QJEEAdv16Che5c8: JEE Advanced 2016 Chemistry - Class 11 JEEAdv16Che5c8 Solution: The diffusion coefficient of an ideal gas is proportional to…

The diffusion coefficient (D) is proportional to the mean free path (λ) and the mean speed (U_mean), so we can write

$$D\propto \lambda U_{\text{mean}}$$

The mean free path (λ) is given by

$$\lambda =\frac{RT}{\sqrt{2}{N}_0\sigma P}$$

hence

$$\lambda \propto \frac{T}{P}$$

The mean speed (U_mean) is given by

$$U_{\text{mean}}=\sqrt{\frac{8RT}{\pi M}}$$

hence

$$U_{\text{mean}}\propto \sqrt{T}$$

Therefore,

$$D\propto \frac{T^{3/2}}{P}$$

The change in the diffusion coefficient would then be given by :

$$\frac{(DC{)}_2}{(DC{)}_1}=\frac{P_1}{P_2}\cdot {\left(\frac{T_2}{T_1}\right)}^{3/2}$$

Substituting $P_2 = 2P_1$ and $T_2 = 4T_1$ into the equation, we get :

$$\frac{(D{)}_2}{(D{)}_1}=\frac{1}{2}\cdot (4{)}^{3/2}=4$$

So, the diffusion coefficient of the gas increases 4 times. Hence, x = 4.