Question

A vessel contains $N$ molecules of a gas at temperature $T$. Now the number of molecules is doubled, keeping the total energy in the vessel constant. The temperature of the gas is.

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

Answer 1

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

We know that,
$E=\frac{3}{2}nRT$
Now,
$E_1=\frac{3}{2}nR{T}_1$...(i)
$E_2=\frac{3}{2}2nR{T}_2$...(ii)
From equation (i) and (ii)
$2E_1=E_2$
Dividing equation (i) and (ii)
$\frac{E_1=\frac{3}{2}nR{T}_1}{E_2=\frac{3}{2}2nR{T}_2}\Rightarrow \frac{E_1}{E_2}=\frac{T_1}{2T_2}$
As the energy remains constant in the vessel, then
$2=\frac{T_1}{T_2}\Rightarrow T_2=\frac{T_1}{2}$
So, if the temperature taken is $T$ instead of $T_1$, then the final temperature becomes $\frac{T}{2}$.