![Kinetic Theory of Gas
Gas
Microscopic [ 𝐾𝐸 ]
Macroscopic [𝑃, 𝑇, 𝑉]
Assumptions of Ideal gas
I. The molecules in the gas can be
considered small hard spheres.
II. All collisions between gas
molecules are elastic and all
motion is frictionless (no energy is
lost in collisions or in motion).
III. Newton’s laws apply .
IV. The distance between molecules
on average is much larger than the
size of the molecules.
V. The gas molecules are constantly
moving in random directions with
a distribution of speeds.
VI. There are no attractive or
repulsive forces between the
molecules or the surroundings
(no intermolecular forces).
Pressure – Average KE
𝑃 =
𝐹
𝐴
Firstly, find F
𝐹𝑥 =
∆𝑝 𝑥
∆𝑡
=
∆(𝑚𝑣 𝑥)
∆𝑡
𝑣 𝑥
−𝑣 𝑥
Since, ∆𝑝 𝑥 = 2𝑚𝑣 𝑥 ; same mass
𝐹𝑥 =
2𝑚𝑣 𝑥
∆𝑡
∆𝑡 =
2𝑑
𝑣 𝑥
𝐹𝑥 =
2𝑚𝑣 𝑥
(2𝑑/𝑣 𝑥)
=
𝑚𝑣 𝑥
2
𝑑
Suppose there are N molecules of the gas
𝐹𝑥 =
𝑁𝑚
𝑑
𝑣 𝑥
2
𝑣 𝑥
2
=
𝑣 𝑥1
2
+ 𝑣 𝑥2
2
+ 𝑣 𝑥3
2
+ ⋯ + 𝑣 𝑥 𝑛
2
𝑛
𝑃 =
𝑁𝑚
𝐴𝑑
𝑣 𝑥
2
=
𝑁𝑚
𝑉
𝑣 𝑥
2
𝐾𝐸 =
1
2
𝑚𝑣2
Since, 𝑣 = 𝑣 𝑥 𝑖 + 𝑣 𝑦 𝑗 + 𝑣𝑧 𝑘
𝑣 = 𝑣 = 𝑣 𝑥
2 + 𝑣 𝑦
2 + 𝑣𝑧
2
𝑣2
= 𝑣 𝑥
2
+ 𝑣 𝑦
2
+ 𝑣𝑧
2
𝐾𝐸 =
1
2
𝑚 𝑣 𝑥
2
+ 𝑣 𝑦
2
+ 𝑣𝑧
2
Assume, 𝑣 𝑥
2
= 𝑣 𝑦
2
= 𝑣𝑧
2
𝐾𝐸 =
3
2
𝑚 𝑣 𝑥
2
𝑑
𝑣 𝑥](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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- 1. VIII Kinetic Theory of Gas
- 2. Kinetic Theory of Gas Gas Microscopic [ 𝐾𝐸 ] Macroscopic [𝑃, 𝑇, 𝑉] Assumptions of Ideal gas I. The molecules in the gas can be considered small hard spheres. II. All collisions between gas molecules are elastic and all motion is frictionless (no energy is lost in collisions or in motion). III. Newton’s laws apply . IV. The distance between molecules on average is much larger than the size of the molecules. V. The gas molecules are constantly moving in random directions with a distribution of speeds. VI. There