Question

The heat absorbed by a system in going through the given cyclic process is :

• A. 61.6 J
• B. 431.2 J
• C. 616 J
• D. 19.6 J

Answer 1

The Correct Option is A

For a cyclic process, the change in internal energy (\(\Delta U\)) is zero. Therefore, the heat absorbed (\(\Delta Q\)) is equal to the work done (\(\Delta W\)), which is the area enclosed by the P-V curve. The area is calculated as:

\[ \Delta Q = \Delta W = \pi \times (140 \times 10^3) \, \text{Pa} \times (140 \times 10^{-6} \, \text{m}^3) \]

\[ \Delta Q = 61.6 \, \text{J} \]

Answer 2

Step 1: Understanding the problem
The graph shows a cyclic process in the PV-plane. The work done by the system in one complete cycle equals the area enclosed by the loop. Since the process is cyclic, the net change in internal energy is zero, so the heat absorbed (Q) equals the work done (W).

Step 2: Identify key parameters from the diagram
From the figure: At maximum pressure, \( P_1 = 340 \, \text{kPa} \) At minimum pressure, \( P_2 = 60 \, \text{kPa} \) At maximum volume, \( V_1 = 340 \, \text{cc} \) At minimum volume, \( V_2 = 60 \, \text{cc} \).

The process path is a circle on the PV-plane.

Step 3: Find the radius of the circular path
Radius in pressure axis: \[ r_P = \frac{340 - 60}{2} = 140 \, \text{kPa}. \] Radius in volume axis: \[ r_V = \frac{340 - 60}{2} = 140 \, \text{cc}. \]

Step 4: Area enclosed by the circle (Work done in one cycle)
The area of a circle: \[ A = \pi \, r_P \, r_V. \] Thus, \[ W = \pi (140 \, \text{kPa}) (140 \, \text{cc}). \]

Step 5: Convert units properly
1 kPa = \(10^3 \, \text{N/m}^2\), 1 cc = \(10^{-6} \, \text{m}^3\). Hence, \[ W = \pi \times 140 \times 10^3 \times 140 \times 10^{-6} = \pi \times 1.96 \times 10^1 = 61.6 \, \text{J}. \]

Step 6: Relation between heat absorbed and work done
For a cyclic process, the change in internal energy \( \Delta U = 0 \), therefore \[ Q = W = 61.6 \, \text{J}. \]

Final answer

61.6 J