Solving P₁/T₁ = P₂/T₂ tells you the new pressure or temperature, but it doesn't say
anything about the energy involved in getting there. That's what the "Advanced: heat
transfer" section above adds. Once the core four values resolve and you enter the
amount of gas n in moles, the calculator applies:
Formula
Q = n × Cv × (T₂ − T₁)
with Cv = (5/2)R ≈ 20.79 J/(mol·K), assuming a diatomic ideal
gas such as air or nitrogen — the standard textbook assumption for this kind of
problem, and the one used throughout this calculator and its worked examples below.
The reason the calculator can stop there, without a separate work term, comes from how
work is defined for an expanding or compressing gas: W = P ΔV.
At constant volume, ΔV = 0, so W = 0 —
no matter how large the pressure or temperature swing is, the gas does no mechanical
work and none is done on it. The first law of thermodynamics, ΔU = Q − W, then collapses to something very clean:
Key result
ΔU = Q — every joule of heat added at constant volume goes
directly into the gas's internal energy. None of it is "spent" pushing back a piston
or expanding a boundary, because there's no boundary movement to spend it on.
It's worth contrasting this with the isobaric case, where pressure — not volume — is
held constant. There, a heated gas is free to expand, so some of the heat you add does
go into work as the gas pushes its surroundings back; only the remainder raises the
internal energy. The isochoric case is the tidy opposite: with the "escape valve" of
expansion removed, heat has nowhere to go but into internal energy, exactly as it does
on the isothermal page's own clean fact that ΔU = 0 forces Q = W. Isochoric and
isothermal end up being mirror-image "one variable does all the work" cases of the first
law.