chapter 03 the first law of thermodynamics (pp 59-81)
Post on 30-Aug-2014
441 Views
Preview:
TRANSCRIPT
CH 03 THE FIRST LAW OF THERMODYNAMICS 59
CHAPTER 03 THE FIRST LAW OF THERMODYNAMICS
3-1 THE FIRST LAW OF THERMODYNAMICS
Problem 3-1
A system receives calories200 of heat and work done by the
system is joules736 . What is change in its internal energy?
)186.41( Jcalorie = . B.U. B.Sc. 2000A
Solution According to first law of thermodynamics
dWdUdQ +=
dWdQdU −=
JdU 2.101736)186.4)(200( =−=
Problem 3-2
A system received joules1254 of heat. Calculate the work
done by the system if the change in its internal energy is
calories200 . B.U. B.Sc. 2007A
Solution According to first law of thermodynamics
dWdUdQ +=
dUdQdW −=
Now JdQ 1254=
JJcaloriesdU 2.837)186.4)(200(200 ===
Hence
JdW 8.4162.8371254 =−=
CH 03 THE FIRST LAW OF THERMODYNAMICS 60 Problem 3-3
A system receives 150 calories of heat and the work done by
the system is 418 joules. What is the change in its internal
energy ( Jcalorie 18.41 = ) B.U. B.Sc. 2009S
Solution According to the first law of thermodynamics
dWdUdQ +=
dWdQdU −=
Now JcaloriesdQ 627)18.4)(150(150 ===
JdU 418=
Therefore
JdU 209418627 =−=
Problem 3-4
An ideal gas expands isothermally, performing J31000.5 ×
of work in the process. Calculate
(a) the change in internal energy of the gas and
(b) the heat absorbed during this expansion.
Solution
(a) 0== INTERNALdEdU because there is no change in
temperature.
(b) dWdUdQ +=
JdQ 33 1000.5)1000.5(0 ×=×+=
Problem 3-5
Let kg00.1 of water be converted to steam by boiling. The
volume changes from an initial value of 331000.1 m−× as a
liquid to 3671.1 m as steam. For this process, find
(a) the work done on the system
(b) the heat added to the system and
(c) the change in internal energy of the system.
(Heat of vapourization 161026.2 −×= kgJ , 1 atmosphere 2310013.1 −× mN ) K.U. B.Sc. 2003
Solution (a) The work done on the system
CH 03 THE FIRST LAW OF THERMODYNAMICS 61
dVpdW =
JdW 535 10692.1)}1000.1(671.1){10013.1( ×=×−×= −
(b) JLmdQ v
66 1026.2)1026.2)(1( ×=×==
(c) dWdQdU −=
JdU 656 10091.2)10692.1()1026.2( ×=×−×=
Problem 3-6
A W40 heat source is applied to a gas sample for s25 ,
during which time the gas expands and does J750 of work
on its surroundings. By how much does the internal energy
of the gas change?
Solution According to the first law of thermodynamics
dWdUdQ +=
dWdQdU −=
Now
JtPdQ 1000)25)(40( ==∆=
JdW 750=
Hence
JdU 2507501000 =−=
CH 03 THE FIRST LAW OF THERMODYNAMICS 62
3-2 HEAT CAPACITIES OF AN IDEAL GAS
Example 3-7
In an experiment, mol35.1 of oxygen ( 2O ) are heated at
constant pressure starting at C011 . How much heat must be
added to the gas to double its volume? Given that 114.29 −−= KmolJCV is for oxygen.
Solution According to Charles’s law
==2
2
1
1
T
V
T
Vconstant
11
1
1
1
1
2
2 22
TTV
VT
V
VT =
=
=
The amount of heat added to the system at constant pressure is
given by
)2()( 1112 TTCnTTCndTCnQ PPP −=−==
JTCnQ P
4
1 10127.1)27311)(4.29)(35.1( ×=+==
Example 3-8
Twelve grams of nitrogen )( 2N in a steel tank are heated
from C025 to C
0125 . (a) How many moles of nitrogen are
present? (b) How much heat is transferred to the nitrogen?
(The specific heat of 2N at constant volume is 118.20 −− kgmolJ ).
Solution
(a) 28 grams of 2N is = 1 mol
12 grams of 2N is nmol === 429.028
12
(b) dTCnQ V=
JQ 892)}27325()273125){(8.20)(429.0( =+−+=
CH 03 THE FIRST LAW OF THERMODYNAMICS 63 Example 3-9
The mass of helium atom is kg271066.6 −× . Compute the
specific heat at constant volume for helium gas
(in 11 −− KkgJ ) from the molar heat capacity at constant
volume. (Given that 115.12 −−= KmolJCV ).
