Question

A box contains $N$ molecules of a perfect 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 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 $P_2=P_1$, $T_2=\frac{T_1}{2}$

$P_{1}V=Nk{T}_1$ ....(1)
$P_{2}V=2Nk{T}_2$ .....(2)
Dividing (2) by (1)
$\frac{P_2}{P_1}=\frac{2T_2}{T_1}$ ....(3)
Since KE remains same, $N\times \frac{3}{2}k{T}_1=2N\times \frac{3}{2}k{T}_2$
$\Rightarrow T_2=\frac{T_1}{2}$ ....(4)
Substituting (4) in (3)
$\frac{P_2}{P_1}=1$
$\Rightarrow P_2=P_1$