Question

An ideal gas in a closed container is slowly heated. As its temperature increases, which of the following statements are true ?
$\left(a\right)$ The mean free path of the molecules decreases
$\left(b\right)$ The mean collision time between the molecules decreases
$\left(c\right)$ The mean free path remains unchanged
$\left(d\right)$ The mean collision time remains unchanged

• A. $\left(b\right)$ and $\left(c\right)$
• B. $\left(a\right)$ and $\left(b\right)$
• C. $\left(c\right)$ and $\left(d\right)$
• D. $\left(a\right)$ and $\left(d\right)$

Answer 1

As we know mean free path
$\lambda =\frac{1}{\sqrt{2}\left(\frac{N}{V}\right)\pi d^2}$--------(1)
Here,
$N=$ no. of molecule
$V=$ volume of container
$d=$ diameter of molecule

We know that,
But $PV=nRT=nNKT$

Or,
$\Rightarrow \frac{N}{V}=\frac{P}{KT}=n$

Putting this value in equation 1

We get,
$\lambda =\frac{1}{\sqrt{2}}\frac{KT}{\pi d^{2}P}$

Now,
For constant volume number of moles (n) of gas molecules $\frac{P}{T}$ is constant. So mean free path remains the same.
As temperature increases no. of collisions increases, so relaxation time decreases.

Hence the correct option is (C)