Physical Chemistry
33. Kinetic Theory of Gases

Ruifeng ZHOU

Faculty of Science &Institute for the Advancement of Higher Education
Hokkaido University

2

1. Velocity distribution
2. Collisions with a wall
3. Collisions between particles

3

1. Velocity distribution• Collision is essential tobimolecular reactions.
• In gas phase, themolecules are treated asclassical particles. They
have sizes, positions andvelocities, rather thanwavefunctions.
• The velocity distributionof molecules complies
with Boltzmann statistics.

𝑟𝑟⃑1 𝑡𝑡𝑟𝑟⃑2 𝑡𝑡4

1. Velocity distribution• In Boltzmann statistics, each velocity state(momentum state) has degeneracy of 1.
• The potential energy of ideal gas molecules is 0.
• The velocity probability distribution function is
𝑓𝑓𝑣𝑣 𝑣𝑣𝑥𝑥 , 𝑣𝑣𝑦𝑦 , 𝑣𝑣𝑧𝑧

𝑚𝑚 𝑣𝑣𝑥𝑥 2 +𝑣𝑣𝑦𝑦 2 +𝑣𝑣𝑧𝑧 2−2𝑘𝑘B 𝑇𝑇∝ 𝑒𝑒

𝑚𝑚𝑣𝑣𝑦𝑦 2𝑚𝑚𝑣𝑣𝑥𝑥 2𝑚𝑚𝑣𝑣𝑧𝑧 2−−−2𝑘𝑘
𝑇𝑇2𝑘𝑘
𝑇𝑇= 𝑒𝑒 B 𝑒𝑒 B 𝑒𝑒 2𝑘𝑘B 𝑇𝑇 ∝ 𝑓𝑓𝑥𝑥 𝑣𝑣𝑥𝑥 𝑓𝑓𝑦𝑦𝑣𝑣𝑦𝑦 𝑓𝑓𝑧𝑧 𝑣𝑣𝑧𝑧

• 𝑓𝑓𝑥𝑥 , 𝑓𝑓𝑦𝑦 and 𝑓𝑓𝑧𝑧 are the speed probability distributionfunction in x, y and z directions, respectively.

5

1. Velocity distribution
𝑚𝑚𝑣𝑣𝑥𝑥 2− 2𝑘𝑘 𝑇𝑇= 𝐵𝐵𝑒𝑒 B

𝑓𝑓𝑥𝑥 𝑣𝑣𝑥𝑥• 𝐵𝐵 is determined by normalization (Gaussianintegral).
∞∞� 𝑓𝑓𝑥𝑥 𝑣𝑣𝑥𝑥 = �−∞−∞𝑚𝑚𝑣𝑣𝑥𝑥 2− 2𝑘𝑘 𝑇𝑇
𝐵𝐵𝑒𝑒 B 𝑑𝑑𝑣𝑣𝑥𝑥 = 𝐵𝐵𝐵𝐵 =𝑚𝑚
2𝜋𝜋𝑘𝑘B 𝑇𝑇

2𝜋𝜋𝑘𝑘B 𝑇𝑇=1𝑚𝑚

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1. Velocity distribution

𝑓𝑓𝑥𝑥 𝑣𝑣 =𝑚𝑚
2𝜋𝜋𝑘𝑘B 𝑇𝑇

𝑚𝑚𝑣𝑣𝑥𝑥 2−𝑒𝑒 2𝑘𝑘B 𝑇𝑇• The mean speed in one direction is𝑣𝑣𝑥𝑥=+∞+∞𝑚𝑚��𝑣𝑣 𝑓𝑓𝑥𝑥 𝑣𝑣 𝑑𝑑𝑑𝑑 = 22𝜋𝜋𝑘𝑘B 𝑇𝑇 0−∞2𝑘𝑘B 𝑇𝑇 1= 𝑣𝑣
𝜋𝜋𝑚𝑚2𝑣𝑣 =8𝑘𝑘B 𝑇𝑇𝜋𝜋𝑚𝑚𝑚𝑚𝑣𝑣 2−2𝑘𝑘 𝑇𝑇𝑣𝑣𝑒𝑒 B 𝑑𝑑𝑣𝑣

