube is 0.5 cm 2 and the atmospheric pressure is 76 cm Hg. What is the mass of air confined in the tube? The density of air at standard temperature and pressure = 310293.1 −× gm/cc. [Ans. 41.5 cm; 31005.21 −× gm] 15. One end of a uniform glass tube is closed; some air is trapped in it by a mercury column. The length between the closed end and the edge of the mercury column is 10 cm at 20 C0 . How far does the mercury column shift when the temperature increases to 70 C0 ? CP

0/00366.0=γ . [Ans. 11.7 cm] 16. The volume of a cylindrical container is 410 cc and it is filled with oxygen gas at a pressure of 6105.2 × dyne/cm 2 and at a temperature of 27 C0 . When some gas comes out, the pressure of the gas inside becomes 3103.1 × dyne/cm 2 . What is the mass of gas that comes out? The molecular weight of oxygen = 32. [Ans. 15.36 gm] 17. At normal condition, 40 cc of oxygen gas is inserted in a one end sealed tube which is then held vertical keeping it immersed in mercury taken in a container. The height of mercury column in the tube is seen to be 15.6 cm above the surface of mercury. If the air pressure is 75.6 cm and the room temperature is 31 C0 , what length of the tube was filled with the gas? [Ans. 47.02 cm] 18. A both side open glass tube is immersed vertically in mercury such that a length of 13 cm remains above mercury surface. Now the tube is pulled upwards by another 35 cm while closing the upper end. Find the length of air column above mercury inside the closed tube. Atmospheric pressure = 76 cm Hg. [Ans. 20.41 cm] 19. A vessel of volume 800 cc is immersed in water holding the side of its opening downward. At what depth the vessel has to be taken so that 300 cc of water enters into it? The barometer height above water = 76 cm Hg and the density of mercury = 13.6 gm/cc. [Ans. 620.2 cm] 20. Some air is trapped in a one end open capillary tube with the help of a 15 cm long mercury thread. If the tube is held vertical with the open end up, the length of the air column becomes 10 cm at 27 C0 . On the other hand if the tube is held upside down, the


length of air column becomes 15 cm. At what temperature the length of air column will become 20 cm in this upside down position? [Ans. 127 C0 ] 21. A glass tube of narrow bore has two sides sealed. A mercury thread is inside the tube such that it divides the tube in a ratio of 3:1. If the temperature of the system is increased from 0 C0 to 273 C0 , what will be the change in air pressure inside the tube? [Ans. increases twice in each half] 22. The volume and temperature of a vessel are V andT , respectively. Three gases are inserted in this. What is the resultant pressure of the gas mixture if the initial volume, temperature and pressure of the gases are ( 1P , 1V , 1T ); ( 2P , 2V , 2T ); ( 3P , 3V , 3T ),

respectively? [Ans.

++

3

33

2

22

1

11

TVP

TVP

TVP

VT ]

23. The length between the sealed end of a uniform cylindrical tube of a barometer and the mercury surface in the tube is l mm. An air bubble is inserted into the tube at Ct 0

1 and at a pressure of P mm Hg. The height of mercury column reduces to 1l mm due to this. If the barometer reading is 2l mm at Ct 0 , what will be the correct reading? Assume

the expansion coefficient of air = α . [Ans. )(..1.1

12

1

12 lP

llll

ttl −

−−

++

+αα mm]

24. A small mercury thread separates some air from surroundings in a tube. The tube can be rotated in the vertical plane. The length of air column in the tube at its horizontal position is 1l and in the vertical position (open face up), the length becomes 2l . Find the length of the air column when the tube makes an angle θ with the vertical.

[Ans. θcos)( 122

21

lllll−−

]

25. The bottom end of a 1 m long vertical tube is closed and an air tight piston is fitted at the other end at top. The piston can move up and down without friction. Some ideal gas is trapped inside the tube. The gas occupies 94 cm in the tube when the piston is in equilibrium. Now mercury is being poured over the piston up to the edge of the tube. The piston now comes down to 30 cm. Find the atmospheric pressure assuming the temperature to be constant. The weight of the piston can be neglected. [Ans. 76.4 cm Hg] 26. A container contains 1m gm gas at 1P pressure. Another container has 2m gm gas at

2P pressure. If the two containers are now joined by a tube, what will be the temperature

of the gas mixture? [Ans. 2112

2121 )(mPmPmmPP

++ ]

27. A closed vessel of volume 0.02 m 3 contains a mixture of Neon and Argon gas at a temperature of 27 C0 and at a pressure of 5101× Newton/m 2 . The total mass of the gas mixture is 28 gm. The molecular weights of Neon and Argon are 20 and 40 gm/mole, respectively. Determine the mass of each gas. Assume the two gases are ideal (universal gas constant = 8.314 joule/ mole K). [I.I.T. ‘94] [Ans. Mass of Neon = 4.08 gm, Mass of Argon = 23.92 gm]


28. An ideal gas is seen to obey an additional law: 2VP = constant. The initial temperature of the gas is T and volume, V . If the volume of the gas becomes V2 as it expands, what is the temperature? [Ans. T2 ] 29. There is a 4 cm long mercury thread in a uniform glass tube which has one end open. The space between the mercury thread and the closed end contains some air at 27 C0 . The length of air column is 9 cm when the tube is held vertical with the open end facing up. The length of the air column becomes 10 cm when the tube is held up side down. (i) Find the atmospheric pressure. (ii) At what temperature the length of air column becomes 9 cm in the inverted position? [Ans. (i) 76 cm Hg, (ii) C03− ] 30. An annulus tube (as shown in the fig.**) contains two ideal gases of equal mass. The relative molar masses of the gases are 321 =M and 282 =M . The two gases are kept separated by a fixed partition and a movable cork. The cork can slide in a frictionless manner inside the tube. Determine the angle α as shown in the figure. [Ans. 192 0 ]

I 31. The lower end of a narrow and uniform vertical tube is sealed. There is some ideal gas in the space between the sealed end and a mercury column. The upper end of the tube is exposed to the atmosphere. The lengths of the mercury column and the confined air column are 20 cm and 43 cm, respectively. The tube is slowly inclined to make an angle of 60 0 with the vertical. What is the length of the column of gas in this situation? Given, atmospheric pressure = 76 cm Hg and the temperature is the gas is constant. [Ans. 48 cm] 32. The lower end of a narrow and uniform 100 cm long vertical tube is sealed while the upper end is open. A movable piston is attached in the tube at 10 cm from the open end such that there is 90 cm long air column under this piston. Now mercury is being poured over the piston until it starts overflowing. The piston comes down to 32 cm. What is the atmospheric pressure? [Ans. 76.125 cm Hg] 33. Show that if the rate of variation of pressure with density and height in Earth’s

atmosphere is given by gdzdP ρ−= , the air pressure at any height h measured from

ground is given by RTMghePP /0

−= , where M is the average molecular weight of air. Assume the atmospheric air is ideal gas and the temperature to be uniform throughout.

Fig. to be included

expansion of gas

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