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One mole of an ideal monoatomaic gas is taken from `A` to `C` along the path `ABC` . The temperature of the gas at `A` is `T_(0)` . For the process `ABC` : A. Work done by the gas is `RT_(0)`B. Change in internal energy of the gas is `(11)/(2)RT_(0)`C. Heat absorbed by the gas is `(11)/(2)RT_(0)`D. Heat absorbed by the gas is `(13)/(2)RT_(0)` |
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Answer» Correct Answer - A::C Work done = Area of `ABC` with `V`- axis =`P_(0)(2V_(0)-V_(0)) + 0 = P_(0)V_(0) = nRT_(0) = RT_(0)` Change in internal energy = `nC_(V)DeltaT` ` 1 xx (3)/(2)R xx (4T_(0)-T_(0)) = (9)/(2)RT_(0)` `therefore` Heat absorbed = `(9)/(2) RT_(0) + RT_(0) = (11)/(2)RT_(0)` |
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