![Kinetic Theory of Gas
Gas
Microscopic [ πΎπΈ ]
Macroscopic [π, π, π]
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. Newtonβs 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 β Average KE
π =
πΉ
π΄
Firstly, find F
πΉπ₯ =
βπ π₯
βπ‘
=
β(ππ£ π₯)
βπ‘
π£ π₯
βπ£ π₯
Since, βπ π₯ = 2ππ£ π₯ ; same mass
πΉπ₯ =
2ππ£ π₯
βπ‘
βπ‘ =
2π
π£ π₯
πΉπ₯ =
2ππ£ π₯
(2π/π£ π₯)
=
ππ£ π₯
2
π
Suppose there are N molecules of the gas
πΉπ₯ =
ππ
π
π£ π₯
2
π£ π₯
2
=
π£ π₯1
2
+ π£ π₯2
2
+ π£ π₯3
2
+ β― + π£ π₯ π
2
π
π =
ππ
π΄π
π£ π₯
2
=
ππ
π
π£ π₯
2
πΎπΈ =
1
2
ππ£2
Since, π£ = π£ π₯ π + π£ π¦ π + π£π§ π
π£ = π£ = π£ π₯
2 + π£ π¦
2 + π£π§
2
π£2
= π£ π₯
2
+ π£ π¦
2
+ π£π§
2
πΎπΈ =
1
2
π π£ π₯
2
+ π£ π¦
2
+ π£π§
2
Assume, π£ π₯
2
= π£ π¦
2
= π£π§
2
πΎπΈ =
3
2
π π£ π₯
2
π
π£ π₯](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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- 1. VIII Kinetic Theory of Gas
- 2. Kinetic Theory of Gas Gas Microscopic [ πΎπΈ ] Macroscopic [π, π, π] 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. Newtonβs 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 β Average KE π = πΉ π΄ Firstly, find F πΉπ₯ = βπ π₯ βπ‘ = β(ππ£ π₯) βπ‘ π£ π₯ βπ£ π₯ Since, βπ π₯ = 2ππ£ π₯ ; same mass πΉπ₯ = 2ππ£ π₯ βπ‘ βπ‘ = 2π π£ π₯ πΉπ₯ = 2ππ£ π₯ (2π/π£ π₯) = ππ£ π₯ 2 π Suppose there are N molecules of the gas πΉπ₯ = ππ π π£ π₯ 2 π£ π₯ 2 = π£ π₯1 2 + π£ π₯2 2 + π£ π₯3 2 + β― + π£ π₯ π 2 π π = ππ π΄π π£ π₯ 2 = ππ π π£ π₯ 2 πΎπΈ = 1 2 ππ£2 Since, π£ = π£ π₯ π + π£ π¦ π + π£π§ π π£ = π£ = π£ π₯ 2 + π£ π¦ 2 + π£π§ 2 π£2 = π£ π₯ 2 + π£ π¦ 2 + π£π§ 2 πΎπΈ = 1 2 π π£ π₯ 2 + π£ π¦ 2 + π£π§ 2 Assume, π£ π₯ 2 = π£ π¦ 2 = π£π§ 2 πΎπΈ = 3 2 π π£ π₯ 2 π π£ π₯
- 3. πΎπΈ = 3 2 π π£ π₯ 2 π π£ π₯ 2 = 2 3 πΎπΈ π = ππ π π£ π₯ 2 ***The reference point of π£ π₯ is CG π = 1 3 ππ π2 π ***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) π = 2 3 π π πΎπΈ πΎπΈ = 1 2 ππ Gas molecule U total Monatomic π = 3 2 πππ Linear π = 5 2 πππ Non-linear π = 3πππ Monatomic molecules *** The rotation of the molecule releases very small energy as πΎπΈ = 1 2 πΌπ2 Translational motion (π₯, π¦, π§) Degree Of Freedom, DOF = 3 Linear molecule Translational motion (π₯, π¦, π§) Rotational motion (π π₯, π π¦) Degree Of Freedom, DOF = 5 ***If include Oscillation motion, DOF = 5+2(KE+PE) =7 Translational motion Rotational motion π§ π₯ π¦ππ‘ππ‘ππ = π·ππΉ 2 πππ
