Class 12 > Unit # 15:Molecular Theory of Gases > Effect of Temperature on Gases

 

Q.5 Prove that the average Translational K.E. of a gas molecule is directly proportional to the absolute temperature and derive an expression for the Kinetic Energy of a molecule.

PROOF:-

As we know from the kinetic theory of gases, the macroscopic pressure exerted by an ideal gas is given by the relation:

$$P = \frac{1}{3}\rho\overline{v^2}$$

Since gas density is defined as total mass divided by internal volume ($\rho = \frac{mN}{V}$), we can rewrite the equation as:

$$P = \frac{1}{3} \left(\frac{mN}{V}\right) \overline{v^2}$$ $$PV = \frac{1}{3} mN\overline{v^2} \quad \text{--- (i)}$$

Also, from the macroscopic ideal gas equation for an ideal state, we know that:

$$PV = nRT$$

Since the number of moles is $n = \frac{N}{N_A}$ (where $N$ is total molecules and $N_A$ is Avogadro's number):

$$PV = \left(\frac{N}{N_A}\right) RT$$ $$PV = N \left(\frac{R}{N_A}\right) T$$

Substituting Boltzmann's constant ($k_B = \frac{R}{N_A}$) gives:

$$PV = Nk_B T \quad \text{--- (ii)}$$

Comparing equation (i) and equation (ii), we get:

$$\frac{1}{3} mN\overline{v^2} = Nk_B T$$

Canceling the total number of molecules ($N$) from both sides yields:

$$\frac{1}{3} m\overline{v^2} = k_B T$$

To explicitly reveal the kinetic energy structural term, we multiply and divide the left side by "$2$":

$$\frac{2}{3} \left( \frac{1}{2}m\overline{v^2} \right) = k_B T$$

Since the average translational kinetic energy of a single molecule is defined as $\text{K.E.}_{\text{avg}} = \frac{1}{2}m\overline{v^2}$, we can write:

$$\frac{2}{3} (\text{K.E.}_{\text{avg}}) = k_B T$$ $$\text{K.E.}_{\text{avg}} = \frac{3}{2} k_B T \quad \text{--- (iii)}$$

Since $\frac{3}{2}$ and $k_B$ are fundamental physical constants, the equation simplifies to:

$$\text{K.E.}_{\text{avg}} = (\text{Constant}) \cdot T$$ $$\text{K.E.}_{\text{avg}} \propto T$$

Hence proved that "The average Translational K.E. of a gas molecule is directly proportional to the absolute temperature."

EXPRESSION FOR KINETIC ENERGY PER MOLE

To determine the macroscopic kinetic energy of one complete mole of gas ($E_m$), we multiply the single-molecule energy equation (iii) by Avogadro's number ($N_A$):

$$E_m = N_A \cdot (\text{K.E.}_{\text{avg}}) = N_A \left( \frac{3}{2} k_B T \right)$$ $$E_m = \frac{3}{2} (N_A k_B) T$$

Since $N_A k_B = R$ (the Universal Gas Constant), the kinetic expression for one mole of gas is:

$$E_m = \frac{3}{2} RT$$