Solution
Now 115.12 −−= KmolJCV
The mass of one mole of helium gas is
(mass of helium atom) AN
kg32327 10011.4)10022.6)(1066.6( −− ×=××=
Hence
113
310116.3
10011.4
5.12 −−
−×=
×= KkgJCV
Example 3-10
Propane gas )( 83HC behaves like an ideal gas with 127.1=γ .
Determine the molar heat capacity at constant volume and
the molar heat capacity at constant volume.
Solution We know that
RCC VP +=
But V
P
C
C=γ or VP CC γ= , therefore above expression
becomes
RCC VV +=γ
RCV =− )1(γ
115.651127.1
314.8
1
−−=−
=−
= KmolJR
CVγ
118.73314.85.65 −−=+=+= KmolJRCC VP
CH 03 THE FIRST LAW OF THERMODYNAMICS 64 Example 3-11
The heat capacity at constant volume of a certain
amount of a monoatomic gas is KJ /8.49 .
(a) Find the number of moles of the gas.
(b) What is the internal energy of the gas at KT 300= ?
(c) What is the heat capacity of the gas at constant
pressure?
Solution (a) For monoatomic gas
RnCV2
3=
molesR
Cn V 4
)314.8(5
)8.49(2
3
2≅==
(b) JTCU V
410494.1)300)(8.49( ×===
(c) VVVP CCRnCC3
2+=+= RnCV
2
3=Θ
KJCC VP /83)8.49(3
5
3
5===
Example 3-12
One mole of a monoatomic ideal gas is initially at K273
and one atmosphere.
(a) What is its initial internal energy?
(b) Find its final internal energy and the work done by the
gas when J500 of heat are added at constant pressure.
(c) Find the same quantities when J500 of heat are added
at constant volume.
Solution (a) The initial internal energy is given by
TRTCU Vi2
3==
JU i 3405)273)(314.8(2
3==
CH 03 THE FIRST LAW OF THERMODYNAMICS 65
(b) Now dTCdQ P= or PC
dQdT =
JCC
dQ
C
dQCdTCdU
VPP
VV 300)3/5(
500
)/(===
==
Therefore the final internal energy and work are given by
JdUUU if 37053003405 =+=+=
JdUdQdW 200300500 =−=−=
(c) JdUdQ 500== and 0)0( === pdVpdW
CH 03 THE FIRST LAW OF THERMODYNAMICS 66
3-2 WORK DONE ON OR BY AN IDEAL GAS 3-2(A) WORK DONE AT CONSTANT VOLUME
Example 3-13
Calculate the increase in internal energy of ten grams of
oxygen whose temperature is increased by C010 at constant
volume. Given that 10182.1 −−= CgJCv .
Solution According to first law of thermodynamics
dWdUdQ +=
dVpdUdTCm v +=
)0()10)(82.1)(10( 0101 pdUCCgJg +=−−
dUJ =182
Increase in internal energy JdUdE INTERNAL 182===
CH 03 THE FIRST LAW OF THERMODYNAMICS 67
3-2(B) WORK DONE AT CONSTANT PRESSURE
Problem 3-14
A gas expands at atmospheric pressure and its volume
increases by 3500 cm . Calculate the work done by the gas.
(1 Atmosphere25 /10013.1 mN×= ) B.U. B.Sc. 1998A
Solution The work done by the gas at constant pressure is given by
dVpW =
JW 65.50)10500)(10013.1( 65 =××= −
Problem 3-15
One kilogram water is converted to steam at standard
atmospheric pressure. The volume changes from 33101 m
−×
as a liquid to 3671.1 m as steam. For this process calculate
the work done on the system when the pressure is 25 /10013.1 mN× . K.U. B.Sc. 1998
Solution The work done by the vapourizing water is
)( if VVpdVpW −==
JW 535 10691.1)101671.1)(10013.1( ×=×−×= −
Problem 3-16
A gas expands at atmospheric pressure and its volume
increases by 3334 cm . Find the work done by the gas. The
atmospheric pressure is 26 /10013.1 cmdynes× .
B.U. B.Sc. (Hons.) 1988A
The work done by the gas, at constant pressure, is given by
)( if VVpdVpW −==
Now
26 /10013.1 cmdynesp ×=
2
22
56
/)10(
10)10013.1(mNp
−
−××= Ndyne 5101 −=Θ
CH 03 THE FIRST LAW OF THERMODYNAMICS 68
25 /10013.1 mNp ×= 343323 1034.3)10)((334(334 mmcmdV −− ×===
Hence
JdW 834.33)1034.3)(10013.1( 45 =××= −
Problem 3-17
A gas is compressed at a constant pressure of atm8.0 from
litres9 to litres2 . In the process, J400 of energy leaves the
gas by heat.