- 4. Non-linear molecule Translational motion (π₯, π¦, π§) Rotational motion (π π₯, π π¦, π π§) 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, π£πππ π£πππ = π£2 π§ π₯ π¦ πΎπΈ = 1 2 π π£2 π£2 = π£1 2 + π£2 2 + β― + π£ π 2 π πΎπΈ = 1 2 ππ£πππ 2 Specific Heat Capacity of gas π = ππβπ π = π πππππππ βπππ‘ πππππππ‘π¦ π½ ππβ Molar Specific Heat Capacity π = ππ πβπ π = ππ’ππππ ππ ππππ π π = πππππ π πππππππ βπππ‘ πππππππ‘π¦ π½ ππππβ When volume is constant, π = βπ ***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 βU πππ¦ = π€πππ ππππ ππ¦ πππ When volume is constant, W=pβV=0 π = βπ β΄βπ = ππ π£βπ πΆπ£ = π πππππππ βπππ‘ πππππππ‘π¦ π€βππ π = 0 π π£ = 1 π βπ βπ = 1 π ππ ππ βπ = π β πππ¦ π π£ = 1 π ππ ππ π = ππ
- 5. For Ideal Gas β΄ π = 2 3 π π 3 2 ππ P-V graph ***Isotherm line; same temperature P-T graph ***Isovolumetric; same volume Monatomic molecule When v=0, π = 3 2 πππ π π£ = 1 π ππ ππ = π π£ = 1 π π ππ 3 2 πππ π π£ = 3 2 ππ π Since π = ππ, π π£ = 3 2 ππ ππ = 3 2 π π Linear molecule Non-linear molecule ***When v=0 and not consider the vibration motion π π- π» graph of Hydrogen gas Hydrogen gas is a linear molecule (diatomic molecule) π π£ = 3 2 π π π π£ = 5 2 π π π π£ = 3 π π π π π» π π π π π π πππππ πππ‘πππ πππππ πππ‘πππ +π ππ‘ππ‘πππ πππππ πππ‘πππ π ππ‘ππ‘πππ +Oscillation π = 2 3 π π πΎπΈ πΎπΈ = 3 2 ππ ππ½ = π΅ππ» π· π½ ππππ² ππππ² ππππ² π»ππππππππππ πππππππππ π· π» π½ π π½ π > π½ π
- 6. Since ππ = πππ and n = π π π΄ π = π ππ΄ Real Gas VS Ideal Gas Because the real gas has intermolecular forces, ππΌππππ > πππππ Pβ = 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. ππΌππππ < πππππ Vβ = volume of Ideal gas V = volume of real gas (by Clausius) ππ = πππ = ππ π ππ‘ππ‘ππ = π·ππΉ 2 πππ = π·ππΉ 2 ππ π πβ² = π + π π π 2 πβ² = π β ππ π·β² π½β² = π· + π π΅ π½ π π½ β ππ = π΅ππ» = ππΉπ» 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), π» π ***We can derive Critical volume and Critical Pressure At π» π, when π½ = π½ π ππ ππ π = 0 and π2 π ππ2 π = 0 π» π is the stationary point of inflexion π½ π π· π½ π· π π» π
