Solution
The problem describes two distinct phases of compression after a mass is placed on the piston.
Phase 1: Quick Settlement ($\Delta h_1$)
The piston settles “quickly”. This implies there is insufficient time for heat exchange with the surroundings. We treat this as an Adiabatic Process ($PV^\gamma = \text{constant}$).
Differentiating $PV^\gamma = C$:
$$ \frac{dP}{P} + \gamma \frac{dV}{V} = 0 \implies \frac{\Delta P}{P} = \gamma \frac{\Delta h_1}{h} $$
Here $\Delta P$ is the pressure increase due to the weight ($mg/A$). So, $\Delta h_1 = \frac{h}{\gamma} \frac{\Delta P}{P}$.
Phase 2: Long Time Settlement ($\Delta h_2$)
After a “long time”, the gas returns to thermal equilibrium with the surroundings. The total process from start to finish is Isothermal ($T = \text{constant}$, $PV = \text{constant}$).
For the total displacement $\Delta h_{total} = \Delta h_1 + \Delta h_2$:
$$ \frac{dP}{P} + \frac{dV}{V} = 0 \implies \frac{\Delta P}{P} = \frac{\Delta h_{total}}{h} $$
So, $\Delta h_{total} = h \frac{\Delta P}{P}$.
Relationship between Displacements:
From the adiabatic equation: $h \frac{\Delta P}{P} = \gamma \Delta h_1$.
Substitute $h \frac{\Delta P}{P}$ with $\Delta h_{total}$:
$$ \Delta h_{total} = \gamma \Delta h_1 $$
Since $\Delta h_{total} = \Delta h_1 + \Delta h_2$, we have:
$$ \Delta h_1 + \Delta h_2 = \gamma \Delta h_1 \implies \Delta h_2 = (\gamma – 1) \Delta h_1 $$
$$ \frac{\Delta h_1}{\Delta h_2} = \frac{1}{\gamma – 1} $$
Multiply both sides by $R$:
$$ \frac{\Delta h_1}{\Delta h_2} R = \frac{R}{\gamma – 1} = C_V $$
Conclusion:
We are given $\frac{\Delta h_1}{\Delta h_2} R = \frac{5}{2} R$.
Therefore, $C_V = \frac{5}{2} R$. This value of molar specific heat corresponds to a Diatomic Gas (rigid rotator).