JEE Mains · Physics · STD 11 - 11. thermodynamics
Using the given \(\mathrm{P}-\mathrm{V}\) diagram, the work done by an ideal gas along the path ABCD is :
- A \(3 \mathrm{P}_0 \mathrm{~V}_0\)
- B \(-4 \mathrm{P}_0 \mathrm{~V}_0\)
- C \(-3 \mathrm{P}_0 \mathrm{~V}_0\)
- D \(4 \mathrm{P}_0 \mathrm{~V}_0\)
Answer & Solution
Correct Answer
(C) \(-3 \mathrm{P}_0 \mathrm{~V}_0\)
Step-by-step Solution
Detailed explanation
Area under graph will be magnitude of graph and being counterclockwise graph it would be negative \(\begin{aligned} & \text { Area }=2 P_0 \times V_0+P_0 V_0=3 P_0 V_0 \\ & \mathrm{~W}=-3 P_0 V_0 \end{aligned}\)
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