The cylindrical box moves through an ideal gas with a constant velocity $\vec{u}$. The drag force arises due to the collisions of gas molecules with the front and back faces of the cylinder. We assume the velocity of the cylinder $u$ is small compared to the thermal velocity (RMS speed, $v_{rms}$) of the gas molecules.

According to the kinetic theory of gases, the pressure exerted by gas molecules on a wall is proportional to the square of the relative velocity with which the molecules hit the wall.

• Front Face: The wall moves towards the gas molecules. The effective relative velocity is $(v_{rms} + u)$. $$ P_{\text{front}} \propto (v_{rms} + u)^2 $$
• Back Face: The wall moves away from the gas molecules. The effective relative velocity is $(v_{rms} – u)$. $$ P_{\text{back}} \propto (v_{rms} – u)^2 $$

The net drag force $F$ is the difference in pressure forces acting on the area $A$: $$ F \propto A [ (v_{rms} + u)^2 – (v_{rms} – u)^2 ] $$ Using the expansion $(a+b)^2 – (a-b)^2 = 4ab$: $$ F \propto A [ 4 v_{rms} u ] $$ $$ F \propto \rho A v_{rms} u $$ Where $\rho$ is the gas density.

For an ideal gas, the root-mean-square speed $v_{rms}$ depends on the absolute temperature $T$ as: $$ v_{rms} = \sqrt{\frac{3RT}{M}} \implies v_{rms} \propto \sqrt{T} $$ Since the chamber is filled with gas (constant volume and mass), the density $\rho$ remains constant. Thus, the relationship for force becomes: $$ F \propto u \sqrt{T} $$

Let the initial state be $(u, T)$ and the final state be $(u’, T’)$. We are given that $T’ = 2T$ and the drag force remains unchanged ($F’ = F$).

Equating the proportionality terms: $$ u’ \sqrt{T’} = u \sqrt{T} $$ Substituting $T’ = 2T$: $$ u’ \sqrt{2T} = u \sqrt{T} $$ $$ u’ \sqrt{2} = u $$ $$ u’ = \frac{u}{\sqrt{2}} $$ $$ u’ \approx 0.707 u $$