Question

A box contains $N$ molecules of a perfect ideal gas at temperature $T_1$ and pressure $P_1$. The number of molecules in the box is doubled keeping the total kinetic energy of the gas same as before. If the new pressure is $P_2$ and new temperature is $T_2$, then

• A. $P_2=P_1,T_2=T_1$
• B. $P_2=P_1,T_2=\frac{T_1}{2}$
• C. $P_2=2P_1,T_2=T_1$
• D. $P_2=2P_1,T_2=\frac{T_1}{2}$

Answer 1

The correct option is B $P_2=P_1,T_2=\frac{T_1}{2}$

Given:
For initial case,
Number of molecules of gas, $N_1=N$
Temperature $=T_1$
Pressure $=P_1$
For final case,
Number of molecules of gas, $N_2=2N$
Temperature, $T_2=?$
Pressure, $P_2=?$
Volume $\left(V\right)$ of the container will be same in both cases.

We know that, average translational kinetic energy is given by,
$KE=\frac{PVf}{2}$ ........(1) and
$KE=\frac{Nf{K}_{B}T}{2}$ ........(2)
where,
$N\to$ number of molecules
$f\to$ no of degrees of freedom [constant]
$K_B\to$ Boltzmann constant [constant]
$T\to$ temperature

As volume of container and average translational kinetic energy has not been changed, so from (1), pressure will not change.
$\Rightarrow P_2=P_1$
Also from (2), $T\propto \frac{1}{N}$
Therefore, temperature will be halved if the number of molecules is doubled.
$\Rightarrow T_2=\frac{T_1}{2}$