Question

during an experiment an ideal gas obeys an addition equation of state $p^{2}V$ constant . The initial temperature and pressure of gas are $T$ and $V$ respectively. When it expends to volume $2V.$ then its temperature will be:

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

Answer 1

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

Given: $P^{2}V=constant$, initial temperature and volume are $T$ and $V$ respectively.

Solution:

Let initial pressure be $P_1$ and final pressure be $P_2$

Final volume is $2V$

Initial temperature be $T$ and final temperature be $T_2$

From ideal gas equation, $PV=RT$ let n=1

$P=\frac{RT}{V}$ and $P^{2}V=constant$

$\therefore V\left({\frac{RT}{V}}^2\right)constant$

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

$\Rightarrow \frac{T_2^2}{2V^2}=\frac{T^2}{V^2}$

$\therefore T_2=\sqrt{2}T$

Hence the correct option is B.