Question

An ideal gas is found to obey an additional law $V{P}^2$ = constant. The gas is initially at temperature T and volume V. When it expands to a volume 2V, the temperature becomes :

• A. $\frac{T}{\sqrt{2}}$
• B. $2T$
• C. $\frac{2T}{\sqrt{2}}$
• D. $4T$

Answer 1

The correct option is A $\frac{T}{\sqrt{2}}$

We are given an ideal gas which is following an additional law,

$P{V}^2=$ a constant ....................(1)

As the gas is ideal, the gas law is

$PV=nRT$

Taking the value of $P$ from the above equation, we get

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

Putting this value in equation (1), we get

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

As $n$ and $R$ are constant, above equation becomes,

$TV=$ a constant .....................(2)

Now, the gas is expanding from initial volume $V_1$ to final volume $V_2$ by $2$ units, it becomes

$V_2=2V_1$

From equation (2), we get

$T_1{V}_1=T_2{V}_2$

$T_1{V}_1=T_2\left(2V_1\right)$

$T_2=\frac{T_1}{2}$

So the correct answer is $\frac{T}{2}$