Thermodynamic Process of Ideal Monoatomic Gas with Spring Loaded Piston

To solve this problem, we analyze the thermodynamic process undergone by the ideal monoatomic gas enclosed in the cylinder with a spring-loaded piston.

1. General Governing Equations

Initially, the gas is in equilibrium at pressure $P_1$, volume $V_1$, and temperature $T_1$, with the spring in its relaxed state. Since the spring is uncompressed initially, the external atmospheric pressure acting on the piston is: $P_{\text{atm}} = P_1$

For a monoatomic ideal gas, the molar heat capacity at constant volume is: $C_v = \frac{3}{2}R$

The ideal gas equation at initial state gives: $P_1 V_1 = n R T_1$

When the piston advances outward by displacement $x$, the change in volume is: $\Delta V = V - V_1 = A x \implies x = \frac{V - V_1}{A}$ where $A$ is the cross-sectional area of the piston.

The force balance on the piston at any volume $V$ gives the pressure of the gas: $P \cdot A = P_{\text{atm}} \cdot A + k x$ $P = P_1 + \frac{k}{A^2} (V - V_1) = P_1 + C(V - V_1)$ where $C = \frac{k}{A^2}$ is a positive constant.

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2. Analysis of Options (A) and (B)

Given conditions: $V_2 = 2V_1 \quad \text{and} \quad T_2 = 3T_1$

Using the ideal gas state equation at state 2: $\frac{P_2 V_2}{P_1 V_1} = \frac{T_2}{T_1}$ $\frac{P_2 (2V_1)}{P_1 V_1} = \frac{3T_1}{T_1} \implies P_2 = \frac{3}{2} P_1$

Using the linear pressure-volume relationship: $P_2 = P_1 + C(V_2 - V_1)$ $\frac{3}{2} P_1 = P_1 + C(2V_1 - V_1) \implies C V_1 = \frac{1}{2} P_1 \implies C = \frac{P_1}{2 V_1}$

Potential energy stored in the spring:

$U_{\text{spring}} = \frac{1}{2} k x^2 = \frac{1}{2} \left(\frac{k}{A^2}\right) (V_2 - V_1)^2 = \frac{1}{2} C (V_2 - V_1)^2$ Substituting $C = \frac{P_1}{2 V_1}$ and $V_2 - V_1 = V_1$: $U_{\text{spring}} = \frac{1}{2} \left(\frac{P_1}{2 V_1}\right) (V_1)^2 = \frac{1}{4} P_1 V_1$

Thus, Statement (A) is correct.

Change in internal energy ($\Delta U$):

$\Delta U = n C_v (T_2 - T_1) = n \left(\frac{3}{2} R\right) (3T_1 - T_1) = 3 n R T_1 = 3 P_1 V_1$

Thus, Statement (B) is correct.

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3. Analysis of Options (C) and (D)

Given conditions: $V_2 = 3V_1 \quad \text{and} \quad T_2 = 4T_1$

Using the ideal gas state equation at state 2: $\frac{P_2 V_2}{P_1 V_1} = \frac{T_2}{T_1}$ $\frac{P_2 (3V_1)}{P_1 V_1} = \frac{4T_1}{T_1} \implies P_2 = \frac{4}{3} P_1$

Using the linear pressure-volume relationship: $P_2 = P_1 + C(V_2 - V_1)$ $\frac{4}{3} P_1 = P_1 + C(3V_1 - V_1) = P_1 + 2 C V_1 \implies 2 C V_1 = \frac{1}{3} P_1 \implies C = \frac{P_1}{6 V_1}$

Work done by the gas ($W$):

Since $P$ varies linearly with $V$, the work done is equal to the area of the trapezoid under the $P-V$ graph: $W = \frac{P_1 + P_2}{2} (V_2 - V_1)$ Substituting $P_2 = \frac{4}{3} P_1$ and $V_2 - V_1 = 2 V_1$: $W = \frac{P_1 + \frac{4}{3} P_1}{2} (2 V_1) = \frac{7}{3} P_1 V_1$

Thus, Statement (C) is correct.

Heat supplied to the gas ($Q$):

By the First Law of Thermodynamics: $Q = \Delta U + W$

First, calculating $\Delta U$: $\Delta U = n C_v (T_2 - T_1) = n \left(\frac{3}{2} R\right) (4T_1 - T_1) = \frac{9}{2} n R T_1 = \frac{9}{2} P_1 V_1$

Now calculating $Q$: $Q = \frac{9}{2} P_1 V_1 + \frac{7}{3} P_1 V_1 = \left(\frac{27 + 14}{6}\right) P_1 V_1 = \frac{41}{6} P_1 V_1$

Thus, Statement (D) is incorrect.

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Conclusion

The correct statements are A, B, and C.