Heat and Thermodynamics 4 Question 36
38. During an experiment, an ideal gas is found to obey an additional law $p^{2} V=$ constant. The gas is initially at a temperature $T$ and volume $V$. When it expands to a volume $2 V$, the temperature becomes
$(1987,2 M)$
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Solution:
$$ V p^{2}=\text { constant } $$
Putting, $p=\frac{n R T}{V}$, we have $\frac{T^{2}}{V}=$ constant
$$ \text { or } \quad T \propto \sqrt{V} $$
So, if $V$ is doubled, $T$ becomes $\sqrt{2}$ times.
