Question

A sample of gas is kept in rigid container. Now the gas is heated such that temperature of gas is doubled, then mean free path for molecular collisions

• A. will be doubled
• B. will become half
• C. will not change
• D. will become $2^{3/2}$ times

Answer 1

The correct option is C will not change

Mean free path:
$\lambda \propto T/P$ for rigid container
As given,
$V=\text{const}$
$PV=nRT$
$V=\frac{nRT}{P}=C$
$\Rightarrow \frac{T}{P}=\text{const}$

Hence, mean free path will not change.