Physics · Thermodynamics · NEET
An isothermal process follows PV = constant. Differentiate both sides: P dV + V dP = 0. So dP/dV = -P/V. This is the slope of the isothermal curve at any point. It is negative, meaning as volume goes up, pressure goes down. The slope depends only on the ratio P/V at that point.
An adiabatic process follows PV^gamma = constant. Differentiate: (dP)V^gamma + P(gamma)V^(gamma-1) dV = 0. Divide through and simplify to get dP/dV = -gamma P/V. So the adiabatic slope is the isothermal slope multiplied by gamma. Because gamma is always bigger than 1, the adiabatic curve is always steeper at the same point.
Take one common point on the P-V graph. At that point P/V is the same for both curves. The only difference is the extra factor gamma in the adiabatic slope. Isothermal slope = -P/V, adiabatic slope = -gamma P/V. Since gamma is greater than 1 (for example 1.67 for a monatomic gas, 1.4 for a diatomic gas), the adiabatic slope has a larger magnitude, so the adiabatic curve drops more sharply.
In an adiabatic process no heat enters or leaves the gas. So when the gas expands, it does work using its own internal energy, and its temperature falls. A drop in temperature pulls the pressure down further than in an isothermal case where temperature stays fixed. That extra pressure drop during expansion (or extra pressure rise during compression) is what makes the adiabatic curve steeper.
Work done equals the area under the P-V curve. Starting from the same point and expanding to the same volume, the adiabatic curve lies below the isothermal, so the isothermal does more work. But in compression to the same smaller volume, the adiabatic rises above the isothermal, so the adiabatic needs more work done on it. Reading the graph correctly is the key to these NEET questions.
Yes. A polytropic process follows PV^n = constant, where n is any number. Its slope is -n P/V. Isothermal is the special case n = 1 and adiabatic is n = gamma. Isobaric (constant pressure) is n = 0 and isochoric (constant volume) is n = infinity. So all the standard processes are just different values of n, and a bigger n always means a steeper curve on the P-V graph.
An ideal gas undergoes four different processes from the same initial state. The processes are adiabatic, isothermal, isobaric and isochoric. Curve 1 is vertical, curve 4 is horizontal, and curves 2 and 3 are falling curves. Which curve represents the adiabatic process?
A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then:
Reversible expansion of an ideal gas under isothermal and adiabatic conditions is shown, where AB is the isothermal expansion and AC is the adiabatic expansion, both from common state A. Which option is NOT correct?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes, at any common point the adiabatic slope is gamma times the isothermal slope, and gamma is always greater than 1 for a real gas. So the adiabatic is always steeper on a P-V graph, both in expansion and in compression.
The ratio of adiabatic slope to isothermal slope at the same point is exactly gamma, the ratio of specific heats Cp/Cv. For a monatomic gas gamma = 1.67, for a diatomic gas gamma = 1.4. So the adiabatic curve is that many times steeper at that point.
Yes. A larger gamma means a larger factor multiplying the slope. A monatomic gas (gamma = 1.67) gives a steeper adiabatic than a diatomic gas (gamma = 1.4) at the same point on the graph.
They can be drawn to start from the same initial state (a common point). After that, on expansion the adiabatic falls below the isothermal, and on compression the adiabatic rises above it, because the adiabatic is steeper.
Work done equals the area under the P-V curve. Because the two curves separate after the common point, the areas differ. In expansion the isothermal does more work; in compression the adiabatic needs more work done on it. NEET questions often test exactly this comparison.
