Chemistry · Thermodynamics · NEET
ΔU is the change in internal energy of the system (final minus initial). q is the heat exchanged: positive when heat flows INTO the system, negative when heat leaves. w is the work: positive when work is done ON the system (gas gets compressed), negative when the system does work (gas expands). So ΔU just adds up the two ways energy can enter or leave: heat and work.
Both exist, but they come from different sign rules. NEET and NCERT use the IUPAC convention where work done ON the system is positive, giving ΔU = q + w. The old physics book form ΔU = q − w uses work done BY the system as positive. Same physics, different bookkeeping. For NEET chemistry, always use ΔU = q + w with w = −p(ext)ΔV for expansion.
Think about what happens to the gas. If the gas is COMPRESSED (volume goes down), the surroundings push on it, so work is done ON the gas: w is positive, energy goes in. If the gas EXPANDS (volume goes up), the gas pushes out and spends energy, so work is done BY the gas: w is negative, energy goes out. For expansion, w = −p(ext)ΔV, and ΔV is positive, so w comes out negative.
q is heat. If the system ABSORBS heat (gets warmer from outside, endothermic), q is positive. If the system RELEASES heat (loses heat to surroundings, exothermic), q is negative. A quick check: 'heat in, q is plus; heat out, q is minus.' In a well-insulated or adiabatic container, no heat can move, so q = 0.
An insulated (adiabatic) container blocks heat flow, so q = 0. Putting q = 0 into ΔU = q + w gives ΔU = w. This is a favourite NEET setup: the only way energy changes is through work. If the gas expands against pressure, w is negative, so ΔU is negative and the gas cools down.
Yes. ΔU depends only on the start and end states, not the path taken, so it is a state function. But q and w each depend on the path. The surprise is that their SUM (q + w) is always path-independent and equals ΔU. So two different paths can have different q and different w, yet the same ΔU.
At a certain temperature T (K), during a process, 500 J is absorbed by the system and work of 200 J is done by the system. Then the change in internal energy of the system 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:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Energy cannot be created or destroyed, only changed from one form to another. For a system, the total energy of the system plus its surroundings stays constant. The maths form is ΔU = q + w.
q (heat) and w (work) are two ways energy enters or leaves a system; both depend on the path. ΔU is the total change in internal energy and depends only on the start and end states. ΔU = q + w always holds.
For an ideal gas, internal energy depends only on temperature. In an isothermal process the temperature does not change, so ΔU = 0. Then the first law gives q = −w: any heat absorbed is fully used to do work.
The joule (J) in SI units. In NEET numericals you may also meet litre-atmosphere (L·atm) or calories; 1 L·atm ≈ 101.3 J and 1 cal ≈ 4.184 J.
No. The first law only balances energy; it never says which direction a change will go. Spontaneity is decided by the second law using entropy and Gibbs free energy.