are no attractive or repulsive forces between the molecules or the surroundings (no intermolecular forces). Pressure – Average KE 𝑃 = 𝐹 𝐴 Firstly, find F 𝐹𝑥 = ∆𝑝 𝑥 ∆𝑡 = ∆(𝑚𝑣 𝑥) ∆𝑡 𝑣 𝑥 −𝑣 𝑥 Since, ∆𝑝 𝑥 = 2𝑚𝑣 𝑥 ; same mass 𝐹𝑥 = 2𝑚𝑣 𝑥 ∆𝑡 ∆𝑡 = 2𝑑 𝑣 𝑥 𝐹𝑥 = 2𝑚𝑣 𝑥 (2𝑑/𝑣 𝑥) = 𝑚𝑣 𝑥 2 𝑑 Suppose there are N molecules of the gas 𝐹𝑥 = 𝑁𝑚 𝑑 𝑣 𝑥 2 𝑣 𝑥 2 = 𝑣 𝑥1 2 + 𝑣 𝑥2 2 + 𝑣 𝑥3 2 + ⋯ + 𝑣 𝑥 𝑛 2 𝑛 𝑃 = 𝑁𝑚 𝐴𝑑 𝑣 𝑥 2 = 𝑁𝑚 𝑉 𝑣 𝑥 2 𝐾𝐸 = 1 2 𝑚𝑣2 Since, 𝑣 = 𝑣 𝑥 𝑖 + 𝑣 𝑦 𝑗 + 𝑣𝑧 𝑘 𝑣 = 𝑣 = 𝑣 𝑥 2 + 𝑣 𝑦 2 + 𝑣𝑧 2 𝑣2 = 𝑣 𝑥 2 + 𝑣 𝑦 2 + 𝑣𝑧 2 𝐾𝐸 = 1 2 𝑚 𝑣 𝑥 2 + 𝑣 𝑦 2 + 𝑣𝑧 2 Assume, 𝑣 𝑥 2 = 𝑣 𝑦 2 = 𝑣𝑧 2 𝐾𝐸 = 3 2 𝑚 𝑣 𝑥 2 𝑑 𝑣 𝑥
- 3. 𝐾𝐸 = 3 2 𝑚 𝑣 𝑥 2 𝑚 𝑣 𝑥 2 = 2 3 𝐾𝐸 𝑃 = 𝑁𝑚 𝑉 𝑣 𝑥 2 ***The reference point of 𝑣 𝑥 is CG 𝑝 = 1 3 𝑁𝑚 𝑐2 𝑉 ***This equation has not considered the Degree Of Freedom, DOF Equipartition Theorem Mode of motions 1. Translational motion 2. Rotational motion 3. Oscillation motion (vibration) 𝑃 = 2 3 𝑁 𝑉 𝐾𝐸 𝐾𝐸 = 1 2 𝑘𝑇 Gas molecule U total Monatomic 𝑈 = 3 2 𝑁𝑘𝑇 Linear 𝑈 = 5 2 𝑁𝑘𝑇 Non-linear 𝑈 = 3𝑁𝑘𝑇 Monatomic molecules *** The rotation of the molecule releases very small energy as 𝐾𝐸 = 1 2 𝐼𝜔2 Translational motion (𝑥, 𝑦, 𝑧) Degree Of Freedom, DOF = 3 Linear molecule Translational motion (𝑥, 𝑦, 𝑧) Rotational motion (𝜔 𝑥, 𝜔 𝑦) Degree Of Freedom, DOF = 5 ***If include Oscillation motion, DOF = 5+2(KE+PE) =7 Translational motion Rotational motion 𝑧 𝑥 𝑦𝑈𝑡𝑜𝑡𝑎𝑙 = 𝐷𝑂𝐹 2 𝑁𝑘𝑇
- 4. Non-linear molecule Translational motion (𝑥, 𝑦, 𝑧) Rotational motion (𝜔 𝑥, 𝜔 𝑦, 𝜔 𝑧) Degree Of Freedom, DOF = 6 ***For oscillation motion or vibration, it is usually not used to calculate as it has a large impact only at very high temperature. Root mean square velocity, 𝑣𝑟𝑚𝑠 𝑣𝑟𝑚𝑠 = 𝑣2 𝑧 𝑥 𝑦 𝐾𝐸 = 1 2 𝑚 𝑣2 𝑣2 = 𝑣1 2 + 𝑣2 2 + ⋯ + 𝑣 𝑛 2 𝑛 𝐾𝐸 = 1 2 𝑚𝑣𝑟𝑚𝑠 2 Specific Heat Capacity of gas 𝑄 = 𝑚𝑐∆𝑇 𝑐 = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝐽 𝑘𝑔℃ Molar Specific Heat Capacity 𝑄 = 𝑛𝑐 𝑚∆𝑇 𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒 𝑐 𝑚 = 𝑚𝑜𝑙𝑎𝑟 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝐽 𝑚𝑜𝑙𝑒℃ When volume is constant, 𝑄 = ∆𝑈 ***When energy is inserted to a system, the system will do work. ***However, Q inserted in is not equal to the Work (w) done ***The remained energy is ∆U 𝑊𝑏𝑦 = 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑏𝑦 𝑔𝑎𝑠 When volume is constant, W=p∆V=0 𝑄 = ∆𝑈 ∴∆𝑈 = 𝑀𝑐 𝑣∆𝑇 𝐶𝑣 = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑤ℎ𝑒𝑛 𝑉 = 0 𝑐 𝑣 = 1 𝑀 ∆𝑈 ∆𝑇 = 1 𝑀 𝑑𝑈 𝑑𝑇 ∆𝑈 = 𝑄 − 𝑊𝑏𝑦 𝑐 𝑣 = 1 𝑀 𝜕𝑈 𝜕𝑇 𝑀 = 𝑚𝑁
- 5. For Ideal Gas ∴ 𝑃 = 2 3 𝑁 𝑉 3 2 𝑘𝑇 P-V graph ***Isotherm line; same temperature P-T graph ***Isovolumetric; same volume Monatomic molecule When v=0, 𝑈 = 3 2 𝑁𝑘𝑇 𝑐 𝑣 = 1 𝑀 𝜕𝑈 𝜕𝑇 = 𝑐 𝑣 = 1 𝑀 𝜕 𝜕𝑇 3 2 𝑁𝑘𝑇 𝑐 𝑣 = 3 2 𝑁𝑘 𝑀 Since 𝑀 = 𝑚𝑁, 𝑐 𝑣 = 3 2 𝑁𝑘 𝑚𝑁 = 3 2 𝑘 𝑚 Linear molecule Non-linear molecule ***When v=0 and not consider the vibration motion 𝒄 𝒗- 𝑻 graph of Hydrogen gas Hydrogen gas is a linear molecule (diatomic molecule) 𝑐 𝑣 = 3 2 𝑘 𝑚 𝑐 𝑣 = 5 2 𝑘 𝑚 𝑐 𝑣 = 3 𝑘 𝑚 𝒄 𝒗 𝑻 𝟑 𝟐 𝟓 𝟐 𝟕 𝟐 𝑇𝑟𝑎𝑛𝑠𝑙𝑎𝑡𝑖𝑜𝑛 𝑇𝑟𝑎𝑛𝑠𝑙𝑎𝑡𝑖𝑜𝑛 +𝑅𝑜𝑡𝑎𝑡𝑖𝑜𝑛 𝑇𝑟𝑎𝑛𝑠𝑙𝑎𝑡𝑖𝑜𝑛 𝑅𝑜𝑡𝑎𝑡𝑖𝑜𝑛 +Oscillation 𝑃 = 2 3 𝑁 𝑉 𝐾𝐸 𝐾𝐸 = 3 2 𝑘𝑇 𝒑𝑽 = 𝑵𝒌𝑻 𝑷 𝑽 𝟑𝟎𝟎𝑲 𝟑𝟐𝟎𝑲 𝟑𝟑𝟎𝑲 𝑻𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝒊𝒏𝒄𝒓𝒆𝒂𝒔𝒆𝒔 𝑷 𝑻 𝑽 𝟏 𝑽 𝟐 > 𝑽 𝟏
- 6. Since 𝑝𝑉 = 𝑁𝑘𝑇 and n = 𝑁 𝑁 𝐴 𝑘 = 𝑅 𝑁𝐴 Real Gas VS Ideal Gas Because the real gas has intermolecular forces, 𝑃𝐼𝑑𝑒𝑎𝑙 > 𝑃𝑟𝑒𝑎𝑙 P’ = pressure of Ideal gas P = pressure of real gas (by Van der Waal) ***When we observe the volume of the real gas, it is the volume of the container. *** But because the real gas has molecular size, the volume of Ideal gas is only the space that is not occupied by the real gas molecules. 