Proportionality of Diffusion Coefficient on Temperature and Pressure Changes

To find the factor $x$ by which the diffusion coefficient increases, we analyze how the diffusion coefficient ($D$) depends on absolute temperature ($T$) and pressure ($P$).

Given that the diffusion coefficient $D$ is directly proportional to the mean free path ($\lambda$) and the mean speed ($\bar{v}$): $D \propto \lambda \cdot \bar{v}$

1. Dependence of Mean Free Path ($\lambda$) on $T$ and $P$: The mean free path of an ideal gas is given by: $\lambda = \frac{1}{\sqrt{2} \pi d^2 n^*}$ where $d$ is the molecular diameter and $n^*$ is the number density of molecules ($n^* = \frac{P}{k_B T}$).

Thus, $\lambda$ is inversely proportional to $n^*$: $\lambda \propto \frac{T}{P}$

1. Dependence of Mean Speed ($\bar{v}$) on $T$: The mean speed of gas molecules is given by: $\bar{v} = \sqrt{\frac{8 R T}{\pi M}}$

Thus, $\bar{v}$ depends on temperature as: $\bar{v} \propto \sqrt{T}$

1. Combined Proportionality for Diffusion Coefficient ($D$): Substituting the proportionalities of $\lambda$ and $\bar{v}$ into $D$: $D \propto \left(\frac{T}{P}\right) \cdot \sqrt{T}$ $D \propto \frac{T^{3/2}}{P}$

2. Calculating the Factor $x$: Let the initial temperature be $T_1$ and initial pressure be $P_1$. The new temperature is $T_2 = 4T_1$, and the new pressure is $P_2 = 2P_1$.

The ratio of the new diffusion coefficient $D_2$ to the initial diffusion coefficient $D_1$ is: $x = \frac{D_2}{D_1} = \frac{\left(\frac{T_2^{3/2}}{P_2}\right)}{\left(\frac{T_1^{3/2}}{P_1}\right)} = \left(\frac{T_2}{T_1}\right)^{3/2} \times \left(\frac{P_1}{P_2}\right)$

Substitute the given ratios: $x = (4)^{3/2} \times \left(\frac{1}{2}\right)$ $x = (2^2)^{3/2} \times \frac{1}{2} = 2^3 \times \frac{1}{2} = \frac{8}{2} = 4$

Thus, the value of $x$ is 4.