JEE Mains · Physics · STD 11 - 11. thermodynamics
The above \(P-V\) diagram represents the thermodynamic cycle of an engine, operating with an ideal monatomic gas. The amount of heat, extracted from the source in a single cycle is
- A \(4P_0V_0\)
- B \(P_0V_0\)
- C \(\left( {\frac{{13}}{2}} \right)\) \( P_0V_0\)
- D \(\;\left( {\frac{{11}}{2}} \right)\) \( P_0V_0\)
Answer & Solution
Correct Answer
(C) \(\left( {\frac{{13}}{2}} \right)\) \( P_0V_0\)
Step-by-step Solution
Detailed explanation
\(Heat\,given\,to\,system = {\left( {n{C_V}\Delta T} \right)_{A \to B}} + {\left( {n{C_P}\Delta T} \right)_{B \to C}}\) \( = {\left[ {\frac{3}{2}\left( {nR\Delta T} \right)} \right]_{A \to B}} + {\left[ {\frac{5}{2}\left( {nR\Delta T} \right)} \right]_{B \to C}}\)…
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