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.

• 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

The correct option is A

$\left(B\right)and\left(C\right)$

Explanation for the correct option(s):

A: $\mathbf{\left(}\mathit{B}\mathbf{\right)}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{\left(}\mathit{C}\mathbf{\right)}$

As Mean Free Path is given as

$\lambda =\frac{1}{\sqrt{2}\pi N{d}^2}$

where

$d$ = Diameter of the molecule,

$N$ = The number of molecules per unit volume.

So Mean Free Path is independent of temperature so it will remain unchanged.

And relaxation time decreases as temperature increases.

Explanation for the incorrect option(s):

B: $\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{\left(}\mathit{B}\mathbf{\right)}$

C: $\mathbf{\left(}\mathit{C}\mathbf{\right)}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{\left(}\mathit{D}\mathbf{\right)}$

D: $\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{\left(}\mathit{D}\mathbf{\right)}$

1. As we know that the mean remains unchanged due to the independent of temperature.
2. And mean collision time is inversely proportional to temperature. Therefore mean collision time decrease with an increase in temperature.
3. Thus, if temperature increase the mean free path remains unchanged, and the mean collision time between the molecules decreases.

Hence, correct option is A: $\left(B\right)and\left(C\right)$