Question

A closed box contains an ideal gas. If temperature of gas increased then
$\left(A\right).$ Mean free path remain same.
$\left(B\right).$ Mean free path decreases.
$\left(C\right).$ Relaxation time decreases.
$\left(D\right).$ Relaxation time remain same.

Which of the following is correct?

• A. $B\&D$
• B. $A\&C$
• C. $A\&D$
• D. $B\&C$

Answer 1

The correct option is B $A\&C$

We know that, mean free path is defined as

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

$N\to$ Number of molecules
$V\to$ Volume of container
$d\to$ diameter of molecule

None of these terms depend on temperature. So, mean free path remains same.

Also, as temperature increases, number of collision increases, So, relaxation time decreases.

Hence, option $\left(b\right)$ is correct.