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 and temperature are $P_2$ are $T_2$ respectively, 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 $P_2=P_1,T_2=\frac{T_1}{2}$

Initial and final conditions

$U_i=U_f$

$N$ | $2N$
$T_1$ | $T_2$
$P_1$ | $P_2$
$v_0$ | $v_0$

$U_i=U_f$ (Given KE remains constant)

$N\frac{f}{2}K{T}_1=2N\frac{f}{2}K{T}_2$

$⟹T_2=0.5T_1$

$Pv=nRT\frac{N_A}{N_A}=NkT$

$⟹PV$ is constant.

so $P_1{v}_1=P_2{v}_2$

$P_2=P_1$