States of Matter - Result 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 $(D C) \propto \lambda U_{\text {mean }}$
But, $\quad \lambda=\dfrac{R T}{\sqrt{2} N_0 \sigma p} \Rightarrow \lambda \propto \dfrac{T}{p}$
and$$ \begin{aligned} & U_{\text {mean }}=\sqrt{\dfrac{8 R T}{\pi M}} \\ & U_{\text {mean }} \propto \sqrt{T} \\ & \therefore \quad \mathrm{DC} \propto \dfrac{(T)^{3 / 2}}{p} \\ & \end{aligned} $$
$\begin{aligned} \dfrac{(\mathrm{DC})_2}{(\mathrm{DC})_1}(x) & =\left(\dfrac{p_1}{p_2}\right)\left(\dfrac{T_2}{T_1}\right)^{3 / 2}=\left(\dfrac{p_1}{2 p_1}\right)\left(\dfrac{4 T_1}{T_1}\right)^{3 / 2} \\ & =\left(\dfrac{1}{2}\right)(8)=4\end{aligned}$
BETA
