Chemistry · Thermodynamics · NEET
It is NEGATIVE. In the IUPAC convention used in NCERT chemistry, work done BY the system (for example, a gas pushing out and expanding) means the system loses energy, so w is negative (w < 0). Only work done ON the system is positive. Remember: the system spends its own energy to do work on the surroundings, so its internal energy drops.
When heat is absorbed (goes INTO the system from the surroundings), q is POSITIVE (+q). The system gains energy, so its internal energy rises. When heat is released (goes OUT of the system to the surroundings), q is NEGATIVE (-q). Simple test: is energy entering or leaving the system?
When a gas expands, it pushes the surroundings outward. The system is doing work ON the surroundings, so it loses energy. Losing energy means the work term is negative: w = -p(ext) x ΔV. If the gas is compressed instead, work is done ON the system, energy enters, and w is positive.
Old physics books wrote the first law as ΔU = q - W, where W is work done BY the system (positive when the gas expands). NCERT chemistry uses the IUPAC rule ΔU = q + w, where w is work done ON the system (positive when the gas is compressed). Both are correct; they just define the sign of work from a different side. For NEET, always use the IUPAC/NCERT rule: ΔU = q + w.
Ask two simple questions. (1) For heat: is heat absorbed by the system? Then +q. Is heat released? Then -q. (2) For work: is work done on the system (compression)? Then +w. Is work done by the system (expansion)? Then -w. After fixing the signs, plug into ΔU = q + w.
In an adiabatic process there is no heat exchange, so q = 0. The first law becomes ΔU = w. So the change in internal energy equals only the work term. This is why insulated-container questions in NEET set q = 0 and use work alone.
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 J) is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Energy entering the system is positive; energy leaving is negative. So heat absorbed = +q, heat released = -q, work done on the system = +w, work done by the system = -w.
Use ΔU = q + w (the IUPAC/NCERT form). Here w is work done ON the system. Do not use ΔU = q - W from old physics books, or you may double-flip the sign.
Yes. Work done ON the system adds energy, so w is positive and internal energy U increases. A common example is compressing a gas.
Positive. Compression means the surroundings do work ON the gas, energy enters the system, so w > 0.
Almost every thermodynamics numerical (first law, ΔU, pressure-volume work, adiabatic problems) starts by fixing the signs of q and w. One wrong sign gives a wrong final answer, and NTA options are designed to catch exactly that mistake.