(a) What is the work done on the gas?
(b) What is the change in its internal energy?
Solution (a) The work done is given by
dVpdW =
JdW 28.567}10)92)}{(10013.1()8.0{( 35 −=×−××= −
(b) The desired change in internal energy of the system is
given by
dUdWdQ +=
dWdQdU −=
JdU 28.167)28.567(400 =−−−=
CH 03 THE FIRST LAW OF THERMODYNAMICS 69
3-2(C) WORK DONE AT CONSTANT TEMPERATURE
Problem 3-18
A gram molecule of a gas at C077 expands isothermally to
its double volume. Calculate the amount of work done.
B.U. B.Sc. (Hons.) 1987A
Solution The work done is given by
=
i
f
V
VnTRnW λ
JV
VnW 2017
2)27377)(314.8)(1(
0
0 =
+= λ
Problem 3-19
Calculate the work done by an external agent in
compressing moles12.1 of oxygen from volume of
litres4.22 and atm32.1 pressure to litres3.15 at the same
temperature. P.U. B.Sc. 2001
Solution The desired work on the gas by the external agent is given by
=
=
i
f
ii
i
f
V
VnVp
V
VnTRnW λλ
×××= −
4.22
3.15)104.22)}(10013.1(32.1( 35
nW λ
kJJW 142.110142.1 3 −=×−=
Problem 3-20
A sample of gas consisting of moles11.0 is compressed
from a volume of 30.4 m to 30.1 m while its pressure
increases from 10 to )/(40 2mNPa . Calculate the work
done. P.U. B.Sc. 2002, K.U. B.Sc. 2008
Solution
CH 03 THE FIRST LAW OF THERMODYNAMICS 70 It may be noted that
mNTRnVpVp ffii •=== 40
which indicates that the given process is isothermal. The work
done at constant temperature is given by
Jnp
pnTRnW
f
i 5.55)40/10()40( −==
= λλ
Problem 3-21
A balloon contains mol30.0 of helium. It rises, while
maintaining at constant K300 temperature, to an altitude
where its volume has expanded five times. How much work
is done by the gas in the balloon during this isothermal
expansion?
Solution The work done at constant temperature is given by
=
i
f
V
VnTRnW λ
JV
VnW
3
0
0 10203.15
)300)(314.8)(30.0( ×=
= λ
Problem 3-22
One mole of nitrogen gas is compressed isothermally from 2/10 mN to
2/20 mN at C027 . Calculate the work done.
Solution The work done during this compression is given by
−=
i
f
p
pnTRnW λ
JnW 172910
20)27327)(314.8)(1( −=
+−= λ
CH 03 THE FIRST LAW OF THERMODYNAMICS 71
3-2(D) WORK DONE IN THERMAL ISOLATION OR UNDER ADIABATIC CONDITIONS
Problem 3-23
One mole of oxygen, initially kept at C017 , is adiabatically
compressed so that its pressure becomes ten times.
Calculate
(a) its temperature after the compression and
(b) the work done on the gas.
Given that 111.21 −−= KmolJCv is for oxygen.
Solution (a) For adiabatic process
γγiiff VpVp =
γγγγ )()( iiiiffff VpppVppp−− =
γγγγ )()( 11
iiff TRnpTRnp−− =
γγγγiiff TpTp
−− = 11
γγ /)1( −
=
f
i
ifp
pTT
Now KKCTi 290)27317(170 =+==
1.010
1==
f
i
p
p , 40.1=γ and
7
2
40.1
40.111−=
−=
−
γ
γ
Hence KT f 560)1.0)(290( 7/2 == −
(b) dTCmdEdW vINTERNAL ==
JdW 5697)290560)(1.21)(1( =−=
Problem 3-24
Calculate the work done to compress adiabatically one gram
mole of air initially at S.T.P. conditions to half its volume if
40.1=γ for air.
Solution
CH 03 THE FIRST LAW OF THERMODYNAMICS 72
The initial volume 0V of air at S.T.P. is given by
111 TRnVp =
3
1
11
p
TRnV =
mVV2
501 1024.210013.1
)2730)(314.8)(1( −×=×
+==
The final pressure of air, under adiabatic conditions, is given by
γ
=
f
i
ifV
Vpp
PaV
Vp f
5
40.1
0
05 10673.25.0
)10013.1( ×=
×=
The desired work done is
1−
−=
γ
iiff VpVpW
40.11
)1024.2)(10013.1()}1024.2()5.0){(10673.2( 2525
−
××−×××=
−−
W
JW 1812−=
Problem 3-25
32 m of a gas at 2/100 mN expands according to law
CVP =2.1 where C is a constant, until volume is doubled.
