Question

During an experiment, an ideal gas obeys an addition equation of state $P^{2}V=$ constant. The initial temperature and volume of the gas are $T$ and $V$ respectively. When it expands to volume $2V$, then its temperature will be :

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

Answer 1

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

Ideal gas: $V{P}^2=$ constant

Again $PV=nRT$ from equation of state,

Hence, $VP\times P=$ constant i.e. $nRT\times P=$ constant

Again, $P=\frac{nRT}{V}$

$\therefore$ $\frac{(nRT{)}^2}{V}=$ constant

$\frac{T^2}{V}=$ constant

Thus volume $V$ when expanded to $2V$, temperature $T_1$

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

Hence, the correct option is $\text{D}$