71. Velocity distribution

f (s/m)• For example, N2, 298K.
• Ideal gas is isotropic. 0.0015• The velocity distribution
0.0010is𝑓𝑓𝑣𝑣 𝑣𝑣𝑥𝑥 , 𝑣𝑣𝑦𝑦 , 𝑣𝑣𝑧𝑧
0.0005= 𝑓𝑓𝑥𝑥 𝑣𝑣𝑥𝑥 𝑓𝑓𝑦𝑦 𝑣𝑣𝑦𝑦 𝑓𝑓𝑧𝑧 𝑣𝑣𝑧𝑧
𝑚𝑚𝑣𝑣 23 −2𝑘𝑘B 𝑇𝑇= 𝐵𝐵 𝑒𝑒

𝑚𝑚=2𝜋𝜋𝑘𝑘B 𝑇𝑇

32𝑒𝑒

𝑚𝑚𝑣𝑣 2−2𝑘𝑘 𝑇𝑇
B0.0000-1000-5000vx(m/s)

5001000

81. Velocity distribution• The speed distribution is2𝜋𝜋𝑓𝑓𝑣𝑣 𝑣𝑣 = �0𝑚𝑚=2𝜋𝜋𝑘𝑘B 𝑇𝑇
2=𝜋𝜋12𝜋𝜋� 𝑓𝑓𝑣𝑣 𝑣𝑣𝑥𝑥 , 𝑣𝑣𝑦𝑦 , 𝑣𝑣𝑧𝑧 𝑣𝑣 2 sin 𝜃𝜃 𝑑𝑑𝜃𝜃𝑑𝑑𝜙𝜙

3202𝜋𝜋 𝜋𝜋𝑚𝑚𝑣𝑣 2−2𝑘𝑘 𝑇𝑇 2𝑒𝑒 B 𝑣𝑣 � � sin 𝜃𝜃 𝑑𝑑𝜃𝜃𝑑𝑑𝜙𝜙

𝑚𝑚𝑘𝑘B 𝑇𝑇

320𝑚𝑚𝑣𝑣 22 −2𝑘𝑘B 𝑇𝑇𝑣𝑣 𝑒𝑒0• It reproduces the Maxwell-Boltzmann distribution ofspeed.

2. Collisions with wall••••

9Assumptions:
The number of particles is very large (𝑁𝑁 ≫ 1).
The volume of particles (points) is negligible.
There is no interaction or collision betweenparticles.
• A particle transits in a constant velocity between 2collisions with walls.
• The velocity distribution of particles follows
Boltzmann statistics.
• The collision duration is negligible.

102. Collisions with wall

• After a collision with the wall, the velocity
component perpendicular to the wall is inversed.
• Suppose a gas particle translates in a cubic box.
• The length of the edges of the cube is 𝑙𝑙.
2𝑙𝑙
𝜏𝜏𝑥𝑥 =𝑣𝑣𝑥𝑥
𝑙𝑙∆𝑝𝑝𝑥𝑥 = 2𝑚𝑚 𝑣𝑣𝑥𝑥

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2. Collisions with wall

• The particle has a of mass 𝑚𝑚 and a velocity of𝑣𝑣⃑ = 𝑣𝑣𝑥𝑥 𝑥𝑥� + 𝑣𝑣𝑦𝑦 𝑦𝑦� + 𝑣𝑣𝑧𝑧 𝑧𝑧̂• The period for the particle to collide with one wall
perpendicular to the x axis is2𝑙𝑙𝜏𝜏 =𝑣𝑣𝑥𝑥• The total frequency of collisions by 𝑁𝑁 particles is𝑁𝑁

𝑁𝑁

𝑖𝑖=1

𝑖𝑖=1

1
1
1
𝜈𝜈 = � = � 𝑣𝑣𝑥𝑥,𝑖𝑖 = 𝑁𝑁 𝑣𝑣𝑥𝑥
𝜏𝜏𝑖𝑖 2𝑙𝑙2𝑙𝑙

2. Collisions with wall12

• Suppose the amount concentration of particles is 𝑐𝑐.
𝑁𝑁 = 𝑁𝑁A 𝑐𝑐𝑙𝑙 31𝜈𝜈 = 𝑁𝑁A 𝑐𝑐𝑙𝑙 2 𝑣𝑣𝑥𝑥
2
• The collision frequency in amount of substance perunit area of all particles is
12𝑐𝑐𝑙𝑙𝑣𝑣𝑥𝑥
𝑁𝑁𝜈𝜈1𝑘𝑘B 𝑇𝑇A
2== 𝑐𝑐 𝑣𝑣𝑥𝑥 = 𝑐𝑐𝑧𝑧 =22𝑁𝑁A 𝑙𝑙
𝑁𝑁A 𝑙𝑙22𝜋𝜋𝑚𝑚
𝑅𝑅𝑇𝑇= 𝑐𝑐
2𝜋𝜋𝑀𝑀

2. Collisions with wall

13

• For ideal gas, 𝑃𝑃 = 𝑐𝑐𝑐𝑐𝑐𝑐.

