Question

The P-V diagram of a Carnot's engine is shown in the graph below. The engine uses 1 mole of an ideal gas as working substance. From the graph, the area enclosed by the P-V diagram is
[ The heat supplied to the gas is 8000 J ]

• A. 2000 J
• B. 3000 J
• C. 1000 J
• D. 1200 J

Answer 1

The Correct Option is B

Given: 

• The engine uses 1 mole of an ideal gas.
• Heat supplied to the gas: \( Q_{in} = 8000 \) J

Step 1: Understanding Work Done

The area enclosed by the P-V diagram represents the work done by the engine:

\[ W = Q_{in} \times \eta \]

For a Carnot engine, efficiency \( \eta \) is given by:

\[ \eta = 1 - \frac{T_C}{T_H} \]

From the problem, we use the given heat supplied and known thermodynamic properties to determine the work done.

Step 2: Calculation

Using the given data:

\[ W = 8000 \times \frac{3}{8} \]

\[ W = 3000 \text{ J} \]

Answer: The correct option is B (3000 J).

Answer 2

1. Identify the Goal: The question asks for the area enclosed by the P-V diagram. For any cyclic process, the area enclosed in the P-V diagram represents the net work done (W_net) by the engine per cycle.

2. Understand the Carnot Cycle: The diagram represents a Carnot cycle, consisting of:

• A → B: Isothermal expansion at high temperature T_H (heat Q_H is absorbed).
• B → C: Adiabatic expansion (temperature drops from T_H to T_C).
• C → D: Isothermal compression at low temperature T_C (heat Q_C is rejected).
• D → A: Adiabatic compression (temperature rises from T_C to T_H).

3. Relate Work, Heat, and Efficiency: The net work done by the engine in a cycle is given by the difference between the heat absorbed from the hot reservoir (Q_H) and the heat rejected to the cold reservoir (Q_C):

W_net = Q_H - Q_C

The efficiency (η) of a heat engine is defined as:

η = W_net / Q_H

For a Carnot engine operating between temperatures T_H and T_C, the efficiency is also given by:

η = 1 - (T_C / T_H)

Combining these, we get:

W_net = η * Q_H = (1 - T_C / T_H) * Q_H

4. Determine Temperatures from the Graph: We are given Q_H = 8000 J. We need to find the ratio T_C / T_H. For an ideal gas, PV = nRT. Since n and R are constant, the temperature T is proportional to the product PV (T ∝ PV).

The high-temperature isotherm (A→B) is at T_H. We can use the coordinates of point A (which lies on this isotherm) to find a value proportional to T_H.

• P_A = 1600 kPa = 1600 × 10³ Pa
• V_A = 2.5 cm³ = 2.5 × 10^-6 m³
• P_AV_A = (1600 × 10³) × (2.5 × 10^-6) = 1600 × 2.5 × 10^-3 = 4000 × 10^-3 = 4 J
• So, T_H ∝ P_AV_A = 4 J (for n=1 mol)

The low-temperature isotherm (C→D) is at T_C. We can use the coordinates of point C (which lies on this isotherm) to find a value proportional to T_C.

• P_C = 400 kPa = 400 × 10³ Pa
• V_C = 6.25 cm³ = 6.25 × 10^-6 m³
• P_CV_C = (400 × 10³) × (6.25 × 10^-6) = 400 × 6.25 × 10^-3 = 2500 × 10^-3 = 2.5 J
• So, T_C ∝ P_CV_C = 2.5 J (for n=1 mol)

5. Calculate Efficiency:

η = 1 - (T_C / T_H) = 1 - (P_CV_C / P_AV_A)

η = 1 - (2.5 J / 4 J) = 1 - (2.5 / 4) = 1 - (25 / 40) = 1 - (5 / 8) = 3 / 8

6. Calculate Net Work Done (Area):

W_net = Area = η * Q_H

W_net = (3 / 8) * 8000 J

W_net = 3 * (8000 / 8) J = 3 * 1000 J = 3000 J

Answer: The area enclosed by the P-V diagram is 3000 J. This corresponds to option (B).