Compressing A Single-Particle Gas

Let the particle hit a wall $n$ times per second, then, the force applied is clearly $2mv n$ .

Also time taken to come back and strike again = $T = \frac{2l}{v}$

Hence, $n = \frac{1}{T} = \frac{v}{2L}$

Hence, force applied on a wall is given by : $\frac{mv^2}{L}$ .

Now, in order to move the piston slowly, $F_{ext}$ is also $\frac{mv^2}{L}$ . Clearly, the work done by this force would add into its kinetic energy. Note that this force is acting at two ends. Say left piston moves $dl_{1}$ and right piston moves $dl_{2}$ , then

$- \frac{mv^2}{L} (dl_{1} + dl_{2}) = - \frac{mv^2}{L} dL = d(\frac{mv^2}{2})$ , where $d(\frac{mv^2}{2}) = mv dv$

Hence, $\displaystyle \boxed{-\frac{dL}{L} = \frac{dv}{v}}$

Or $\displaystyle -\int_{L_{0}}^{\frac{L_{0}}{3}} \frac{dL}{L} = \int_{v_{0}}^{v'} \frac{dv}{v}$ , $\Rightarrow v' = 3v_{0} = 900m/s$