An ideal gas in a closed container is slowly heated.

(1) As we know mean free path

$\lambda=\frac{1}{\sqrt{2}\left(\frac{N}{V}\right) \pi d^{2}}$

Here, $\quad N=$ no. of molecule

$V=$ volume of container

$d=$ diameter of molecule

But $P V=n R T=n N K T$

$\Rightarrow \frac{N}{V}=\frac{P}{K T}=n$

$\lambda=\frac{1}{\sqrt{2}} \frac{K T}{\pi d^{2} P}$

For constant volume and hence constant number density

$n$ of gas molecules $\frac{P}{T}$ is constant.

So mean free path remains same.

As temperature increases no. of collision increases so relaxation time decreases.