NEETPhysicsThermodynamics
If one mole of an ideal gas at P 1 , V 1 is allowed to expand reversibly and isothermally ( A to B ) its pressure is reduced to one-half of the original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value B → C . Then it is restored to its initial state by a reversible adiabatic compression ( C to A ). The net workdone by the gas is equa
Options
- A0
- BR T ln 2
- C- R T 2 γ - 1
- DRT ln 2 - 1 2 γ - 1
Correct answer
D. RT ln 2 - 1 2 γ - 1
Step-by-step solution
A - B = isothermal process W A B = P 1 V 1 ln 2 V 1 V 1 = P 1 V 1 ln 2 B - C → Isochoric process W B C = 0 C - A → Adiabatic process W C A = P 1 V 1 - P 1 4 × 2 V 1 1 - γ = P 1 V 1 1 - 1 2 1 - γ = P 1 V 1 2 1 - γ W net = W A B + W B C + W C A P 1 V 1 = R T = P 1 V 1 ln 2 + 0 + P 1 V 1 2 1 - γ W net = R T ln 2 - 1 2 γ - 1