Physics · Thermodynamics · NEET
Look at the shape of the line starting from the common point. A vertical straight line means volume is not changing, so it is isochoric. A horizontal straight line means pressure is not changing, so it is isobaric. A falling curve means both P and V change; of the two falling curves the less steep one is isothermal and the steeper one is adiabatic. So: vertical = isochoric, horizontal = isobaric, gentle curve = isothermal, steep curve = adiabatic.
The slope of a curve on a P-V graph tells how fast pressure falls as volume grows. Isothermal follows PV = constant, so its slope is proportional to minus P over V. Adiabatic follows PV^gamma = constant, so its slope is proportional to minus gamma times P over V. Since gamma (Cp/Cv) is always greater than 1, the adiabatic slope is gamma times larger in size. At the same point the adiabatic line drops faster, so it is steeper. The ratio of slopes is exactly gamma.
It gets colder. In an adiabatic process no heat enters or leaves (Q = 0). When the gas expands it does work on the surroundings, and that work energy can only come from the internal energy of the gas. Losing internal energy means the temperature falls. This is why in the NEET 2023 PYQ the statement T_C > T_A after adiabatic expansion is wrong; the correct result is T_C < T_A.
Work done by the gas equals the area under the P-V curve between the two volumes. Because the isobaric line stays high (constant pressure), then the isothermal falls a bit, and the adiabatic falls fastest, the areas rank as: isobaric > isothermal > adiabatic > isochoric (isochoric does zero work because volume does not change). So for equal expansion, isobaric gives the most work and isochoric gives none.
All four processes here start from one common point, so they all meet there. The isothermal and adiabatic both leave that point going down-right, but the adiabatic dives below the isothermal because it is steeper. After the start point they separate and do not meet again in normal expansion. The common start point is just the shared initial state (same P, V, T) before the gas is taken along four different paths.
An ideal gas undergoes four different processes from the same initial state. The processes are adiabatic, isothermal, isobaric and isochoric. The curve which represents the adiabatic process among curves 1, 2, 3, 4 is (curve 1 is vertical, curve 4 is horizontal, curves 2 and 3 are falling curves with 2 steeper than 3):
Reversible expansion of an ideal gas is shown, where AB is isothermal expansion and AC is adiabatic expansion, both starting from common state A. Which option is NOT correct?
A P-V diagram shows three isotherms at 700 K, 500 K and 300 K. Four processes I, II, III, IV start from a common state (I is a vertical line, II is a steep falling curve crossing to a lower-temperature isotherm, III runs along an isotherm, IV is a horizontal line). Match each process with its type: (a) Adiabatic, (b) Isobaric, (c) Isochoric, (d) Isothermal.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
From least steep to most steep for the changing-volume processes: isobaric (horizontal, slope zero), then isothermal (gentle falling curve), then adiabatic (steeper falling curve), then isochoric (vertical, slope infinite). So isochoric is the steepest possible line and isobaric is the flattest.
Work done by a gas equals pressure times change in volume, and it is the area under the P-V line. In an isochoric process volume does not change, so there is no area under a vertical line. Therefore work done is zero and all the heat goes into changing internal energy (Q = dU).
Yes. At the same point on the graph, the adiabatic slope equals gamma multiplied by the isothermal slope in magnitude. For a monatomic gas gamma is about 1.67 and for a diatomic gas about 1.40, so the adiabatic is roughly 1.4 to 1.7 times steeper at that point.
The adiabatic expansion. Isothermal keeps temperature constant, isobaric expansion actually raises temperature (V rises at constant P so T rises), and adiabatic expansion lowers temperature because the gas spends internal energy doing work with no heat coming in.
This page is about reading and comparing the shapes on the graph. The next concept, work done in different processes, gives the exact formulas: W = P.dV for isobaric, W = nRT ln(V2/V1) for isothermal, W = (P1V1 - P2V2)/(gamma - 1) for adiabatic, and W = 0 for isochoric.
