Question

During an experiment, an ideal gas is found to obey an additional law $V{p}^2$ = constant. The gas is initially at temperature $T$ and volume $V$. What will be the temperature of the gas when it expand to a volume $2V$?

• A. $\sqrt{2}T$
• B. $2T$
• C. $\frac{T}{\sqrt{2}}$
• D. $\frac{T}{2}$

Answer 1

Answer: (A)

$\sqrt{2}T$

Here it is given that $V{p}^2=$constant k (say). Hence, we may write the gas equation as,
$pV=nRT$
$\Rightarrow \sqrt{\frac{k}{V}}V=nRT$
$\Rightarrow \sqrt{V}=\frac{nR}{\sqrt{k}}T$
So, $\sqrt{\frac{V_1}{V_2}}=\frac{T_1}{T_2}$
$\therefore T_2=T_1\sqrt{\frac{V_2}{V_1}}$
$=T\sqrt{\frac{2V}{V}}=T\sqrt{2}$