Question

During an experiment an ideal gas is found to obey an additional law $V{P}^2=constant$. The initial volume and temperature of gas are V and T respectively. The temperature of gas when it expands to a volume 2V is :

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

Answer 1

The correct option is B $\sqrt{2}T$

As we know,
For gas $PV=nRT$

Also, $P=\sqrt{\frac{K}{V}}(\because V^{2}P=K)$

$\therefore \sqrt{\frac{K}{V}}\cdot V=nRT$

or $\sqrt{V}=\frac{nR}{\sqrt{K}}\times T$

$\therefore \sqrt{V}\propto T$

$\sqrt{\frac{V}{2V}}=\frac{T}{T_1}$

$\therefore T_1=\sqrt{2}\times T$