Chemistry · Thermodynamics · NEET
POSITIVE. In the new IUPAC (chemistry) convention, work done ON the system is +w. Compression means the surroundings push in and do work on the gas, so w is positive and the internal energy rises. In expansion the gas pushes out and does work, so w is negative. Rule of thumb: compression = +w, expansion = −w in chemistry.
They just define 'work' oppositely. Chemistry (IUPAC): w = work done ON the gas, so adding energy = q + w. Physics: W = work done BY the gas, which removes energy from the gas, so ΔU = q − W. Note w = −W (they are negatives of each other). The stored energy result is identical; only the bookkeeping of the sign differs. For NEET use the chemistry/IUPAC form ΔU = q + w.
w = −p_ext ΔV = −p_ext(V_f − V_i). The minus sign is what makes the convention 'new'. On expansion V_f > V_i, so ΔV is positive and w is negative (gas did work, lost energy). On compression V_f < V_i, so ΔV is negative and w becomes positive (work done on gas, energy gained). For a reversible isothermal ideal gas: w = −nRT ln(V_f/V_i).
No — heat is the same in both books. Heat added TO the system is positive (+q), heat released BY the system is negative (−q). Only WORK changed between old and new conventions. So the only thing you must be careful about when switching between physics and chemistry is the sign of work.
NEET follows NCERT chemistry, which uses the NEW IUPAC convention: ΔU = q + w with work done ON the system positive. NCERT itself says the old sign (work by system positive) 'is still followed in physics books, although IUPAC has recommended the use of new sign convention.' In a chemistry NEET question, always use ΔU = q + w.
The correct option for free expansion of an ideal gas under adiabatic condition is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The final internal energy will be correct if you are consistent, but the numerical sign of w you report may not match the given options. NEET options follow chemistry (IUPAC) signs, so always use ΔU = q + w with work done on the gas positive to match the answer key.
They are exact negatives: w = −W. Chemistry w is work done on the gas; physics W is work done by the gas. That single sign flip is the whole 'old vs new' difference.
Negative. The gas pushes the surroundings and gives out energy, so w = −p_ext·ΔV is negative because ΔV is positive. Its internal energy tends to drop unless heat is supplied.
To make the first law read as a simple sum, ΔU = q + w, where BOTH energy inputs (heat added and work done on the system) carry a plus sign. It is more consistent: anything that adds energy to the system is positive.