Calculate the work done.
Solution
Now CVP ii =2.1
7.229)2)(100( 2.1 ==C
Hence
7.2292.1 =VP
Vp
7.229=
The work done by the gas is given by
CH 03 THE FIRST LAW OF THERMODYNAMICS 73
dVVdVpW
V
V
∫ ∫−==
2
1
4
2
2.17.229
[ ] JV
W 43.129242.0
7.229
2.07.229 2.02.0
4
2
2.0
=−
−=
−= −−
−
Problem 3-26
Five litres of argon at C00 are compressed to one litre
adiabatically and reversibly. What will be the final
temperature if 3/5=γ for argon?
Solution For an adiabatic process
γγiiff VpVp =
γγ1i
i
i
f
f
fV
V
TRnV
V
TRn
=
TRnVp =Θ
11 −− = γγiiff VTVT
1−
=
γ
f
i
ifV
VTT
KT f 798)1/5)(273( 1)3/5( == −
Problem 3-27
Calculate the final temperature of a sample of carbon
dioxide of mass g0.16 that is expanded reversibly and
adiabatically from 3500 cm at K15.298 to
32000 cm .
Solution For an adiabatic process
1−
=
γ
f
i
ifV
VTT
KT f 71.1962000
500)15.298(
130.1
=
=
−
Problem 3-28
CH 03 THE FIRST LAW OF THERMODYNAMICS 74
By how much must the volume of a gas with 40.1=γ be
changed to an adiabatic process if the Kelvin temperature is
to double?
Solution For an adiabatic process
γγiiff VpVp =
γγ1i
i
i
f
f
fV
V
TRnV
V
TRn
=
TRnVp =Θ
11 −− = γγiiff VTVT
f
i
i
f
T
T
V
V=
−1γ
)1/(1 −
=
γ
f
i
i
f
T
T
V
V
177.02
)140.1/(1
=
=
−
T
T
V
V
i
f
if VV 177.0=
Problem 3-29
A mol00.1 sample of an ideal diatomic gas originally at
atm00.1 and C020 , expands adiabatically to twice its
volume. What are final pressure and temperature for the
gas? Assume no molecular vibrations.
Solution For an adiabatic process
γγiiff VpVp =
γ
=
f
i
ifV
Vpp
CH 03 THE FIRST LAW OF THERMODYNAMICS 75
40.1
0
0
2)1(
=
V
Vp f 40.1=γΘ for a diatomic gas.
atmp f 379.0=
Further
11 −− = γγiiff VTVT
140.1
0
0
1
2)27320(
−−
+=
=
V
V
V
VTT
f
i
if
γ
KT f 222=
Problem 3-30
An ideal gas initially at atm00.8 and K300 is permitted to
expand adiabatically until its volume doubles. Find the final
pressure and temperature if the gas is
(a) Monoatomic (b) Diatomic
Solution The expressions for final pressure and temperature of the gas
are given by γ
=
f
i
ifV
Vpp
and
1−
=
γ
f
i
ifV
VTT
(a) For monoatomic gas 3
5=γ therefore
atmV
Vp
i
i
f 52.22
)00.8(
3/5
=
=
KV
VT
i
if 189
2)300(
1)3/5(
=
=
−
CH 03 THE FIRST LAW OF THERMODYNAMICS 76
(b) For diatomic gas 40.15
7==γ therefore
atmV
Vp
i
i
f 03.32
)00.8(
40.1
=
=
KV
VT
i
i
f 2272
)300(
140.1
=
=
−
Problem 3-31
A volume of dry air at S.T.P. is expanded to three times its
original volume under adiabatic conditions. Calculate the
final temperature and pressure if 40.1=γ for air.
