Question

The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The absolute temperature of an ideal gas is increased 4 times and its pressure is increased 2 times. As a result, the diffusion coefficient of this gas increases $x$ times. Given that mean free path is proportional to T/P, the value of $x$ is
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Answer 1

Diffusion coefficient $D$ $\propto \lambda \mu$,
Where, $\lambda$ is the mean free path and $\mu$ is the mean speed.
Since $\lambda \propto \frac{T}{P}$ and $\mu \propto \sqrt{T}$
Diffusion coefficient $\propto \frac{T\sqrt{T}}{P}$
Let $D_i$ = initial value and $D_f$ = final value
Thus, $\frac{D_i}{D_f}=\frac{\frac{T\sqrt{T}}{P}}{\frac{4T\sqrt{4T}}{2P}}=\frac{1}{\frac{(4\times 2)}{2}}=\frac{1}{4}$ or $\frac{D_f}{D_i}=\frac{4}{1}$
$D_f=4D_i$