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 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

Answer: (B)

$P_2=P_1,T_2=\frac{T_1}{2}$

Total KE for N molecules at temp T is $KE=\frac{1}{2}m{v}^2\ast N=N\ast \frac{f}{2}{k}_{B}T$ from kinetic theory of gases

where,
$m=$ the mass of each molecule
$f=$ no of degrees of freedom
$k_B=$ Boltzman constant
$T=$ Temp of the gas
If KE remains same but N is doubled, to keep the KE same, temp will reduce to half.

Using the ideal gas equation,
$PV=nRT$

When no of molecules is doubled, no of moles is also doubled. Hence, the product $nRT$ remains same.

Since the volume of the box is constant, Pressure does not change.