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Two moles of helium gas are taken over the cycle ABCDA, as shown in the P-T diagram The work done on the gas in taking it from D to A is :A. `-414 R`B. `+414R`C. `-690 R`D. `+690R` |
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Answer» Correct Answer - B `D` to `A` is isothermal process Work done by the gas in `SD` to `A` is `W_(DA) = nRT In (V_(2))/(V_(1)) = nRT In (P_(1))/(P_(2))` `= (2)(R )(300)In (10^(5))/(2xx10^(5)) = (600R)[-In2]` `=- (600R) (0.693) =- 414 R` `W_(DA) = - 414R`, it is work done by the gas So work done on the gas is `+414 R` |
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