JEE MAIN - Physics (2020 - 2nd September Evening Slot - No. 15)
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) 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.
(C) and (D)
(A) and (D)
(B) and (C)
(A) and (B)
Explanation
The mean free path of molecules of an ideal gas is given as:
$$\lambda $$ = $${V \over {\sqrt 2 \pi {d^2}N}}$$
where : V = Volume of container
N = No of molecules
Mean free path is independent of temperature hence with increasing temp since volume of container does not change (closed container), so mean free path is unchanged.
Average collision time =
and Vav $$ \propto $$ $$\sqrt T $$
$$ \therefore $$ Average collision time $$ \propto $$ $${1 \over {\sqrt T }}$$
Hence with increase in temperature the average collision time decreases.
$$\lambda $$ = $${V \over {\sqrt 2 \pi {d^2}N}}$$
where : V = Volume of container
N = No of molecules
Mean free path is independent of temperature hence with increasing temp since volume of container does not change (closed container), so mean free path is unchanged.
Average collision time =
$$\lambda $$
Vav
and Vav $$ \propto $$ $$\sqrt T $$
$$ \therefore $$ Average collision time $$ \propto $$ $${1 \over {\sqrt T }}$$
Hence with increase in temperature the average collision time decreases.
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