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

Mean free path $\left(l\right)\propto \frac{T}{P}$

Mean speed $\left(C_{uv}\right)\propto \sqrt{T}$

Diffusion coefficient (D) is proportional to both mean free path (l) and mean speed.

$\therefore D\propto \frac{T^{\frac{3}{2}}}{P}$

At temperature $T_1$ and $P_1$

$D_1=\frac{K{T}^{\frac{3}{2}}}{P_1}\dots \dots \left(1\right)$

If $T_1$ is increased 4 times and $P_1$ is increased 2 times.

$D_2=\frac{K(4T_1{)}^{\frac{3}{2}}}{2P_1}\dots \dots \left(2\right)$

$\frac{D_2}{D_1}=\frac{(4{)}^{\frac{3}{2}}}{2}=\frac{2^3}{2}=4$