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Two moles of helium gas are taken over the cycle ABCDA, as shown in the P-T diagram The net work done on the gas in the cycle ABCDA is :A. ZeroB. `276R`C. `1076R`D. `1904R` |
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Answer» Correct Answer - B `W_(ABCDA) = W_(AB) +W_(BC) +W_(CD) +W_(DA)` `= nR (DeltaT)_(AB) +nR(T_(B)) In (P_(B))/(P_(C))` `+nR(DeltaT)_(CD) +nR(T_(D)) In (P_(D))/(P_(A))` `= nR (200) +500 nR In 2 +nR (-200)` `+300 nR In (1)/(2)` `= 2 In 2[500R - 300R]` `= (400R)(In2) - (400R) (0.693) = 276R` `W_(ABCDA) = 276R` |
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