Chemistry · Thermodynamics · NEET
Isothermal expansion does more work (for the same start and end volume). In an isothermal process the temperature is held constant, so the gas keeps pulling in heat and keeps its pressure high. Higher pressure at every volume means a bigger push, so more work. In an adiabatic process no heat enters (q = 0), so the gas cools as it expands, pressure falls faster, and less work is done. This is exactly why NEET 2023 states W(isothermal) > W(adiabatic).
Start both from the same point A. As volume increases, the adiabatic gas cools (temperature drops), so its pressure falls more steeply. The isothermal gas keeps the same temperature, so its pressure falls more gently. At any given volume the isothermal pressure is higher, so its curve sits above the adiabat. The adiabatic curve is steeper because pV^gamma = constant (gamma > 1) drops faster than pV = constant.
Work in expansion is w = -integral of p dV. Graphically, that integral is just the area under the curve between the initial and final volume. So the process whose curve sweeps the LARGER area between the same two volumes does the maximum work. This is the whole idea behind NEET 2022, which asks you to pick the p-V curve with the biggest area.
In an adiabatic process no heat is exchanged, so q = 0. By the first law, delta U = q + w = w. During expansion the gas does work on the surroundings, so w is negative, which makes delta U negative. For an ideal gas delta U = nCv(delta T), so a negative delta U means the temperature FALLS. The gas spends its own internal energy to do the work, so it cools.
During any expansion the gas pushes outward and does work ON the surroundings, so the work done ON the gas is NEGATIVE (w < 0), using the IUPAC convention. NEET frames this both ways: 'work done by the gas' is a negative value too in their answer keys (see NEET 2019 answer -30 J). Always check whether the question asks work done BY the gas or ON the gas, but the magnitude comparison Iso > Adiabat stays the same.
No. Adiabatic means no heat flow (q = 0); the temperature usually CHANGES. Isothermal means constant temperature; heat usually flows. They only look the same in one special case: adiabatic FREE expansion into vacuum, where p(external) = 0 so w = 0, q = 0, and delta T = 0 (NEET 2020). In ordinary expansions against a real pressure, adiabatic and isothermal give different work.
Reversible expansion of an ideal gas under isothermal and adiabatic conditions is shown (A to B isothermal, A to C adiabatic on a p-V plot). Which statement is NOT correct?
Which of the following p-V curves represents the maximum work done?
The work done during the reversible isothermal expansion of one mole of hydrogen gas at 25 C from a pressure of 20 atm to 10 atm is (R = 2.0 cal/K/mol):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes, for expansion between identical initial and final volumes. The isothermal curve stays higher (constant T keeps pressure up), so it encloses more area and does more work. Adiabatic cools and drops faster, doing less work.
w = -nRT ln(V2/V1) = -2.303 nRT log(V2/V1). You can also write it in pressures as w = -2.303 nRT log(P1/P2), since at constant T pressure and volume are inversely related.
Because q(rev) = 0 and delta S = q(rev)/T. With no heat exchanged reversibly, the entropy change of the system is zero, so a reversible adiabat is called isentropic. An isothermal expansion instead has delta S > 0.
No. Only ISOTHERMAL expansion of an ideal gas has delta U = 0 (because delta T = 0). In adiabatic expansion the temperature falls, so delta U = nCv(delta T) is negative, not zero.
NEET repeatedly compares isothermal vs adiabatic work, p-V areas, and the sign of delta T. Knowing 'isothermal curve above, more work; adiabatic cools, less work' lets you solve 2022, 2023 and 2024 questions in seconds without long calculation.
