Chemistry · Thermodynamics · NEET
At constant volume the gas cannot expand, so ΔV = 0 and no pV work is done. The first law ΔU = q − pΔV then gives qv = ΔU. At constant pressure the gas CAN expand, so some heat leaves as work (pΔV). That heat is not lost from the accounting — we roll U and the pV term into one new function H = U + pV, and the heat you added equals its change: qp = ΔH. So the difference is simply whether expansion work happens or not.
Start with the first law at constant pressure: ΔU = qp − pΔV. Write it as U2 − U1 = qp − p(V2 − V1). Rearrange to group things: qp = (U2 + pV2) − (U1 + pV1). Now define enthalpy H = U + pV. Then U2 + pV2 = H2 and U1 + pV1 = H1. So qp = H2 − H1 = ΔH. That is the full NCERT derivation — no extra assumptions except constant pressure.
Most lab reactions happen in open beakers, so pressure is constant (atmospheric), not volume. Under those conditions the useful measured quantity is the heat at constant pressure, qp. Instead of always writing qp = ΔU + pΔV, chemists defined ONE state function H = U + pV so that qp = ΔH directly. It is just a shortcut that makes constant-pressure heat easy to track.
Heat q by itself IS path dependent. But at constant pressure, qp happens to equal ΔH. Since H = U + pV is built only from state functions (U, p, V), ΔH is path independent. So qp inherits that: under the fixed condition of constant pressure, qp is path independent. Remove the constant-pressure condition and this no longer holds.
Yes. The derivation never assumed a gas. qp = ΔH holds for any system at constant pressure. For solids and liquids the volume change ΔV is tiny, so ΔH is almost equal to ΔU, but the equation qp = ΔH is still exactly true at constant pressure.
A negative ΔH means qp is negative, so the system releases heat to the surroundings — the reaction is exothermic. A positive ΔH means the system absorbs heat (qp positive) — endothermic. So the sign of ΔH at constant pressure directly tells you if heat goes out or in.
Which amongst the following options is the correct relation between change in enthalpy and change in internal energy?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
qp is the heat absorbed or released by a system when the pressure is kept constant. At constant pressure it equals the enthalpy change: qp = ΔH.
It is true whenever the pressure is constant, for any system (gas, liquid or solid). If pressure is not constant, qp = ΔH does not apply.
qv = ΔU applies at constant volume (no expansion work). qp = ΔH applies at constant pressure (expansion work allowed, absorbed into H = U + pV).
So that the heat measured in open, constant-pressure conditions (qp) equals a single state function change ΔH, making calculations simpler.
It depends on Δn_g. Since ΔH = ΔU + Δn_g RT, if moles of gas increase (Δn_g > 0) then ΔH > ΔU; if gas moles decrease, ΔH < ΔU.