𝑉𝐼𝑑𝑒𝑎𝑙 < 𝑉𝑟𝑒𝑎𝑙 V’ = volume of Ideal gas V = volume of real gas (by Clausius) 𝑝𝑉 = 𝑁𝑘𝑇 = 𝑛𝑅𝑇 𝑈𝑡𝑜𝑡𝑎𝑙 = 𝐷𝑂𝐹 2 𝑁𝑘𝑇 = 𝐷𝑂𝐹 2 𝑛𝑅𝑇 𝑃′ = 𝑃 + 𝑎 𝑁 𝑉 2 𝑉′ = 𝑉 − 𝑛𝑏 𝑷′ 𝑽′ = 𝑷 + 𝒂 𝑵 𝑽 𝟐 𝑽 − 𝒏𝒃 = 𝑵𝒌𝑻 = 𝒏𝑹𝑻 Real Gas equation Van der Waals equation a and b are Van der Waals constants. a and b are vary for different gases How to determine a and b values? ***Van der Waals Curves ***Choose the isotherm line that first not curve down (Critical temperature) ***Choose the stationary point of inflexion (Critical point), 𝑻 𝒄 ***We can derive Critical volume and Critical Pressure At 𝑻 𝒄, when 𝑽 = 𝑽 𝒄 𝜕𝑃 𝜕𝑉 𝑇 = 0 and 𝜕2 𝑃 𝜕𝑉2 𝑇 = 0 𝑻 𝒄 is the stationary point of inflexion 𝑽 𝒄 𝑷 𝑽 𝑷 𝒄 𝑻 𝒄
- 7. 𝑷′ 𝑽′ = 𝑷 + 𝒂 𝑵 𝑽 𝟐 𝑽 − 𝒏𝒃 = 𝑵𝒌𝑻 = 𝒏𝑹𝑻 At 𝑻 𝒄, when 𝑽 = 𝑽 𝒄 𝜕𝑃 𝜕𝑉 𝑇 = 0 and 𝜕2 𝑃 𝜕𝑉2 𝑇 = 0 From Van der Waals equation, 𝑃 + 𝑎 𝑁 𝑉 2 𝑉 − 𝑛𝑏 = 𝑁𝑘𝑇 𝑃 = 𝑁𝑘𝑇 𝑉 − 𝑛𝑏 − 𝑎 𝑁 𝑉 2 𝜕𝑃 𝜕𝑉 = − 𝑁𝑘𝑇 𝑉 − 𝑛𝑏 2 + 2𝑎𝑁2 𝑉3 = 0 ⋯ (1) 𝜕2 𝑃 𝜕𝑉2 = 2𝑁𝑘𝑇 𝑉 − 𝑛𝑏 3 − 6𝑎𝑁2 𝑉4 = 0 ⋯ (2) (1) (2) ; 𝑉 − 𝑁𝑏 = 2𝑉 3 𝑏 = 𝑉 3𝑁 = 𝑉𝑐 3𝑁 Sub 𝑏 = 𝑉 3𝑁 into (1) and (2) 𝑇 = 9𝑉𝑘𝑇 8𝑁 𝑃 = 𝑎 27𝑏2 Hence, 𝑎 = 9𝑉𝑘𝑇 8𝑁 𝑏 = 𝑉 3𝑁 ***To find a and b, an experiment is conducted 𝒂 = 𝟗𝑽 𝒄 𝒌𝑻 𝒄 𝟖𝑵 , 𝒃 = 𝑽 𝒄 𝟑𝑵 Speed/Velocity of gas molecule 1. Average speed, 𝑣 𝑎𝑣 𝑣 𝑎𝑣 = 𝑣1 + 𝑣2 + 𝑣3 + ⋯ + 𝑣 𝑛 𝑛 2. Root mean square velocity, 𝑣𝑟𝑚𝑠 𝑣𝑟𝑚𝑠 = 𝑣2 = 𝑣1 2 + 𝑣2 2 + 𝑣3 2 + ⋯ + 𝑣 𝑛 2 𝑛 3. Most probable velocity, 𝑣 𝑚𝑝 𝑣 𝑚𝑝 = 𝑚𝑜𝑑𝑒 Distribute of Molecular Speeds Maxwell speed distribution 𝑒𝑥𝑝 − 𝑚𝑣2 2𝑘𝑇 = 𝑒− 𝑚𝑣2 2𝑘𝑇 ***m = mass of 1 molecule Probability density function, 𝑷(𝒗) 𝑝 = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑃 𝑣 = 4 𝜋 𝑚 2𝑘𝑇 3 2 𝑣2 𝑒𝑥𝑝 − 𝑚𝑣2 2𝑘𝑇 𝑣 𝑁𝑢𝑚𝑏𝑒𝑟𝑜𝑓𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝑉 𝑣 + ∆𝑥 ∆𝑥 𝑃 𝑣 = 𝑝 𝑣 + ∆𝑣 − 𝑝(𝑣) ∆𝑣
