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.
Solution
The mean free path of molecules of an ideal gas is given as:
<br><br>$\lambda$ = ${V \over {\sqrt 2 \pi {d^2}N}}$
<br><br>where : V = Volume of container
<br>N = No of molecules
<br><br>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.
<br><br>Average collision time = <span style="display: inline-block;vertical-align: middle;">
<div style="text-align: center;border-bottom: 1px solid black;">$\lambda$</div>
<div style="text-align: center;">V<sub>av</sub></div>
</span>
<br><br>and V<sub>av</sub> $\propto$ $\sqrt T$
<br><br>$\therefore$ Average collision time $\propto$ ${1 \over {\sqrt T }}$
<br><br>Hence with increase in temperature the average collision time decreases.
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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