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An ideal monoatomic gas undergoes a cyclic process `ABCA` as shown in the figure. The ratio of heat absorbed during `AB` to the work done on the gas during `BC` id A. `(5)/(2ln2)`B. `(5)/(3)`C. `(5)/(4ln2)`D. `(5)/(6)` |
Answer» Correct Answer - C `W_(AB) = (2V_(0) - V_(0))P_(0) = P_(0)V_(0)` [isobaric process] `W_(BC) = |nRT(2T_(0))ln"(V_(0))/(2V_(0))| = 2P_(0)V_(0)ln2` [Isothermal process] `therefore (Q_(AB))/(W_(BC)) = ((3)/(2)P_(0)V_(0)+ P_(0)V_(0))/(2P_(0)V_(0)ln2) = (5)/(4ln2)` |
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