Physics · Thermodynamics · NEET
The work done by the gas is W = (P1 V1 - P2 V2) / (gamma - 1). Using the ideal gas law PV = nRT, this becomes W = nR (T1 - T2) / (gamma - 1). Here P1, V1, T1 are the starting values and P2, V2, T2 are the final values. Gamma is the ratio of specific heats Cp/Cv. Both forms give the same answer, so pick whichever data the question gives you.
Start with W = integral of P dV from V1 to V2. In an adiabatic process P V^gamma = constant = P1 V1^gamma, so P = P1 V1^gamma / V^gamma. Put this inside the integral and integrate V^(-gamma) dV. The integral of V^(-gamma) is V^(1-gamma)/(1-gamma). After you put in the limits and simplify using P1 V1^gamma = P2 V2^gamma, you get W = (P1 V1 - P2 V2)/(gamma - 1). Replacing PV with nRT gives W = nR(T1 - T2)/(gamma - 1).
Adiabatic means the gas is insulated, so no heat is exchanged: Q = 0. From the first law Q = dU + W, this gives 0 = dU + W, so W = -dU. The gas can only do work by using up its own internal energy. That is why an adiabatic expansion cools the gas and an adiabatic compression heats it.
In an adiabatic expansion the gas pushes out, so the gas does positive work (W > 0). Because W = -dU, the internal energy drops, so dU is negative and the temperature falls (T2 < T1). In an adiabatic compression the opposite happens: W is negative, internal energy rises, and the temperature goes up (T2 > T1).
In an isothermal process heat can flow in or out to hold the temperature fixed, so T stays constant. In an adiabatic process no heat can flow, so when the gas expands it must spend its own internal energy to do work, and internal energy of an ideal gas depends only on temperature. Less internal energy means lower temperature, so T must change.
Gamma is always greater than 1 (for a monatomic gas 5/3, for a diatomic gas 7/5), so (gamma - 1) is always positive. Never make it negative. The sign of the work comes from the temperature difference (T1 - T2), not from gamma. If the gas expands, T1 > T2 so W is positive.
In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas (gamma = 5/3) decreases from 60 K to 50 K. The work done by the gas in the process is: [Take R = 8.3 J per mol per K]
Try the real previous-year questions from this chapter — each with the answer and a full solution.
W = (P1 V1 - P2 V2)/(gamma - 1) = nR(T1 - T2)/(gamma - 1), where gamma = Cp/Cv.
No. An isothermal process obeys PV = constant. An adiabatic process obeys P V^gamma = constant, which is why its P-V curve is steeper.
Use gamma = 5/3 for a monatomic ideal gas (like helium or argon). For a diatomic gas (like oxygen or nitrogen) use gamma = 7/5 = 1.4.
Yes. If the question gives you the two temperatures and gamma, this form is the fastest. If it gives pressures and volumes instead, use W = (P1 V1 - P2 V2)/(gamma - 1).
During adiabatic expansion the gas cools, so its pressure drops faster and the area under the P-V curve is smaller. For the same compression, however, adiabatic work is larger because the gas heats up and pressure rises faster.
