Physics · Thermodynamics · NEET
An adiabatic process is any change where the gas exchanges no heat with its surroundings, so Q = 0 by definition. This happens when the gas is inside insulating (adiabatic) walls, OR when the process is so fast that heat has no time to flow in or out (like a sudden compression). The adiabatic wall is just one way to achieve it. Do not confuse the two: the wall is the insulating barrier, the process is the actual pressure-volume change that occurs while Q = 0.
PV = constant (Boyle's law) only holds when temperature is fixed, which needs heat to flow in or out. In an adiabatic process no heat can flow, so temperature is NOT fixed. Starting from the first law with Q = 0 and using dU = nCv dT for an ideal gas, you can show P V^gamma = constant, where gamma = Cp/Cv is greater than 1. Since gamma > 1, this curve falls faster than the PV = constant curve. Related equations: T V^(gamma-1) = constant and P^(1-gamma) T^gamma = constant.
No. That is the isothermal process, not adiabatic. In an adiabatic process temperature changes. During adiabatic expansion the gas does work using its own internal energy, so temperature falls (that is why an aerosol can feels cold when sprayed). During adiabatic compression, work is done on the gas, so temperature rises (diesel engines ignite fuel this way). Only U depends on T, and here U changes, so T changes.
On a P-V diagram the slope of an isothermal curve is proportional to -P/V, while the slope of an adiabatic curve is proportional to -gamma P/V. Because gamma > 1, the adiabatic slope is steeper (more negative) at the same point. So from a common starting point, the adiabatic curve drops down faster. This is the single most-tested visual fact for NEET: steeper falling curve = adiabatic, gentler falling curve = isothermal.
A very fast (sudden) process is approximately adiabatic because heat has little time to flow, so Q is close to 0. A very slow (quasi-static) process inside insulating walls can also be adiabatic. So speed is not the definition; the definition is Q = 0. Speed is just a common practical way to make Q approximately 0.
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]
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 1, 2, 3, 4 is (curve 1 vertical, curve 4 horizontal, curves 2 and 3 falling with 2 steeper):
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:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is a process where the gas exchanges no heat with its surroundings (Q = 0). The gas is either insulated or the change is so fast that heat cannot flow. Any work done then changes the gas temperature.
The main relation is PV^gamma = constant, where gamma = Cp/Cv. Two more useful forms are T V^(gamma-1) = constant and P^(1-gamma) T^gamma = constant.
Yes. Since Q = 0, the first law gives delta U = -W. In expansion the gas does work, U falls and temperature drops. In compression work is done on the gas, U rises and temperature increases.
The adiabatic curve is steeper. Its slope is proportional to -gamma P/V while the isothermal slope is proportional to -P/V, and gamma > 1.
Spraying a deodorant can (rapid expansion, gas cools), and compression of air in a diesel engine (rapid compression, air heats up enough to ignite fuel).
