An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the figure). Initially the gas is at temperature $T_1$, pressure $P_1$ and volume $V_1$ and the spring is in its relaxed state. The gas is then heated very slowly to temperature $T_2$, pressure $P_2$ and volume $V_2$. During this process the piston moves out by a distance $x$. Ignoring the friction between the piston and the cylinder, the correct statement(s) is (are)

Step 1: Formulate the relation between pressure and volume.

At any position $x$, the force balance on the piston of area $A$ is given by the equilibrium of forces.

$$PA = P_1 A + kx$$

The displacement $x$ of the piston is related to the volume change.

$$x = \frac{V - V_1}{A}$$

Substitute $x$ into the force balance equation.

$$P = P_1 + \frac{k}{A^2}(V - V_1)$$

Define the constant $C = \frac{k}{A^2}$.

$$P = P_1 + C(V - V_1)$$

Step 2: Analyze Case I where $V_2 = 2V_1$ and $T_2 = 3T_1$.

Apply the ideal gas law to find the final pressure $P_2$.

$$\frac{P_2 V_2}{T_2} = \frac{P_1 V_1}{T_1}$$ $$\frac{P_2 (2V_1)}{3T_1} = \frac{P_1 V_1}{T_1}$$ $$P_2 = \frac{3}{2}P_1$$

Substitute $P_2$ and $V_2$ into the pressure-volume relation to find $C$.

$$P_2 = P_1 + C(V_2 - V_1)$$ $$\frac{3}{2}P_1 = P_1 + C(2V_1 - V_1)$$ $$C = \frac{P_1}{2V_1}$$

Step 3: Calculate the spring energy and change in internal energy for Case I.

The energy stored in the spring is given by the work done on the spring.

$$U_{\text{spring}} = \frac{1}{2}kx^2$$ $$U_{\text{spring}} = \frac{1}{2}C(V_2 - V_1)^2$$ $$U_{\text{spring}} = \frac{1}{2}\left(\frac{P_1}{2V_1}\right)(V_1)^2$$ $$U_{\text{spring}} = \frac{1}{4}P_1 V_1$$

For a monoatomic ideal gas, the change in internal energy is calculated using $C_V = \frac{3}{2}R$.

$$\Delta U = \frac{3}{2}(P_2 V_2 - P_1 V_1)$$ $$\Delta U = \frac{3}{2}\left(\frac{3}{2}P_1 (2V_1) - P_1 V_1\right)$$ $$\Delta U = 3P_1 V_1$$

Step 4: Analyze Case II where $V_2 = 3V_1$ and $T_2 = 4T_1$.

Apply the ideal gas law to find the final pressure $P_2$.

$$\frac{P_2 (3V_1)}{4T_1} = \frac{P_1 V_1}{T_1}$$ $$P_2 = \frac{4}{3}P_1$$

Substitute $P_2$ and $V_2$ into the pressure-volume relation to find $C$.

$$\frac{4}{3}P_1 = P_1 + C(3V_1 - V_1)$$ $$C = \frac{P_1}{6V_1}$$

Step 5: Calculate the work done and heat supplied for Case II.

The work done by the gas is the area under the $P\text{-}V$ curve.

$$W = P_1(V_2 - V_1) + \frac{1}{2}C(V_2 - V_1)^2$$ $$W = P_1(2V_1) + \frac{1}{2}\left(\frac{P_1}{6V_1}\right)(2V_1)^2$$ $$W = \frac{7}{3}P_1 V_1$$

Calculate the change in internal energy for Case II.

$$\Delta U = \frac{3}{2}(P_2 V_2 - P_1 V_1)$$ $$\Delta U = \frac{3}{2}\left(\frac{4}{3}P_1 (3V_1) - P_1 V_1\right)$$ $$\Delta U = \frac{9}{2}P_1 V_1$$

Apply the First Law of Thermodynamics to find the heat supplied.

$$Q = \Delta U + W$$ $$Q = \frac{9}{2}P_1 V_1 + \frac{7}{3}P_1 V_1$$ $$Q = \frac{41}{6}P_1 V_1$$

Answer: The correct statements are:

• If $V_2 = 2V_1$ and $T_2 = 3T_1$, the energy stored in the spring is $\dfrac{1}{4}P_1 V_1$.
• If $V_2 = 2V_1$ and $T_2 = 3T_1$, the change in internal energy of the gas is $3P_1 V_1$.
• If $V_2 = 3V_1$ and $T_2 = 4T_1$, the heat supplied to the gas is $\dfrac{41}{6}P_1 V_1$.