$5.6.31.$ In the vacuum space, a cylindrical vessel is placed vertically, closed from above by a movable piston of mass $M$. Inside the vessel is a monoatomic gas at a pressure of $P$. The inner section of the cylinder is $S$, and the piston is located at a height $H$ above its bottom. The plunger was released. After a brief hesitation, it stops. At what distance from the initial position will the piston stop if the heat capacity of the gas at a constant volume is much greater than the heat capacity of the piston and cylinder? The entire system is heat-insulated.
Applying Energy Conservation Law:
$$MgH + nC_v T = MgH' + nC_vT' \quad(1)$$
where $T'$ is final temperature at equilibrium, moreover, for final position, $Mg = P'S$, so
$$P' = \frac{Mg}{S} \quad(2)$$
From State equation, before: $PV = nRT$, since $V=SH$ and separating $T$,
$$T = \frac{PSH}{nR} \quad(3)$$
After: $P'V' = nRT'$, since $V'=SH'$ and sparating $T'$,
$$T' = \frac{PSH'}{nR} \quad(4)$$
Substituting (3) and (4) into (1):
$$MgH + \frac{C_vPSH}{R} = MgH' + \frac{C_vP'SH'}{R}$$
According to (2), the Mayer relation ($C_p = R+C_v$) and separating $H'$,
$$H' = H \frac{C_vPS+MgR}{MgC_p}$$
As the gas is monoatomic, $C_v = 3/2R$ and $C_p = 5/2R$,then
$$H' = \frac{H}{5}\left(2+\frac{3PS}{Mg}\right) \quad(5)$$
The position variation is given by
$$\Delta H = H-H' \quad(6)$$
Putting (5) into (6):
$$\boxed{\Delta H = \frac{3H}{5}\left(1-\frac{PS}{Mg}\right)}$$
$$\Delta H = \frac{3H}{5}\left(1-\frac{PS}{Mg}\right)$$