InterviewSolution
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InterviewSolution
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For an ideal gas, an illustration of the different paths, A `(B + C)` and `(D + E)` from an intial state `P_(1), V_(1), T_(1)` to a final state `P_(2), V_(2), T_(1)` as shown in the given figure. Path (A) represents a reversible isothernal expanison from `P_(1)V_(1)` to `P_(2)V_(2)`. Path `(B + C)` represents a reversible adiabatic expansion (B) from `P_(1), V_(1), T_(1)` to `P_(3), V_(2), T_(2)` followed by reversible heating the gas at constant volume (C) from `P_(3), V_(2), T_(2)` to `P_(2), V_(2) T_(1)` to `P_(1) , V_(2), T_(3)` followed by reversible cooling at a constant volume `V_(2)` (E) from `P_(1), V_(2), T_(3)` to `P_(2), V_(2), T_(1)` What is `Delta S` for path `(D + E)` ?A. zeroB. `underset(T_(1))overset(T_(2))int (C_(V)(T))/(T) dT`C. `-nR ln. (V_(2))/(V_(1))`D. `nR ln. (V_(2))/(V_(1))` |
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Answer» Correct Answer - D path (A) : isothermal `(Delta U = 0)` `q_(rev) = w = nRT_(1) ln. (v_(2))/(v_(1))` path `(B + C) : q_(rev) = 0 + (-nR ln.(v_(2))/(v_(1)))` path `(D + E)`: `(Delta S)_("sys") = nR ln.(v_(2))/(v_(1)) + .^(n)C_(V) ln. (T_(2))/(T_(1)) = nR ln. (v_(2))/(v_(1))` |
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