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One mole of an ideal gas undergoes a process `P = P_(0) [1 + ((2 V_(0))/(V))^(2)]^(-1)`, where `P_(0) V_(0)` are constants. Change in temperature of the gas when volume is changed from `V = V_(0) to V = 2 V_(0)` is:A. `(4)/(5) (P_(0) V_(0))/(nR)`B. `(3)/(4) (P_(0) V_(0))/(nR)`C. `(2)/(3) (P_(0) V_(0))/(nR)`D. `(9)/(7) (P_(0) V_(0))/(nR)` |
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Answer» Correct Answer - A `P=P_(0) [1+(2V_(0)//V)^(2)T^(-1)` at `V =V_(0), P = P_(0)//5` `T_(i) = (PV)/(nR) = ((P_(0)//5)V_(0))/(nR) = (P_(0)V_(0))/(5nR)` At `V = 2V_(0), P=P_(0)//2` `T_(f) = (PV)/(nR) = ((P_(0)//2)(2V_(0)))/(nR) = (P_(0)V_(0))/(nR)` `Delta T = T_(f)-T_(i) = (4P_(0)V_(0))/(5nR)`. |
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