- 7. π·β² π½β² = π· + π π΅ π½ π π½ β ππ = π΅ππ» = ππΉπ» At π» π, when π½ = π½ π ππ ππ π = 0 and π2 π ππ2 π = 0 From Van der Waals equation, π + π π π 2 π β ππ = πππ π = πππ π β ππ β π π π 2 ππ ππ = β πππ π β ππ 2 + 2ππ2 π3 = 0 β― (1) π2 π ππ2 = 2πππ π β ππ 3 β 6ππ2 π4 = 0 β― (2) (1) (2) ; π β ππ = 2π 3 π = π 3π = ππ 3π Sub π = π 3π into (1) and (2) π = 9πππ 8π π = π 27π2 Hence, π = 9πππ 8π π = π 3π ***To find a and b, an experiment is conducted π = ππ½ π ππ» π ππ΅ , π = π½ π ππ΅ Speed/Velocity of gas molecule 1. Average speed, π£ ππ£ π£ ππ£ = π£1 + π£2 + π£3 + β― + π£ π π 2. Root mean square velocity, π£πππ π£πππ = π£2 = π£1 2 + π£2 2 + π£3 2 + β― + π£ π 2 π 3. Most probable velocity, π£ ππ π£ ππ = ππππ Distribute of Molecular Speeds Maxwell speed distribution ππ₯π β ππ£2 2ππ = πβ ππ£2 2ππ ***m = mass of 1 molecule Probability density function, π·(π) π = ππππππππππ‘π¦ π π£ = 4 π π 2ππ 3 2 π£2 ππ₯π β ππ£2 2ππ π£ ππ’ππππππππππππ’ππ π π£ + βπ₯ βπ₯ π π£ = π π£ + βπ£ β π(π£) βπ£
- 8. Use Integrate by part π π£1 β π£2 = π π£ ππ£ π£2 π£1 Speed/Velocity of gas molecule 1. Average speed, π£ ππ£ π£ ππ£ = π£π π£ ππ£ β 0 π ππ = π£ 4 π π 2ππ 3 2 π£2 ππ₯π β ππ£2 2ππ ππ£ β 0 = 4 π π 2ππ 3 2 π£3 ππ₯π β ππ£2 2ππ ππ£ β 0 From formula, π₯3 πβππ₯2 ππ₯ β 0 = 1 2π2 π = ππππππ’πππ πππ π (ππ) 2. Root mean square velocity, π£πππ π£πππ = π£2 = π£2 π π£ ππ£ β 0 With formula, π₯4 πβππ₯2 ππ₯ β 0 = 3 8 π π3 3. Most probable velocity, π£ ππ π£ = π£ ππ, ππ ππ£ π£=π£ ππ = 0 Probability density function, π·(π) ππππππππππ‘π¦ π‘π ππππ ππππππ’πππ π€ππ‘β π£ππππππ‘π¦ π£1 β π£2 ***the areas under 3 graphs of the same gas but different temperature is the SAME ***π = ππ Γ 1.66 Γ 10β27 ***π = 1.38 Γ 10β23 ***π ππ ππ πΎπππ£ππ π£1 π(π£) π π£2 π π£1 β π£2 = π π£ ππ£ π£2 π£1 π π£ ππ£ β 0 = 1 π£ ππ1 π(π£) π π£ ππ2 π£ ππ3 π3 π2 π1 π1 < π2< π3 π£ ππ£ = 8ππ ππ = 8π π ππ π£2 = π£2 = 3ππ π π£πππ = 3ππ π π£ ππ = 2ππ π
- 9. π£ ππ < π£ ππ£ < π£πππ = 1: 1.128: 1.224 Mean Free Path, π [π] free path ***1 assumption is that the other molecules are static When 2 molecules of π diameter collide, we use a cross section of 2π to calculate as it is more convenient. π 2π The collision occurs in βπ‘ second(s) Thus, the volume which the collision takes place is βπ = (ππ2 )(π£βπ‘) The number of the collisions between gas molecules is approximately equal to the number of molecules inside the volume which the collision takes place (βπ) Suppose in π volume, there are π numbers of gas molecules Therefore, in βπ, there will be π π Γ βπ = π π (ππ2 )(π£βπ‘) Ξ» = πππ π‘ππππ ππ βπ‘ number of the collisions = π£βπ‘ π π (ππ2)(π£βπ‘) = 1 π π (ππ2) However, in reality, every molecule moves randomly all the time. Thus, from the experiments. 2π βππ‘ Ξ» = 1 2 ππ2 π π