1. Relation between pressure and volume during slow heating

Since the piston is frictionless and the spring is initially relaxed, at the initial state the spring force is zero, so $P_1 = P_{\text{ext}}.$

As the gas is heated slowly, the process is quasi-static. If the piston moves out by distance $x$, then spring force is proportional to $x$, hence extra pressure due to spring is proportional to $x$.

Since cylinder has constant cross-sectional area $A$, $V - V_1 = Ax \quad \Rightarrow \quad x \propto (V-V_1).$

Therefore pressure varies linearly with volume: $P = P_1 + \alpha (V-V_1).$

So, between initial and final states, the $P$-$V$ curve is a straight line, and hence the work done is $W = \int_{V_1}^{V_2} P\,dV = \frac{P_1+P_2}{2}(V_2-V_1).$

Also, for the ideal gas, $P_1V_1=nRT_1, \qquad P_2V_2=nRT_2.$ Thus, $\frac{P_2V_2}{P_1V_1}=\frac{T_2}{T_1}.$

2. Case 1: $V_2=2V_1$, $T_2=3T_1$

Using ideal gas law: $\frac{P_2(2V_1)}{P_1V_1}=3 \Rightarrow P_2=\frac{3}{2}P_1.$

(A) Energy stored in spring

The extra pressure due to spring at final state is $P_2-P_1=\frac{1}{2}P_1.$

Since pressure increases linearly from $P_1$ to $P_2$, the spring energy equals the triangular area under excess-pressure vs volume graph: $U_s=\frac{1}{2}(P_2-P_1)(V_2-V_1).$

Now, $V_2-V_1=V_1,$ so $U_s=\frac{1}{2}\left(\frac{1}{2}P_1\right)(V_1)=\frac{1}{4}P_1V_1.$

So A is correct.

(B) Change in internal energy

For monoatomic ideal gas, $\Delta U = \frac{3}{2}nR(T_2-T_1).$ Using $nRT_1=P_1V_1$ and $T_2=3T_1$, $\Delta U = \frac{3}{2}nR(2T_1)=3nRT_1=3P_1V_1.$

So B is correct.

3. Case 2: $V_2=3V_1$, $T_2=4T_1$

Using ideal gas law: $\frac{P_2(3V_1)}{P_1V_1}=4 \Rightarrow P_2=\frac{4}{3}P_1.$

(C) Work done by gas

Since process is linear in $P$-$V$, $W=\frac{P_1+P_2}{2}(V_2-V_1).$ Substitute values: