Question

One mole of an ideal monoatomic gas is taken round the cyclic process ABCA as shown in figure.

• A. $\frac{25P_0V_0}{8R}$
• B. $\frac{P_0V_0}{R}$
• C. $P_0V_0$
• D. $3P_0V_0$
• E. $2P_0V_0$

Answer 1

P → 3, Q → 4, R → 2, S → 1

Answer 2

The given cyclic process ABCA for one mole of an ideal monoatomic gas is shown in the P-V diagram. The coordinates of the vertices are A: $(V_0, P_0)$, B: $(V_0, 3P_0)$, and C: $(2V_0, P_0)$.
For an ideal gas, $PV = nRT$. Here $n=1$, so $PV = RT$.

(P) The work done by the gas during the complete cycle:
The work done during a cyclic process is the area enclosed by the cycle on the P-V diagram. The cycle ABCA forms a triangle. The vertices are A, B, and C.
The work done is the area of the triangle ABC. The base of the triangle can be taken as the segment CA along the constant pressure line $P=P_0$. The length of the base is $V_C - V_A = 2V_0 - V_0 = V_0$.
The height of the triangle is the perpendicular distance from B to the line CA. This is the difference in pressure between B and the line CA (which is at pressure $P_0$), so the height is $P_B - P_A = 3P_0 - P_0 = 2P_0$.
The area of the triangle is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times V_0 \times 2P_0 = P_0 V_0$.
The cycle is traced in the clockwise direction (A $\to$ B $\to$ C $\to$ A) on the P-V diagram, which means the net work done by the gas is positive.
So, the work done by the gas during the complete cycle is $W_{cycle} = P_0 V_0$.
Matching with List-II, (P) corresponds to (3) $P_0V_0$.

(Q) Heat absorbed by the gas in the path AB:
Path AB is an isochoric process at volume $V_0$. The pressure changes from $P_0$ to $3P_0$.
For an isochoric process, the work done is $W_{AB} = 0$.
The change in internal energy is $\Delta U_{AB} = n C_V \Delta T$. For a monoatomic gas, $C_V = \frac{3}{2}R$.
The temperatures at A and B are $T_A = \frac{P_A V_A}{R} = \frac{P_0 V_0}{R}$ and $T_B = \frac{P_B V_B}{R} = \frac{3P_0 V_0}{R}$.
$\Delta U_{AB} = 1 \times \frac{3}{2}R (T_B - T_A) = \frac{3}{2}R (\frac{3P_0 V_0}{R} - \frac{P_0 V_0}{R}) = \frac{3}{2}R (\frac{2P_0 V_0}{R}) = 3P_0 V_0$.
According to the first law of thermodynamics, $Q_{AB} = \Delta U_{AB} + W_{AB}$.
$Q_{AB} = 3P_0 V_0 + 0 = 3P_0 V_0$.
Since $Q_{AB}$ is positive, heat is absorbed by the gas in the path AB.
Matching with List-II, (Q) corresponds to (4) $3P_0V_0$.

