If $V_2=2V_1\;\text{and}\;T_2=3T_1$ ,then the energy stored in the spring is $\frac{1}{4}P_1V_1$
If $V_2=2V_1\;\text{and}\;T_2=3T_1$ ,then the charge in internal energy is $3P_1V_1$
If $V_2=3V_1\;\text{and}\;T_2=4T_1$ ,then the work done by the gas is $\frac{7}{3}P_1V_1$
If $V_2=3V_1\;\text{and}\;T_2=4T_1$ ,then the heat supplied to the gas is is $\frac{17}{6}P_1V_1$

Solution:

B or A,B,C

P(pressure of gas)=

P_1+\frac{kx}{A}

=\int P dV = P_1(V_2-V_1)+\frac{kx^2}{2}

=P_1(V_2-V_1)+\frac{(P_2-P_1)(V_2-V_1)}{2}

\Delta U = nC_v \Delta T = \frac{3}{2}(P_2V_2-P_1V_1)

Q=W+ ΔU

Case 1: ΔU

=3P_1V_1,\;W=\frac{5P_1V_1}{4},

\;Q=\frac{17P_1V_1}{4},

U_{\text{spring}}=\frac{P_1V_1}{4}

Case 2: ΔU

=\frac{9P_1V_1}{2},\;W=\frac{7P_1V_1}{3},

\;Q=\frac{41P_1V_1}{6},\;U_{\text{spring}}=\frac{P_1V_1}{3}

Note: A and c will be true after assuming pressure to the right of piston has constant value of

P_1

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