States of Matter 1 Question 63
63. 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. The value of $x$ is …
(2016 Adv.)
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Answer:
Correct Answer: 63. (4)
Solution:
- (DC) Diffusion coefficient $\propto \lambda$ (mean free path) $\propto U_{\text {mean }}$ Thus (DC) $\propto \lambda U_{\text {mean }}$
But, $\quad \lambda=\frac{R T}{\sqrt{2} N_{0} \sigma p} \Rightarrow \lambda \propto \frac{T}{p}$
and $\quad U_{\text {mean }}=\sqrt{\frac{8 R T}{\pi M}}$
$U_{\text {mean }} \propto \sqrt{T}$
$\therefore \quad \mathrm{DC} \propto \frac{(T)^{3 / 2}}{p}$
$$ \begin{aligned} \frac{(\mathrm{DC}){2}}{(\mathrm{DC}){1}}(x) & =\left(\frac{p_{1}}{p_{2}}\right)\left(\frac{T_{2}}{T_{1}}\right)^{3 / 2}=\left(\frac{p_{1}}{2 p_{1}}\right)\left(\frac{4 T_{1}}{T_{1}}\right)^{3 / 2} \ & =\left(\frac{1}{2}\right)(8)=4 \end{aligned} $$