𝑃𝑃𝑅𝑅𝑅𝑅
𝑃𝑃=𝑧𝑧 =𝑅𝑅𝑇𝑇 2𝜋𝜋𝑀𝑀2𝜋𝜋𝑀𝑀𝑀𝑀𝑀𝑀• For example, the collision frequency per unit areafor nitrogen at 25℃ and 1bar is1 × 105 Pa
𝑧𝑧 =2𝜋𝜋 × 0.028kg � mol−1 × 8.314J � mol−1 � K −1 × 298K= 4.79 × 103 mol � m−2 � s −1

142. Collisions with wall

• The Newton’s second law says
𝑑𝑑𝑝𝑝
𝐹𝐹 =𝑑𝑑𝑡𝑡• Its integral form is• In one cycle,𝐼𝐼 = � 𝐹𝐹𝐹𝐹𝐹𝐹 = � 𝑑𝑑𝑑𝑑 = ∆𝑝𝑝
𝐼𝐼 = 𝐹𝐹𝑥𝑥 𝜏𝜏 = ∆𝑝𝑝𝑥𝑥
2𝑙𝑙
2𝑃𝑃 𝑙𝑙= 2𝑚𝑚 𝑣𝑣𝑥𝑥𝑣𝑣𝑥𝑥

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2. Collisions with wall

𝑚𝑚 𝑣𝑣𝑥𝑥 2 𝑚𝑚𝑣𝑣𝑥𝑥 2𝑃𝑃 ==3𝑙𝑙
𝑉𝑉• The total pressure by 𝑁𝑁 identical particles is𝑁𝑁

𝑁𝑁𝑖𝑖=1𝑖𝑖=1+∞+∞
𝑚𝑚𝑁𝑁𝑚𝑚� 𝑓𝑓𝑥𝑥 𝑣𝑣𝑥𝑥 𝑣𝑣𝑥𝑥 2 𝑑𝑑𝑣𝑣𝑥𝑥
𝑃𝑃 = � 𝑃𝑃 𝑖𝑖 = � 𝑣𝑣𝑥𝑥,𝑖𝑖 2 =𝑉𝑉𝑉𝑉 −∞𝑁𝑁𝑁𝑁𝑚𝑚�=𝑉𝑉 2𝜋𝜋𝑘𝑘B 𝑇𝑇 −∞𝑚𝑚𝑣𝑣 2
𝑁𝑁𝑘𝑘B 𝑇𝑇 𝑛𝑛𝑅𝑅𝑇𝑇−2𝑘𝑘 𝑇𝑇 2
𝑒𝑒 B 𝑣𝑣𝑥𝑥 𝑑𝑑𝑣𝑣𝑥𝑥 ==𝑉𝑉𝑉𝑉• The pressure of gas is caused by the collisions ofparticles with the wall.

3. Collisions between particles

••
•
•

16

Assumptions:
The number of particles is very large (𝑁𝑁 ≫ 1).
The particles are spherical.
There is no interaction between particles except
their collisions.
• A particle translates in a constant velocity between
2 collisions.
• The velocity distribution of particles follows
Boltzmann statistics.
• The collision duration is negligible.

3. Collisions between particles

17• To simplify the problem, suppose there are only 2
particles in the system.
• Their masses and velocities are 𝑚𝑚1 , 𝑣𝑣⃑1 and 𝑚𝑚2 , 𝑣𝑣⃑2 ,respectively.
𝜌𝜌⃑𝑚𝑚1𝑚𝑚2• The total mass is
𝑅𝑅
𝑀𝑀 = 𝑚𝑚1 + 𝑚𝑚2𝑂𝑂• The mass centre velocity is𝑚𝑚1 𝑣𝑣⃑1 + 𝑚𝑚2 𝑣𝑣⃑2
𝑈𝑈 =𝑀𝑀• The relative velocity is𝜐𝜐⃑ = 𝑣𝑣⃑1 − 𝑣𝑣⃑2

