12

TOPIC 3- KINETIC THEORY OF GASES

Examples of Solved Problems

1. Calculate the mean speed, <v> and the most probable speed, vmp for CO

molecules at 100 oC.

Solution:

1

2/1

3

22

2/1

13

1132/1

13

11

2/1

531)1028(

)373)(314.8(8

)1028(

)373)(314.8(8

)1028(

)373)(314.8(8

8

−

−

−

−−

−−

−−

−−

=

×=

×=

×=

=><

smkg

smkg

molkg

KmolKmPa

molkg

KmolJK

M

RTv

π

ππ

π

The unit of the average speed is m s-1

, so to calculate it, we used gas constant,

R in unit J K-1

mol-1

(1 J = 1 kg m2

s-2

) or Pa m3

K-1

mol-1

( 1 Pa = 1 kg m-1

s-2

)

and the molar mass M was expressed in kg mol-1

.

1

2/1

13

1122

2/1

4711028

)373)(314.8(2

2

−

−−

−−−

=

×=

=

smmolkg

KmolKsmkg

M

RTvmp

(To remember the formula of various speed: most probable speed , vmp which

has 2 words before speed , so the formula involve, 2RT,…..for the root-mean-

square speed, which has 3 words, so in the formula involve 3RT.... The most

important formula that you will always use is the mean speed, <v>).

Other formula for speeds

1. Relative mean speed between 2 molecule of same type

2/1

1

1

2

1

2

111

822

×>=<=><+><>==<

M

RTvvvvC rel

π

2. Relative mean speed between 2 molecule of different type

+=

+

=

><+><>==<

2121

2

2

2

112)12(

11888

MM

RT

M

RT

M

RT

vvvC rel

πππ

13

2. For 1.00 mol of CH4(g) at 400 K and 1.20 atm, calculate:

i) the probability density for vy at 90.000 m s-1

ii) the number of molecules whose speed lies in range 90.000 to 90.001 m s-1

.

iii) the probability that a molecules’s speed lies between 90.000 to 100.000 m s-1

.

Solution:

i) The probability density in y direction with vy , is equal to

RT

Mv

y

y

eRT

Mvg 2

2

1 2

2)(

−

=

π

Calculate part by part carefully:

smKmolKsmkg

molkg

RT

M 14

2/1

1122

132

1

1075.8)400)(314.8(2

1016

2

−−

−−−

−−

×=

×=

ππ

01948.0)400)(314.8(2

)90(1016

2 1122

21132

=

×=

−−−

−−−

KmolKsmkg

smmolkg

RT

Mv y

smsmvg

RT

Mv

y

1414

2

1058.8)9807.0)(1075.8()(

9807.0)01948.0exp(2

exp

−−−−×=×=

=−=

−

The probability density in y direction = g(vy)= 8.58×10-4

m-1

s

ii) The number of molecules = dN = G(v)dv N

dveRT

MvdvvG RT

Mvy

22

3

2

2

24)(

−

=

ππ

4πv2= 4π (90.000 ms

-1)2 = 1.0178×10

5 m

2 s

-2

3310342

3

107.6)1075.8(2

smRT

M −−−×=×=

π

813310225 107.6)001.0)(9807.0)(107.6)(1002.1()( −−−−−

×=××= smsmsmdvvG

The number of molecules , dN = G(v)dv×N

= (6.7×10-8

)(1.00×6.02×1023

) = 4.03×1016

molecules.

14

N = nNA , where n is number of mol and NA is Avogadro number.

dN/N = molecules fraction = mol fraction= probability

iii) the probability that a molecules’s speed lies between 90.000 m s-1

to

100.000 m s-1

= dN/N = G(v)dv

The speed that will involve in calculation, v =(v1 + v2) /2= 95.000 m s-1

.

