Question:

Heat energy absorbed by a system in going through the cyclic process shown in the figure is

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For any cyclic process, \[ \Delta U=0 \] and therefore \[ Q=W. \] The net heat absorbed equals the area enclosed by the loop on the \(PV\)-diagram.
Updated On: Jul 9, 2026
  • \(10^{7}\pi\,\text{J}\)
  • \(10^{4}\pi\,\text{J}\)
  • \(10^{2}\pi\,\text{J}\)
  • \(10^{-3}\pi\,\text{J}\) \bigskip
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The Correct Option is C

Solution and Explanation

Concept: For a cyclic process, \[ \Delta U=0. \] Hence, from the first law of thermodynamics, \[ Q=W. \] Therefore, the heat absorbed by the system is equal to the area enclosed by the cycle on the \(PV\)-diagram.

Step 1:
Identify the shape of the cycle. The cycle is a circle on the \(PV\)-diagram. From the figure, \[ P_{\max}=30\,\text{kPa}, \qquad P_{\min}=10\,\text{kPa}. \] Hence the radius along the pressure axis is \[ r_P=\frac{30-10}{2}=10\,\text{kPa}. \] Similarly, \[ V_{\max}=30\,\text{L}, \qquad V_{\min}=10\,\text{L}. \] Thus, \[ r_V=\frac{30-10}{2}=10\,\text{L}. \]

Step 2:
Calculate the area enclosed by the cycle. \[ W=\text{Area} = \pi r_Pr_V. \] \[ W = \pi(10\,\text{kPa})(10\,\text{L}). \] \[ W = 100\pi\,(\text{kPa}\cdot\text{L}). \]

Step 3:
Convert into joules. Since \[ 1\,\text{kPa}\cdot\text{L} = 10^3\,\text{Pa}\times10^{-3}\,\text{m}^3 = 1\,\text{J}, \] we get \[ W=100\pi\,\text{J}. \] \[ W=10^2\pi\,\text{J}. \]

Step 4:
Write the final answer. \[ \boxed{Q=W=10^2\pi\,\text{J}} \] \[ \boxed{\text{Answer = (C)}} \]
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