Consider one mole of perfect gas in a cylinder of unit cross-section with a piston attached as shown in the figure. $A$ spring (spring constant $k$) is attached (unstretched length $L$) to the piston and to the bottom of the cylinder. Initially,the spring is unstretched and the gas is in equilibrium. $A$ certain amount of heat $Q$ is supplied to the gas,causing an increase in volume from $V_0$ to $V_1$.
$(a)$ What is the initial pressure of the system?
$(b)$ What is the final pressure of the system?
$(c)$ Using the first law of thermodynamics,write down a relation between $Q, V_0, V_1, P_a$ and $k$.

(N/A) Initially,the system is in equilibrium,hence the pressure on the piston is the atmospheric pressure.
$\therefore P_{i} = P_{a}$
$(b)$ On the supply of heat,the volume of the gas increases from $V_{0}$ to $V_{1}$.
Since the cross-sectional area $A = 1$ unit,the displacement of the piston is $x = V_{1} - V_{0}$.
The force exerted by the spring on the piston is $F = kx = k(V_{1} - V_{0})$.
The final pressure $P_{f}$ on the gas is the sum of atmospheric pressure and the pressure due to the spring force:
$P_{f} = P_{a} + \frac{F}{A} = P_{a} + k(V_{1} - V_{0})$
$(c)$ According to the first law of thermodynamics,$Q = \Delta U + \Delta W$.
The change in internal energy for one mole of an ideal gas is $\Delta U = C_{V}(T_{f} - T_{0})$.
Using the ideal gas law $PV = RT$,we have $T = \frac{PV}{R}$,so $\Delta U = C_{V} \left( \frac{P_{f}V_{1}}{R} - \frac{P_{a}V_{0}}{R} \right)$.
The work done by the gas is $\Delta W = P_{a}(V_{1} - V_{0}) + \frac{1}{2}k(V_{1} - V_{0})^{2}$.
Thus,the relation is: $Q = C_{V} \left( \frac{(P_{a} + k(V_{1} - V_{0}))V_{1} - P_{a}V_{0}}{R} \right) + P_{a}(V_{1} - V_{0}) + \frac{1}{2}k(V_{1} - V_{0})^{2}$.