Question

Prove that the average kinetic energy of a molecule of an ideal gas is directly proportional to the absolute temperature of the gas.

Answer 1

Pressure (P) = m n v² / 3

m = mass , v = rms speed, n = number of molecules per unit volume

n = N/v where N = Number of molecules

Substituting in Pressure equation we get,

Pv = m n v ² / 3    ... (2)

m v² / 2 = E (Kinetic Energy)

Substituting that in (2)

Pv = 2 N E / 3  .... (3)

For Ideal gas, Pv = μ R T , μ = N / Na  

Substituting in (3) we get,

N R T /Na = 2 N E / 3 ..(4)

k = R/Na -  Boltzmann constant

Substitute in 4, we get

E = 3kT/2

Hence  E α T i.e. Kinetic Energy Proportional to Absolute Temperature.

Answer 2

Answer:

According to kinetic Molecular theory, as the temperature rises, the average kinetic energy of the molecules rises as well. It demonstrates that the molecule's average kinetic energy is related to the absolute temperature of the gas.

Explanation:

According to the kinetic pressure energy equation,

P = $\frac{1}{3}$ρc²

Where, P - the pressure of the gas

ρ- density of the gas particles

c - velocity of the gas particles

Formula of density,

ρ = $\frac{M}{V}$

Where, M- the mass of all particles

V - the volume of the gas particles

Substituting the above density equation in the kinetic pressure equation gives,

P = $\frac{1}{3}$ X $\frac{M}{V}$ X c²

Total number of particles, M = (m X N)

Where, m-mass of every individual particle

N-number of particles

Therefore, the kinetic pressure equation becomes,

P = $\frac{1}{3}$ X $\frac{m X N}{V}$ X c²

P.V = $\frac{1}{3}$ X N X mc²

The kinetic energy equation is given by,

K.E = $\frac{1}{2}$ X mc²

Substituting the K.E equation in kinetic pressure equation,

P.V = $\frac{1}{3}$ X N X 2 K.E

According to ideal gas law,

P.V = nRT

Where, n-amount of substance

R - ideal gas constant

T - temperature

Substituting the ideal gas equation,

nRT = $\frac{2}{3}$ X N X K.E

The amount of substance,

n = $\frac{N}{NA}$

Where, N- total number of particles

NA - Avagadro constant (NA = 6.022×1023mol−1)

Substituting the value of n,

$\frac{N}{NA}$ X RT = $\frac{2}{3}$ X N X K.E

⇒ T = $\frac{NA}{R}$ X $\frac{2}{3}$ X K.E

In the above equation, the value NA and R are the constant.

Therefore T ∝ K.E, that means the absolute temperature of the gas is directly proportional to the average kinetic energy of the molecule of an ideal gas.

Hence it is proved.