3. Collisions between particles

18• Therefore,𝑚𝑚2
𝜐𝜐⃑𝑣𝑣⃑1 = 𝑈𝑈 +𝑀𝑀𝑚𝑚1
𝜐𝜐⃑𝑣𝑣⃑2 = 𝑈𝑈 −𝑚𝑚𝑣𝑣
𝑚𝑚𝑀𝑀−𝑒𝑒 2𝑘𝑘 𝑇𝑇𝑓𝑓𝑥𝑥,𝑚𝑚 𝑣𝑣 =2𝜋𝜋𝑘𝑘B 𝑇𝑇• The distribution of 𝜐𝜐𝑥𝑥 is∞
𝑚𝑚2𝑚𝑚1𝜐𝜐𝑥𝑥 𝑓𝑓𝑥𝑥,𝑚𝑚2 𝑈𝑈𝑥𝑥 −𝜐𝜐𝑥𝑥 𝑑𝑑𝑈𝑈𝑥𝑥
𝑓𝑓 𝜐𝜐𝑥𝑥 = � 𝑓𝑓𝑥𝑥,𝑚𝑚1 𝑈𝑈𝑥𝑥 +𝑀𝑀𝑀𝑀−∞
2B

22𝑚𝑚2𝑚𝑚1𝑚𝑚1 𝑈𝑈𝑥𝑥 + 𝑀𝑀 𝜐𝜐𝑥𝑥 +𝑚𝑚2 𝑈𝑈𝑥𝑥 − 𝑀𝑀 𝜐𝜐𝑥𝑥∞
𝑚𝑚1 𝑚𝑚2−2𝑘𝑘B 𝑇𝑇� 𝑒𝑒=
𝑑𝑑𝑈𝑈𝑥𝑥2𝜋𝜋𝑘𝑘B 𝑇𝑇 −∞3. Collisions between particles

𝑓𝑓 𝜐𝜐𝑥𝑥

2 𝑚𝑚1 𝑚𝑚2 2𝑀𝑀𝑈𝑈∞
𝑥𝑥 + 𝑀𝑀 𝜐𝜐𝑥𝑥
𝑚𝑚1 𝑚𝑚2−2𝑘𝑘B 𝑇𝑇� 𝑒𝑒=
𝑑𝑑𝑈𝑈𝑥𝑥2𝜋𝜋𝑘𝑘B 𝑇𝑇 −∞• Introduce the reduced mass.
𝑚𝑚1 𝑚𝑚2𝜇𝜇 =𝑀𝑀• Then,
22∞𝜇𝜇𝜐𝜐
𝑀𝑀𝑈𝑈𝑥𝑥𝑥𝑥
𝑚𝑚1 𝑚𝑚2 −2𝑘𝑘 𝑇𝑇− 2𝑘𝑘 𝑇𝑇BB 𝑑𝑑𝑈𝑈
� 𝑒𝑒𝑒𝑒𝑓𝑓 𝜐𝜐𝑥𝑥 =𝑥𝑥
2𝜋𝜋𝑘𝑘B 𝑇𝑇−∞=𝜇𝜇2𝜋𝜋𝑘𝑘B 𝑇𝑇𝜇𝜇𝜐𝜐𝑥𝑥 2−2𝑘𝑘 𝑇𝑇
𝑒𝑒 B

19

20

3. Collisions between particles

 The relative velocity distribution between 2 particles
in ideal gas is the same as the velocity distributionof one particle of the reduced mass.
• The relative speed distribution is2𝑓𝑓 𝜐𝜐 =𝜋𝜋12𝜇𝜇𝑘𝑘B 𝑇𝑇

32𝜇𝜇𝜐𝜐22 −2𝑘𝑘B 𝑇𝑇𝜐𝜐 𝑒𝑒• The mean relative speed is given by𝜐𝜐 =8𝑘𝑘B 𝑇𝑇
𝜋𝜋𝜇𝜇213. Collisions between particles• Another way tounderstand 𝜐𝜐 .
𝑣𝑣1 =𝑣𝑣2 =8𝑘𝑘B 𝑇𝑇
𝜋𝜋𝑚𝑚18𝑘𝑘B 𝑇𝑇
𝜋𝜋𝑚𝑚2