4πv2= 4π (95.000 ms

-1)2 = 1.134×10

5 m

2 s

-2

3310342

3

107.6)1075.8(2

smRT

M −−−×=×=

π

0217.0)400)(314.8(2

)95(1016

2 1122

21132

=

×=

−−−

−−−

KmolKsmkg

smmolkg

RT

Mv

9785.0)0217.0exp(2

exp2

=−=

−

RT

Mv

)10)(9785.0)(107.6)(10134.1()( 13310225 −−−−××= smsmsmdvvG

=7.4×10-4

The probability that a molecules’s speed lies between 90.000 m s-1

to

100.000 m s-1

= dN /N= G(v)dv = 0.00074

15

Exersice 3a

1. A 1.0 dm3 glass bulb contains 1.0×10

23 molecules of N2. If temperature of

the gas is 100 oC , what is the vrms of the molecules? (576 m s

-1)

2. Calculate the mean speed for the following set of molecules: 50 molecules

moving 5 ×102 m s

-1, 80 molecules moving 15 ×10

3 m s

-1, and 30 molecules

moving 2 ×102 m s

-1. (88 m s

-1)

3. What is the speed of of a molecules with vx = 20 ms

-1, vy =30 ms

-1, and vz =

60 ms-1

? (70 m s-1

)

4. Calculate the vrms and vmp of H2 having a kinetic energy of 40 J mol

-1? (200

m s-1

, 5164 m s-1

)

5. A 0.85 L container is filled with 3.79×10

18 molecules of CO2 and the

pressure is 750 torr. Calculate the kinetic energy of the gas. (127.50 J)

6. 0.85 L container is filled with 3.79×10

18 molecules of O2. Calculate the

kinetic energy for O2 molecules at 75 oC. (0.0273 J)

7. For 1.00 mol of N2 at 400 K and 1.20 atm, calculate the number of

molecules with vz in the range 140.000 to 140.001 m s-1

? (6.41×1017

)

8. Calculate the probability density for vy of O2 molecules at 400 K at 400

and 600 m s-1

.(0.0824 m-1

s, 0.0315 m-1

s)

9. What is the probability that a molecules’s speed of oxygen lies between

400 m s-1

and 450 m s-1

at 550 K? (0.071)

10. What fraction of nitrogen molecules at 300 K have velocities between 400

and 405 m s-1

? (0.0098)

11. Calculate the most probable speed, the mean speed, and the root-mean-

square speed for CO2 molecules at 298 K. (336,378,411 ms-1

)

12. Plot the probability density of the x component of the velocity of oxygen

molecules at 300 K and 500 K.

13. Plot the probability density G(v) of moleculear speeds versus speed of

nitrogen molecules at 300 K and 500 K.

16

Exercise 3b

1. What is t the ratio of the probability that gas molecules have 3 times the mean speed to

the probability that they have the mean speed? (3.39×10-4

)

2. For a certain gas, calculate the ratio of the fraction of molecules that has the most-

probable speed, vmp at 35 oC in range v to dv, to the fraction that has vmp at 25

oC in the

same range.(0.95)

3. A 0.10 m3 glass bulb contains 1.00 g of an

O2 gas. If the pressure exerted by the gas is

1 kPa, what are the temperature of the gas and the most-probable speed of the

molecules? (385 K, 252 m s-1

)

4. For 1.00 g of N2 at 400 K and 1.20 atm, calculate the number of molecules that

simultaneously have vz in the range 140.000 to 140.001 m s-1

and have vx in the range

140.000 to 140.001 m s-1

? (2.44×1010

)

5. At what temperature do ethane molecules have the same vrms as methane molecules

at 27 oC? (563 K)

6. For CH4(g), at 27 oC and 1 bar calculate the probability that a molecule of CH4(g)

picked at random has its speed in the range of 400 to 400.001 m s-1

.( 1.24 x 10-4

)

7. Determine the ratios of (a) the mean speeds (b) the mean kinetic energies of H2

molecules and He atoms at 20 oC. (1.414, 0.9996)

8. At a certain temperature 1 mole of gas G has a distribution function of speed,

G(v) = 1.52 x 10-3

s m-1

. Calculate the number of molecules having the speed in

the range of v → v + 0.001 m s-1

(1.52 x 10-6

NA)

9. For H2 gas at 25 oC, calculate the ratio of the fraction of the molecules that have

an average speed of 2v to the fraction that have the average speed of v. How

does this ratio depend on the mass of the molecules and the temperature?

( 8.77 x 10-2

, does not depend on M or T)

Problems 3

1. Use the Maxwell distribution to verify that <v>= (8RT/πM)1/2

and

show that vmp= (2RT/M)1/2

2. For O2 at 1 atm and 298 K, what fraction of molecules has speed is

greater than vrms? (0.288)

3. The kinetic energy of a molecule is given by 22/1 mv=ε . Using the

Maxwell equation, prove that the probability that a molecule has the

kinetic energy in the range ε to ε + dε is

( )

εεπ

πε ε

dkT

GkT/2/1

2/3e

2)( −

=