(R) The minimum temperature attained by the gas during the cycle:
The temperature at any point $(P, V)$ is $T = \frac{PV}{R}$.
Temperature at A: $T_A = \frac{P_0 V_0}{R}$.
Temperature at B: $T_B = \frac{3P_0 V_0}{R}$.
Temperature at C: $T_C = \frac{P_0 (2V_0)}{R} = \frac{2P_0 V_0}{R}$.
Along path AB (isochoric $V=V_0$), $T = \frac{P V_0}{R}$. As $P$ goes from $P_0$ to $3P_0$, $T$ goes from $T_A$ to $T_B$. Minimum temperature is $T_A$.
Along path CA (isobaric $P=P_0$), $T = \frac{P_0 V}{R}$. As $V$ goes from $2V_0$ to $V_0$, $T$ goes from $T_C$ to $T_A$. Minimum temperature is $T_A$.
Along path BC, the line passes through $(V_0, 3P_0)$ and $(2V_0, P_0)$. The equation of the line is $P - P_0 = \frac{3P_0 - P_0}{V_0 - 2V_0}(V - 2V_0) = \frac{2P_0}{-V_0}(V - 2V_0) = -\frac{2P_0}{V_0}(V - 2V_0)$.
$P = P_0 - \frac{2P_0}{V_0}V + 4P_0 = 5P_0 - \frac{2P_0}{V_0}V$.
The temperature along BC is $T(V) = \frac{PV}{R} = \frac{1}{R} (5P_0 - \frac{2P_0}{V_0}V)V = \frac{P_0}{R} (5V - \frac{2}{V_0}V^2)$.
To find the minimum temperature, we check the endpoints and any critical points.
$T(V_0) = \frac{P_0}{R}(5V_0 - \frac{2}{V_0}V_0^2) = \frac{P_0}{R}(5V_0 - 2V_0) = \frac{3P_0 V_0}{R} = T_B$.
$T(2V_0) = \frac{P_0}{R}(5(2V_0) - \frac{2}{V_0}(2V_0)^2) = \frac{P_0}{R}(10V_0 - \frac{2}{V_0}4V_0^2) = \frac{P_0}{R}(10V_0 - 8V_0) = \frac{2P_0 V_0}{R} = T_C$.
To find critical points, $\frac{dT}{dV} = \frac{P_0}{R} (5 - \frac{4}{V_0}V)$. Setting $\frac{dT}{dV} = 0$, we get $5 = \frac{4}{V_0}V$, so $V = \frac{5V_0}{4}$.
The second derivative is $\frac{d^2T}{dV^2} = \frac{P_0}{R} (-\frac{4}{V_0}) < 0$, so this is a maximum.
Thus, the minimum temperature along BC occurs at the endpoints, which are $T_B$ and $T_C$.
Comparing the temperatures at the vertices $T_A = \frac{P_0 V_0}{R}$, $T_B = \frac{3P_0 V_0}{R}$, $T_C = \frac{2P_0 V_0}{R}$, the minimum temperature is $T_A = \frac{P_0 V_0}{R}$.
Matching with List-II, (R) corresponds to (2) $\frac{P_0V_0}{R}$.

(S) The maximum temperature attained by the gas during the cycle:
From the previous calculation, the temperatures at the vertices are $T_A = \frac{P_0 V_0}{R}$, $T_B = \frac{3P_0 V_0}{R}$, $T_C = \frac{2P_0 V_0}{R}$.
Along path AB, temperature increases from $T_A$ to $T_B$. Maximum is $T_B$.
Along path CA, temperature decreases from $T_C$ to $T_A$. Maximum is $T_C$.
Along path BC, the temperature has a maximum at $V = \frac{5V_0}{4}$.
The maximum temperature along BC is $T_{max, BC} = T(\frac{5V_0}{4}) = \frac{P_0}{R} (5(\frac{5V_0}{4}) - \frac{2}{V_0}(\frac{5V_0}{4})^2) = \frac{P_0}{R} (\frac{25V_0}{4} - \frac{2}{V_0}\frac{25V_0^2}{16}) = \frac{P_0 V_0}{R} (\frac{25}{4} - \frac{50}{16}) = \frac{P_0 V_0}{R} (\frac{100-50}{16}) = \frac{50P_0 V_0}{16R} = \frac{25P_0 V_0}{8R}$.
Now compare the temperatures $T_A$, $T_B$, $T_C$, and $T_{max, BC}$.
$T_A = \frac{P_0 V_0}{R} = \frac{8P_0 V_0}{8R}$
$T_B = \frac{3P_0 V_0}{R} = \frac{24P_0 V_0}{8R}$
$T_C = \frac{2P_0 V_0}{R} = \frac{16P_0 V_0}{8R}$
$T_{max, BC} = \frac{25P_0 V_0}{8R}$.
Comparing these values, the maximum temperature is $T_{max, BC} = \frac{25P_0 V_0}{8R}$.
Matching with List-II, (S) corresponds to (1) $\frac{25P_0V_0}{8R}$.

Summary of matches:
(P) $\to$ (3)
(Q) $\to$ (4)
(R) $\to$ (2)
(S) $\to$ (1)