Chemistry · Thermodynamics · NEET
In an isothermal process the temperature stays the same (ΔT = 0), so heat can flow in or out to keep T fixed. In an adiabatic process no heat flows at all (q = 0), so the temperature usually changes. Simple line: isothermal keeps T constant BY letting heat move; adiabatic blocks heat, so T moves instead. For an ideal gas, isothermal means ΔU = 0, while adiabatic means ΔU = w.
Adiabatic means the system is thermally insulated from its surroundings. No heat can pass through the boundary, so q = 0 by definition. NEET clue words for adiabatic: 'well insulated container', 'thermally isolated', or 'no heat exchange'. Whenever you see those, immediately write q = 0 and use ΔU = q + w = w.
For an ideal gas the internal energy depends only on temperature: ΔU = nCᵥΔT. In an isothermal process ΔT = 0, so ΔU = 0. This is a very common NEET fact. Careful: this is true only for an ideal gas, and only because T is constant, not because heat is zero. In fact isothermal expansion still exchanges heat (q = −w).
Isobaric = constant pressure (P fixed); the gas can still change its volume, so it does P-V work: w = −PΔV. Isochoric (also called isometric) = constant volume (V fixed); since ΔV = 0, the work is w = −PΔV = 0. So in isochoric the whole heat goes into internal energy: qᵥ = ΔU. In isobaric, qₚ = ΔH (heat at constant pressure equals enthalpy change).
P-V work is zero in two important cases. First, an isochoric (constant volume) process: ΔV = 0, so w = −PΔV = 0. Second, free expansion into a vacuum: the external pressure is zero, so w = −pₑₓₜΔV = 0 even though the volume changes. NEET loves the free-expansion case (see the 2020 PYQ below) where q = 0, w = 0, and ΔT = 0 all at once.
No. This is the most common mix-up. In adiabatic expansion the gas does work using its own internal energy (because no heat comes in), so it cools down and T falls. In adiabatic compression T rises. Only the ISOTHERMAL process keeps T constant. Trap: students see 'adiabatic' and wrongly set ΔT = 0; that is wrong unless it is also a free expansion of an ideal gas.
The correct option for free expansion of an ideal gas under adiabatic condition is:
A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ΔU of the gas in joules will be:
Reversible expansion of an ideal gas under isothermal (A→B) and adiabatic (A→C) conditions is shown on a p–V plot. Which of the following options is NOT correct?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Isothermal (constant temperature, ΔT = 0), adiabatic (no heat, q = 0), isobaric (constant pressure, P fixed) and isochoric (constant volume, V fixed, so w = 0). These four names appear directly in NEET questions, so learn what stays constant in each.
Isothermal: ΔT = 0, and for an ideal gas ΔU = 0. Adiabatic: q = 0. Isobaric: ΔP = 0. Isochoric: ΔV = 0, so w = 0. Writing these down first turns a hard problem into a one-line first-law calculation.
No. Isothermal expansion happens slowly against a real external pressure, so it does work and exchanges heat (q = −w). Free expansion is into a vacuum, so external pressure = 0, w = 0, and if adiabatic, q = 0 and ΔT = 0 too. They only share ΔU = 0 for an ideal gas.
No heat can enter (q = 0), but the gas still does work pushing outward. That work energy must come from the gas's own internal energy, so U falls. Since ΔU = nCᵥΔT for an ideal gas, a drop in U means a drop in T. That is why adiabatic expansion cools the gas.
The process type tells you which term drops out of ΔU = q + w. Adiabatic → q = 0 → ΔU = w. Isochoric → w = 0 → ΔU = q. Isothermal (ideal gas) → ΔU = 0 → q = −w. Isobaric → q = ΔH. Recognising the process first is the key NEET skill, and it leads straight into internal energy (U).
