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The state of an ideal gas is changed through an isothermal process at temperature `T_0` as shown in figure. The work done by the gas in going from state B to C is double the work done by gas in going from state A to B. If the pressure in the state B is `(p_0)/(2)`, then the pressure of the gas in state C is A. `(P_(0))/(2)`B. `(P_(0))/(4)`C. `(P_(0))/(6)`D. `(P_(0))/(8)` |
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Answer» Work done by gas in going isothermally from state `A` to `B` is `DeltaW_(AB)=nRT"ln"(P_(A))/(P_(C))=nRTln2` .....(1) Work done by gas in going isothermally from state `B` to `C` is `DeltaW_(BC)=nRT"ln"(P_(B))/(P_(C))=nRT"ln"(P_(0))/(2P_(C))`.....(2) It is given that `DeltaW_(BC)=2 DeltaW_(AB)` `therefore "ln"(P_(0))/(2P_(C))=ln(2)^(2) therefore P_(c)=(P_(0))/(8)` |
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