$A$ cylinder with a piston of unit cross-sectional area contains one mole of an ideal gas as shown in the figure. $A$ spring of unstretched length $L$ and spring constant $K$ is connected between the piston and the bottom of the cylinder. Initially, the spring is unstretched and the gas is in equilibrium. When an amount of heat $Q$ is supplied to the gas, its volume changes from $V_0$ to $V_1$. Find:
$(a)$ The initial pressure of the system.
$(b)$ The final pressure of the system.
$(c)$ Using the first law of thermodynamics, find the relation between $Q, V_0, V_1, P_a$ and $K$.

(N/A) Initially, the spring is unstretched, so the force exerted by the spring is zero. The piston is in equilibrium under the atmospheric pressure $P_a$ and the gas pressure $P_i$. Thus, $P_i = P_a$.
$(b)$ When the volume increases from $V_0$ to $V_1$, the piston moves by a distance $x = V_1 - V_0$ (since the area is unit). The spring is now stretched by $x$. The final pressure $P_f$ is balanced by atmospheric pressure, spring force, and gas pressure: $P_f = P_a + \frac{Kx}{A} = P_a + K(V_1 - V_0)$.
$(c)$ According to the first law of thermodynamics, $Q = \Delta U + W$.
For an ideal gas, $\Delta U = C_v \Delta T = \frac{C_v}{R} (P_f V_1 - P_i V_0) = \frac{f}{2} (P_f V_1 - P_i V_0)$.
The work done by the gas is $W = \int_{V_0}^{V_1} P \, dV$. Since the pressure varies linearly with volume, $W = \frac{P_i + P_f}{2} (V_1 - V_0)$.
Substituting the values: $Q = \frac{f}{2} [ (P_a + K(V_1 - V_0)) V_1 - P_a V_0 ] + \frac{P_a + P_a + K(V_1 - V_0)}{2} (V_1 - V_0)$.