Discussion Forum

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 time. The value of x is .........

Answer:

Explanation:

 (DC) Diffusion coefficient  $\propto$  $\lambda$ (means free path )  $\propto$  $U_{mean}$ , thus

  (DC) $\propto$  $\lambda$ $U_{mean}$

  but, $\lambda = \frac{RT}{\sqrt{2}N_{0}\sigma p}\Rightarrow \lambda \propto\frac{T}{p}$,

  and    $U_{mean}=\sqrt{\frac{8RT}{\pi M}}\Rightarrow U_{mean}\propto\sqrt{T}$

$\therefore$     $DC\propto \frac{(T)^{\frac{3}{2}}}{p}$

 $\frac{(DC)_{2}}{(DC)_{1}}(x)=(\frac{p_{1}}{p_{2}}) (\frac{T_{2}}{T_{1}})^{\frac{3}{2}}$

$= (\frac{p_{1}}{2p_{1}})(\frac{4T_{1}}{T_{1}})^{\frac{3}{2}}=(\frac{1}{2})(8)=4$