Question

An ideal gas undergoes a cyclic transformation starting from the point \( A \) and coming back to the same point by tracing the path \( A \to B \to C \to A \) as shown in the diagram. The total work done in the process is \( \_\_\_\_ \) J.

Answer 1

Correct Answer: 200

This problem requires the calculation of total work performed by an ideal gas during a cyclic process, denoted as A → B → C → A, as depicted in the provided V-P diagram.

Concept Used:

The total work in a cyclic thermodynamic process is the aggregate of work done in each individual stage. Work done by a gas during a volume change from an initial state to a final state is quantified by the integral:

\[ W = \int_{V_i}^{V_f} P \, dV \]

Specific processes are defined as:

1. Isobaric process (constant pressure): \( W = P \Delta V = P(V_f - V_i) \)
2. Isochoric process (constant volume): \( W = 0 \) due to \( dV = 0 \).

Alternatively, the net work in a cycle can be determined from the area enclosed by the cycle on a P-V diagram. However, this solution will calculate the work for each step individually from the given V-P diagram and then sum these values.

The necessary unit conversion is: \( 1 \text{ kPa} \times 1 \text{ dm}^3 = (10^3 \text{ Pa}) \times (10^{-3} \text{ m}^3) = 1 \text{ J} \).

Step-by-Step Solution:

Step 1: Extract the Pressure (P) and Volume (V) coordinates for points A, B, and C from the V-P diagram. The y-axis represents Volume (V) in dm³ and the x-axis represents Pressure (P) in kPa.

• Point A: \( P_A = 10 \text{ kPa}, V_A = 10 \text{ dm}^3 \)
• Point B: \( P_B = 10 \text{ kPa}, V_B = 30 \text{ dm}^3 \)
• Point C: \( P_C = 30 \text{ kPa}, V_C = 10 \text{ dm}^3 \)

The total work done in the cycle is expressed as \( W_{total} = W_{A \to B} + W_{B \to C} + W_{C \to A} \).

Step 2: Compute the work done for the path A → B. This is an isobaric process with constant pressure \( P_A = P_B = 10 \) kPa. The volume increases from 10 dm³ to 30 dm³, signifying expansion.

\[ W_{A \to B} = P_A (V_B - V_A) \] \[ W_{A \to B} = (10 \text{ kPa}) \times (30 \text{ dm}^3 - 10 \text{ dm}^3) = 10 \times 20 \text{ kPa} \cdot \text{dm}^3 \] \[ W_{A \to B} = 200 \text{ J} \]

Step 3: Calculate the work done for the path C → A. This is an isochoric process, as the volume remains constant at \( V_C = V_A = 10 \) dm³. Work done in any isochoric process is zero.

\[ W_{C \to A} = 0 \text{ J} \]

Step 4: Determine the work done for the path B → C. In this stage, both pressure and volume undergo changes. The path is a linear segment on the V-P diagram connecting B(10 kPa, 30 dm³) and C(30 kPa, 10 dm³). The work done is represented by the area beneath this segment on a P-V diagram. As the volume decreases from 30 dm³ to 10 dm³, this is a compression, resulting in negative work done.

The area beneath the line segment BC on a P-V diagram forms a trapezoid. The magnitude of the work is equivalent to the area of this trapezoid.

\[ |W_{B \to C}| = \text{Area of Trapezoid} = \frac{1}{2} (P_B + P_C) (V_B - V_C) \] \[ |W_{B \to C}| = \frac{1}{2} (10 \text{ kPa} + 30 \text{ kPa}) \times (30 \text{ dm}^3 - 10 \text{ dm}^3) \] \[ |W_{B \to C}| = \frac{1}{2} (40) \times (20) \text{ kPa} \cdot \text{dm}^3 = 400 \text{ J} \]

Given that this is a compression process (\(V_f < V_i\)), the work done by the gas is negative.

\[ W_{B \to C} = -400 \text{ J} \]

Step 5: Calculate the total work done throughout the cycle by summing the work from each individual step.

\[ W_{total} = W_{A \to B} + W_{B \to C} + W_{C \to A} \] \[ W_{total} = 200 \text{ J} + (-400 \text{ J}) + 0 \text{ J} \]

Final Computation & Result:

The total work performed during the process is determined by aggregating the work from each stage:

\[ W_{total} = 200 - 400 + 0 = -200 \text{ J} \]

The negative sign signifies that the net work is performed on the gas during the cycle.

The total work done in the process is -200 J.