- 8. Use Integrate by part 𝑝 𝑣1 → 𝑣2 = 𝑃 𝑣 𝑑𝑣 𝑣2 𝑣1 Speed/Velocity of gas molecule 1. Average speed, 𝑣 𝑎𝑣 𝑣 𝑎𝑣 = 𝑣𝑃 𝑣 𝑑𝑣 ∞ 0 𝒗 𝒂𝒗 = 𝑣 4 𝜋 𝑚 2𝑘𝑇 3 2 𝑣2 𝑒𝑥𝑝 − 𝑚𝑣2 2𝑘𝑇 𝑑𝑣 ∞ 0 = 4 𝜋 𝑚 2𝑘𝑇 3 2 𝑣3 𝑒𝑥𝑝 − 𝑚𝑣2 2𝑘𝑇 𝑑𝑣 ∞ 0 From formula, 𝑥3 𝑒−𝑎𝑥2 𝑑𝑥 ∞ 0 = 1 2𝑎2 𝑀 = 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑘𝑔) 2. Root mean square velocity, 𝑣𝑟𝑚𝑠 𝑣𝑟𝑚𝑠 = 𝑣2 = 𝑣2 𝑃 𝑣 𝑑𝑣 ∞ 0 With formula, 𝑥4 𝑒−𝑎𝑥2 𝑑𝑥 ∞ 0 = 3 8 𝜋 𝑎3 3. Most probable velocity, 𝑣 𝑚𝑝 𝑣 = 𝑣 𝑚𝑝, 𝑑𝑃 𝑑𝑣 𝑣=𝑣 𝑚𝑝 = 0 Probability density function, 𝑷(𝒗) 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡𝑜 𝑓𝑖𝑛𝑑 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑤𝑖𝑡ℎ 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑣1 → 𝑣2 ***the areas under 3 graphs of the same gas but different temperature is the SAME ***𝑚 = 𝑀𝑟 × 1.66 × 10−27 ***𝑘 = 1.38 × 10−23 ***𝑇 𝑖𝑠 𝑖𝑛 𝐾𝑒𝑙𝑣𝑖𝑛 𝑣1 𝑃(𝑣) 𝑉 𝑣2 𝑝 𝑣1 → 𝑣2 = 𝑃 𝑣 𝑑𝑣 𝑣2 𝑣1 𝑃 𝑣 𝑑𝑣 ∞ 0 = 1 𝑣 𝑚𝑝1 𝑃(𝑣) 𝑉 𝑣 𝑚𝑝2 𝑣 𝑚𝑝3 𝑇3 𝑇2 𝑇1 𝑇1 < 𝑇2< 𝑇3 𝑣 𝑎𝑣 = 8𝑘𝑇 𝜋𝑚 = 8𝑅𝑇 𝜋𝑀 𝑣2 = 𝑣2 = 3𝑘𝑇 𝑚 𝑣𝑟𝑚𝑠 = 3𝑘𝑇 𝑚 𝑣 𝑚𝑝 = 2𝑘𝑇 𝑚
- 9. 𝑣 𝑚𝑝 < 𝑣 𝑎𝑣 < 𝑣𝑟𝑚𝑠 = 1: 1.128: 1.224 Mean Free Path, 𝜆 [𝑚] free path ***1 assumption is that the other molecules are static When 2 molecules of 𝑑 diameter collide, we use a cross section of 2𝑑 to calculate as it is more convenient. 𝑑 2𝑑 The collision occurs in ∆𝑡 second(s) Thus, the volume which the collision takes place is ∆𝑉 = (𝜋𝑑2 )(𝑣∆𝑡) The number of the collisions between gas molecules is approximately equal to the number of molecules inside the volume which the collision takes place (∆𝑉) Suppose in 𝑉 volume, there are 𝑁 numbers of gas molecules Therefore, in ∆𝑉, there will be 𝑁 𝑉 × ∆𝑉 = 𝑁 𝑉 (𝜋𝑑2 )(𝑣∆𝑡) λ = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 ∆𝑡 number of the collisions = 𝑣∆𝑡 𝑁 𝑉 (𝜋𝑑2)(𝑣∆𝑡) = 1 𝑁 𝑉 (𝜋𝑑2) However, in reality, every molecule moves randomly all the time. Thus, from the experiments. 2𝑑 ∆𝑉𝑡 λ = 1 2 𝜋𝑑2 𝑁 𝑉