This document provides examples and exercises related to the kinetic theory of gases. It begins with an example calculating the mean and most probable speeds of CO molecules at 100°C. It then provides formulas for various gas molecule speeds and kinetic energies. The exercises provide additional examples calculating gas properties like speeds, energies, and probability distributions using the kinetic theory equations.

1. 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
 8RT 
< v >= 
 πM 
1/ 2 1/ 2
 8(8.314 JK −1mol −1 )(373 K )   8(8.314 Pa m 3 K −1mol −1 )(373 K ) 
= −3 −1  = 
 π (28 × 10 kg mol )   π (28 × 10 −3 kg mol −1 ) 
1/ 2
 8(8.314 kg m 2 s −2 )(373 ) 
=  = 531 m s −1
 π (28 × 10 −3 kg ) 
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-1mol-1 ( 1 Pa = 1 kg m-1s-2 )
and the molar mass M was expressed in kg mol-1.
1/ 2
 2 RT 
v mp = 
 M  
1/ 2
 2(8.314 kg m 2 s −2 K −1 mol −1 )(373 K ) 
=  = 471 m s −1
 28 × 10 −3 kg mol −1 
(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
1/ 2
2 2  8 RT 
C rel =< v11 >= < v1 > + < v1 > = 2 < v1 >= 2 ×  
 πM 1 
2. Relative mean speed between 2 molecule of different type
C rel (12 ) =< v12 >= < v1 > 2 + < v 2 > 2
 8 RT   8 RT  8 RT  1 1 
= 
 πM  +  πM
  =
 
M + 

 1   2  π  1 M2 
12

2. 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
1
Mv 2
 M  2 − 2 RT
y
g (v y ) =   e
 2πRT 
Calculate part by part carefully:
1 1/ 2
 M 2  16 × 10 −3 kg mol −1 
  = 2 −2 −1 −1  = 8.75 × 10 − 4 m −1 s
 2πRT   2π (8.314 kg m s K mol )(400 K ) 
 Mv y   16 × 10 −3 kg mol −1 (90 m s −1 ) 2
2

 = = 0.01948
 2 RT   2(8.314 kg m s K mol )(400 K ) 
2 −2 −1 −1

 
 Mv 2 

exp −  = exp(−0.01948) = 0.9807
 2 RT 
g (v y ) = (8.75 × 10 − 4 m −1 s )(0.9807) = 8.58 × 10 −4 m −1 s
The probability density in y direction = g(vy)= 8.58×10-4 m-1s
ii) The number of molecules = dN = G(v)dv N
3
Mv 2
 M 2 −
y
2
G (v)dv = 4πv   e 2 RT
dv
 2πRT 
4πv2= 4π (90.000 ms-1)2 = 1.0178×105 m2 s-2
3
 M 2 −4 3 −10 −3 3
  = (8.75 × 10 ) = 6.7 × 10 m s
 2πRT 
G (v)dv = (1.02 × 10 5 m 2 s −2 )(6.7 × 10 −10 m −3 s 3 )(0.9807)(0.001 m s −1 ) = 6.7 × 10 −8
The number of molecules , dN = G(v)dv×N
= (6.7×10-8 )(1.00×6.02×1023) = 4.03×1016 molecules.
13

3. 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 = 1.134×105 m2 s-2
3
 M 2 −4 3 −10 −3 3
  = (8.75 × 10 ) = 6.7 × 10 m s
 2πRT 
 Mv 2   16 × 10 −3 kg mol −1 (95 m s −1 ) 2 
 = = 0.0217
 2 RT   2(8.314 kg m s K mol )(400 K ) 
2 −2 −1 −1

 
 Mv 2 
exp −
 2 RT  = exp(−0.0217) = 0.9785

 
G (v)dv = (1.134 × 10 5 m 2 s −2 )(6.7 × 10 −10 m −3 s 3 )(0.9785)(10 m s −1 )
=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
14

4. Exersice 3a
1. A 1.0 dm3 glass bulb contains 1.0×1023 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 ×103 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 3.79×1018 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 3.79×1018 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-1s, 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.
15

5. 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)1/2and
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 ε = 1 / 2mv 2 . Using the
Maxwell equation, prove that the probability that a molecule has the
kinetic energy in the range ε to ε + dε is
2π
G (ε ) = 3/ 2
ε 1 / 2 e −ε / kT dε
(π kT )
16