Solution
The final temperature of air is given by
1−
=
γ
f
i
ifV
VTT
KV
VT f 176
3)2730(
40.1
0
0 =
+=
The final pressure of the gas is given by
γ
=
f
i
ifV
Vpp
PaV
Vp f
4
40.1
0
05 10176.23
)10013.1( ×=
×=
Problem 3-32
An ideal monoatomic gas for which 3/5=γ undergoes an
adiabatic expansion to one third of its initial pressure. Find
the ratio of final volume to initial volume if the process is
(a) Isothermal
(b) Adiabatic. K.U. B.Sc. 2000
CH 03 THE FIRST LAW OF THERMODYNAMICS 77 Solution
(a) For an isothermal process
iiff VpVp =
3)3/1(
===i
i
f
i
i
f
p
p
p
p
V
V
(b) For an adiabatic process
γγiiff VpVp =
f
i
i
f
p
p
V
V=
γ
933.1)3()3/1(
6.0
)3/5/(1/1
==
=
=
i
i
f
i
i
f
p
p
p
p
V
Vγ
Example 3-33
A volume of argon gas at C027 expands adiabatically until
its volume is increased four times. Find the resulting fall in
temperature. Given that 67.1=γ is for argon gas.
Solution
Now
1−
=
γ
f
i
ifV
VTT
CorKV
VT f
0
167.1
0
0 5.1545.1184
)27327( −=
+=
−
Example 3-34
An ideal gas at K300 is compressed adiabatically to half
its initial volume.
(a) What is the final temperature of the gas if it is
monoatomic?
(b) What is the final temperature of the gas if it is diatomic?
CH 03 THE FIRST LAW OF THERMODYNAMICS 78 Solution
Now
1−
=
γ
f
i
ifV
VTT
(a) For monoatomic gas 3
5=γ therefore
KV
VT f 476
5.0)300(
1)3/5(
0
0 =
=
−
(b) For diatomic gas 40.15
7==γ therefore
KV
VT f 396
5.0)300(
140.1
0
0 =
=
−
Example 3-35
Calculate the rise in temperature when a gas at C027 is
compressed to eight times its original pressure. The value of
γ is 1.5 for the given gas.
Solution The final temperature of the gas is given by the expression
γγ /)1( −
=
f
i
ifp
pTT
Kp
pT
i
i
f 6008
)27327(
5.1/)5.11(
=
+=
−
The rise in temperature of the gas will be
KTTT if 300300600 =−=−=∆
Example 3-36
A given mass of gas at C00 is suddenly compressed to a
pressure twenty times the initial pressure. What will be the
final temperature of the gas if γ is 42.1 .
CH 03 THE FIRST LAW OF THERMODYNAMICS 79 Solution The final temperature of the gas is given by the expression
γγ /)1( −
=
f
i
ifp
pTT
Kp
pT
i
i
f 2.66220
)2730(
42.1/)42.11(
=
+=
−
Example 3-37
One mol of an ideal monoatomic gas at K300 and atm0.3
expands adiabatically to a final pressure of atm0.1 . How
much work does the gas do in the expansion?
Solution For adiabatic process
γγ /)1( −
=
f
i
ifp
pTT
KT f 1930.1
0.3)300(
67.1/)67.11(
=
=
−
The work done in adiabatic process is given by
JTTR
W fi 1328167.1
)193300(314.8)(
1=
−
−=−
−=
γ
Example 3-38
An ideal monoatomic gas, consisting of mol6.2 of volume 3084.0 m , expands adiabatically. The initial and final
temperatures are C025 and C
068− respectively. What is
the final volume of the gas?
Solution For adiabatic process
11 −− = γγiiff VTVT
CH 03 THE FIRST LAW OF THERMODYNAMICS 80
11 −−
= γγ
i
f
i
f VT
TV
)1/(1 −
=
γ
f
i
fT
TV
Now
3084.0 mVi =
KKCTi 298)27325(250 =+==
KKCT f 205)27368(680 =+−=−=
67.1=γ for monoatomic gas.
Hence
3
)167.1/(1
2 147.0)084.0(205
298mV =
=
−
CH 03 THE FIRST LAW OF THERMODYNAMICS 81 ADDITIONAL PROBLEMS
(1) A system receives 150 calories of heat and the change
in its internal energy is 209 joules. Calculate the
work done by the system. ( Jcalorie 18.41 = )
B.U. B.Sc. 2007A
(2) The atmospheric pressure is 26 /10013.1 cmdynes× . A
gas expands at this pressure and the increase in its
volume is ..668 cc Find the work done by the gas.
B.U. B.Sc. (Hons.) 1989A
(3) A gas expands at atmospheric pressure and its
volume increases by3400 cm . Find the work done by
the gas. B.U. B.Sc. (Hons.) 1991A
(4) A gas is suddenly compressed to one fourth of its
original volume. Calculate rise in temperature, the
original at C027 and 5.1=γ . F.P.S.C. 1978
Answers
(1) 418 J (2) 67.67 J (3) 40.52 J
(4) 600 K
top related