Physics · Thermodynamics · NEET
Constant temperature does not mean no heat. When the gas expands it does work and would normally cool down. To keep T fixed, heat flows IN from the reservoir at exactly the rate the gas loses energy by doing work. So temperature stays the same while heat keeps moving. In an isothermal expansion, Q > 0 (heat in). In an isothermal compression, Q < 0 (heat out).
Only the change is zero: dU = 0. The gas still has internal energy (it is a hot gas). For an ideal gas, U depends only on temperature. Since T does not change in an isothermal process, U does not change either. So dU = 0, NOT U = 0. This is the single most tested point.
Start from the ideal gas equation PV = nRT. Here n (moles) and R are fixed, and T is held constant. So the whole right side nRT is a constant number. That forces PV = constant. This is exactly Boyle's Law: pressure varies inversely with volume at fixed temperature.
Yes. The First Law says Q = dU + W. Since dU = 0 for an isothermal process, we get Q = W. Every joule of heat added leaves the gas as work done by the gas. So heat in fully becomes work out (expansion), or work in fully becomes heat out (compression).
To keep the temperature truly equal to the reservoir at every step, the gas must be in thermal equilibrium at all times. That only happens if the change is very slow, so heat has time to flow in or out. A fast (sudden) expansion would let the gas cool before heat arrives, so T would not stay constant. That is why the ideal isothermal process is quasi-static.
It is a smooth curve called an isotherm, of the form P = constant / V (a rectangular hyperbola). It falls from high P, low V to low P, high V. A higher temperature gives an isotherm that sits farther from the origin. The isothermal curve is LESS steep than the adiabatic curve through the same point.
An ideal gas expands isothermally from 10^-3 m^3 to 10^-2 m^3 at 300 K against a constant external pressure of 10^5 N/m^2. The work done ON the gas is
In which of the following processes, heat is neither absorbed nor released by a system?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is a change in a gas where the temperature stays the same from start to end. The gas is kept inside a large heat bath (reservoir), so heat can move in or out to hold T fixed. Example: slow expansion of gas in a thin metal cylinder placed in a big water bath.
For a reversible isothermal process, W = nRT ln(V2/V1), which can also be written as W = 2.303 nRT log(V2/V1) or W = nRT ln(P1/P2). For expansion V2 > V1 so W is positive; for compression W is negative.
For an ideal gas, internal energy U depends only on temperature. In an isothermal process the temperature does not change, so U does not change. Therefore dU = 0, which then makes Q = W by the First Law.
Isothermal: temperature constant, heat flows (Q = W), PV = constant, curve is less steep. Adiabatic: no heat flow (Q = 0), temperature changes, PV^gamma = constant, curve is steeper. The adiabatic line is always steeper than the isothermal line through the same point.
Boyle's Law (PV = constant at fixed T) is exactly the relation followed by a gas during an isothermal process. So an isothermal change of an ideal gas obeys Boyle's Law.
