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. The value of $x$ is.

Answer 1

Answer: 4

Diffusion constant $=\lambda v$
$=\frac{T}{P}\times \sqrt{\frac{8RT}{\pi M}}$
$\lambda \propto \frac{T^{3/2}}{P}$
$\propto \frac{(4{)}^{3/2}}{2}$
$\propto 4$
$\therefore$ It will become four times