A box containing N molecules of a perfect gas at temperature T1 and pressure P1. 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 P2 and temperature T2 , then

The correct option is P₂ = P₁, T₂ = T₁/2.

Let us analyze the problem step-by-step using the principles of kinetic theory of gases and the ideal gas equation.

Step 1: Relate the temperature and the total kinetic energy of the gas
The total translation kinetic energy ($E$) of $N$ molecules of a perfect (ideal) gas at absolute temperature $T$ is given by the formula:

$E=\frac{3}{2}NkT$

where $k$ is the Boltzmann constant.

Let the initial state parameters be:

• Number of molecules = $N$
• Temperature = $T_1$
• Initial kinetic energy:

$E_1=\frac{3}{2}Nk{T}_1$

In the final state, the number of molecules is doubled, so $N_2=2N$, and the final temperature is $T_2$. The final kinetic energy is:

$E_2=\frac{3}{2}\left(2N\right)k{T}_2$

Since the total kinetic energy remains the same ($E_2=E_1$), we can equate the two expressions:

$\frac{3}{2}\left(2N\right)k{T}_2=\frac{3}{2}Nk{T}_1$

Simplifying the equation by canceling common terms on both sides:

$2T_2=T_1$

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

Step 2: Relate the initial and final pressures
Using the ideal gas equation in terms of the number of molecules, $PV=NkT$, the pressure $P$ inside a box of constant volume $V$ is:

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

The initial pressure $P_1$ is:

$P_1=\frac{Nk{T}_1}{V}$

The final pressure $P_2$ with $2N$ molecules at temperature $T_2$ is:

$P_2=\frac{\left(2N\right)k{T}_2}{V}$

Substitute $T_2=\frac{T_1}{2}$ into the equation for $P_2$:

$P_2=\frac{2Nk\left(\frac{T_1}{2}\right)}{V}$

$P_2=\frac{Nk{T}_1}{V}$

Comparing this with the expression for $P_1$, we find:

$P_2=P_1$

Therefore, the new pressure remains unchanged while the temperature is halved.