Question

During an experiment, an ideal gas is found to obey a law $V{P}^2=$ constant. The gas is initially at temperature $T$ and volume $V$. When it expands to volume $2V$, the resulting temperature will be

• A. $\sqrt{5}T$
• B. $\sqrt{3}T$
• C. $\sqrt{2}T$
• D. $\sqrt{6}T$

Answer 1

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

Given that,
Initial temperature, $T_i=T$
Initial volume, $V_i=V$
Final volume, $V_f=2V$
Final temperature, $T_f=?$

$V{P}^2=$ constant ..............$\left(1\right)$
From the ideal gas equation,
$PV=nRT$
$\Rightarrow P=\frac{nRT}{V}$ ..............$\left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$
$V{\left(\frac{nRT}{V}\right)}^2=$ constant
$\Rightarrow \frac{n^2{R}^2{T}^{2}V}{V^2}=$ constant
$\Rightarrow \frac{T^2}{V}=$ constant
$[n$ and $R$ are constant $]$
So,
$\frac{T_f^2}{V_f}=\frac{T_i^2}{V_i}$
$\Rightarrow \frac{T_f^2}{2V}=\frac{T^2}{V}$
$\Rightarrow T_f=\sqrt{2}T$