# An ideal gas in a closed container is slowly heated.

Question:

An ideal gas in a closed container is slowly heated. As its temperature increases, which of the following statements are true?

(a) The mean free path of the molecules decreases

(b) The mean collision time between the molecules decreases

(c) The mean free path remains unchanged

(d) The mean collision time remains unchanged

1. (1) (b) and (c)

2. (2) (a) and (b)

3. (3) (c) and (d)

4. (4) (a) and (d)

Correct Option: 1

Solution:

(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.