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

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