Heat and Thermodynamics 4 Question 40
42. The volume $V$ versus temperature $T$ graphs for a certain amount of a perfect gas at two pressure $p _1$ and $p _2$ are as shown in figure. It follows from the graphs that $p _1$ is greater than $p _2 .(1982,2 M)$
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Solution:
$$ p V=n R T $$
$$ \therefore \quad V=\left(\frac{n R}{p}\right) T $$
Comparing this with $y=m x, V-T$ graph is a straight line passing through origin with slope $=\frac{n R}{p}$.
$$ \begin{aligned} & \text { or } \quad \text { slope } \propto \frac{1}{p}(\text { slope }) _1>(\text { slope }) _2 \\ & \therefore \quad p _1<p _2 \\ & \text { 43. } v _{\text {rms }}=\sqrt{\frac{3 R T}{M}} \end{aligned} $$
It not only depends on $T$ but also on $M$.