𝜐𝜐 2 = 𝑣𝑣1 2 + 𝑣𝑣2 28𝑘𝑘B 𝑇𝑇 1
18𝑘𝑘B 𝑇𝑇=+=𝜋𝜋𝑚𝑚1 𝑚𝑚2𝜋𝜋𝜋𝜋𝑣𝑣1

𝜐𝜐

𝑣𝑣2

22

3. Collisions between particles• The 2-body system can be treated in such a way:
one of the particles is stationary and the othermoves in the relative speed.
• Suppose the 2 particles have radii 𝑟𝑟1 and 𝑟𝑟2 ,respectively. When their distance falls below 𝑅𝑅 =𝑟𝑟1 + 𝑟𝑟2 , they collide.
𝑟𝑟2𝑟𝑟1 𝑟𝑟2

𝑟𝑟1Collision

No collision

𝑟𝑟1

23

3. Collisions between particles• The condition of collision is the same as if thetranslating particle has radius 𝑅𝑅 while the stationary
particle has 0 radius (a point).
No collision

𝑑𝑑
𝑑𝑑Collision𝑑𝑑

3. Collisions between particles

24

𝑅𝑅 = 𝑟𝑟1 + 𝑟𝑟2

𝑉𝑉No collision

3. Collisions between particles

25• The ball swaps a cylinder when translating.
• The cross section of the cylinder is the collisioncross section.
𝜎𝜎 = 𝜋𝜋𝑅𝑅2• The number of collisions is the ratio of the volumeof the cylinder ( 𝜐𝜐 𝑡𝑡𝜎𝜎) to the whole volume.
• The collision frequency is number of collisions perunit time.
𝜐𝜐 𝑡𝑡𝜎𝜎 𝜎𝜎 8𝑘𝑘B 𝑇𝑇=𝜈𝜈 =𝑉𝑉𝑡𝑡𝑉𝑉
𝜋𝜋𝜇𝜇3. Collisions between particles

26• Suppose the amount concentrations of 2 types ofparticles are 𝑐𝑐1 and 𝑐𝑐2 .
• The collision frequency in amount of substance perunit volume between 2 types of particles is1 𝜈𝜈
8𝑘𝑘B 𝑇𝑇𝑐𝑐1 𝑐𝑐2
𝑁𝑁A 𝑉𝑉𝑐𝑐1 𝑁𝑁A 𝑉𝑉𝑐𝑐2 = 𝑁𝑁A 𝜎𝜎
𝑍𝑍 =𝑁𝑁A 𝑉𝑉𝜋𝜋𝜇𝜇• It suggests that the rate of bimolecular reaction ingas phase is proportional to the product of theirconcentrations (i.e., 2nd order reaction).

3. Collisions between particles

27• If there is only one type of particles of concentration
𝑐𝑐, the collision frequency in mole per unit volumeper unit time is (𝑚𝑚 = 2𝜇𝜇)1 1 𝜈𝜈
𝑘𝑘B 𝑇𝑇 22𝑐𝑐
𝑁𝑁A 𝑉𝑉𝑐𝑐 = 2𝑁𝑁A 𝜎𝜎
𝑍𝑍 =2 𝑁𝑁A 𝑉𝑉
𝜋𝜋𝑚𝑚• For example, the collision frequency of 0.8bar
nitrogen at 25℃. 𝜎𝜎 = 0.45nm2 .
𝑍𝑍 = 9.45 × 107 M � s −13. Collisions between particles

28• The mean free path, 𝜆𝜆, is the average distance thata particle travels between 2 collisions.

• Suppose there is one type of particles with molar
concentration 𝑐𝑐.
8𝑘𝑘B 𝑇𝑇1 𝜐𝜐
11𝜋𝜋𝑚𝑚=𝜆𝜆 ==𝑁𝑁A 𝑉𝑉𝑐𝑐 𝜈𝜈
𝑁𝑁A 𝑉𝑉𝑐𝑐 𝜎𝜎 8𝑘𝑘B 𝑇𝑇
2𝑁𝑁A 𝑐𝑐𝜎𝜎
𝑉𝑉 𝜋𝜋𝜇𝜇3. Collisions between particles

29• Mean free path depends on concentration (distancebetween particles).
• For ideal gas,
𝑃𝑃 = 𝑐𝑐𝑐𝑐𝑐𝑐• Thus,𝑅𝑅𝑅𝑅
𝑘𝑘B 𝑇𝑇
𝜆𝜆 ==
2𝑁𝑁A 𝜎𝜎𝑃𝑃2𝜎𝜎𝑃𝑃• For example, the mean free path of 0.8bar nitrogenat 25℃ is𝜆𝜆 ≈ 80nm