Solution

Assuming the gas is pumped out (or the state changes) slowly, the system is in mechanical equilibrium at any given moment. The pressure force on the piston of area $S$ is balanced by the spring force:
$$PS = kx$$

Since the spring is undeformed when the piston is at the right wall, the distance $x$ relates to the gas volume as $V = xS$. Therefore, $x = \frac{V}{S}$, and the pressure is:
$$P = \frac{kV}{S^2}$$

Using the ideal gas law $PV = \nu RT$, we substitute $P$:
$$\left(\frac{kV}{S^2}\right)V = \nu RT \implies V^2 = \frac{\nu R S^2}{k} T$$

For a small change in state, the new parameters are $V + dV$ and $T + dT$:
$$(V + dV)^2 = \frac{\nu R S^2}{k} (T + dT)$$

Expanding the square and neglecting the second-order infinitesimal $(dV)^2$:
$$V^2 + 2VdV = \frac{\nu R S^2}{k} T + \frac{\nu R S^2}{k} dT$$

Since $V^2 = \frac{\nu R S^2}{k} T$, this simplifies to:
$$2VdV = \frac{\nu R S^2}{k} dT \implies dV = \frac{\nu R S^2}{2kV} dT$$

The work done by the gas $dA$ is:
$$dA = P dV = \left(\frac{kV}{S^2}\right) \left(\frac{\nu R S^2}{2kV} dT\right) = \frac{1}{2} \nu R dT$$

The change in internal energy for a monatomic gas is:
$$dU = \frac{3}{2} \nu R dT$$

According to the First Law of Thermodynamics, the heat transferred $dQ$ is:
$$dQ = dU + dA = \frac{3}{2} \nu R dT + \frac{1}{2} \nu R dT = 2 \nu R dT$$

The heat capacity of the system $C$ is:
$$C = \frac{dQ}{dT} = 2 \nu R$$

From the initial state equation $P_0 V_0 = \nu R T_0$, we know that $\nu R = \frac{P_0 V_0}{T_0}$. Substituting this into our result:
$$C = \frac{2 P_0 V_0}{T_0}$$