- Home
- Standard 11
- Physics
During an experiment an ideal gas is found to obey an additional law $VP^2 =$ constant. The gas is initially at temperature $T$ and volume $V$. What will be the temperature of the gas when it expands to a volume $2V$?
$T' = \sqrt 4 \,T$
$T' = \sqrt 2 \,T$
$T' = \sqrt 5 \,T$
$T' = \sqrt 6 \,T$
Solution
According to the given problems
$\mathrm{VP}^{2}=$ constant
From the gas law
$\mathrm{PV}=\mathrm{nRT}$
$\Rightarrow \quad\left(\frac{\mathrm{k}}{\sqrt{\mathrm{V}}}\right) \mathrm{V}=\mathrm{nRT}$
$\Rightarrow \quad \sqrt{\mathrm{V}}=\left(\frac{\mathrm{nR}}{\mathrm{K}}\right) \mathrm{T}$
$\therefore \quad \sqrt{\frac{\mathrm{V}_{1}}{\mathrm{V}_{2}}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{T}_{2}}\right),$ i.e, $\sqrt{\frac{\mathrm{V}}{2 \mathrm{V}}}=\frac{\mathrm{T}}{\mathrm{T}^{\prime}}$
$\Rightarrow \quad \mathrm{T}^{\prime}=\sqrt{2} \mathrm{T}$
