How this instrument works
Enthalpy is internal energy with one addition: the product of pressure and volume, H = U + PV. U is the energy locked in a system's molecular motion and bonds; the PV term is not extra energy the system contains, but the work the system would do, or have done to it, pushing back the surrounding pressure to occupy its volume. Add the two together and the result tracks how much energy a system trades with its surroundings when pressure, not volume, is held fixed.
The shape of the formula comes straight from the first law. At constant pressure, the work done by an expanding system is W = PΔV, so the heat added, Q, splits as Q = ΔU + PΔV. Because P does not change, PΔV is just Δ(PV), so Q = Δ(U + PV) = ΔH. That single substitution is the entire reason chemists and engineers bother defining H at all: it turns a heat measurement taken in an open beaker into a state-function difference you can add, invert, and look up in a table, rather than a path-dependent number you recompute for every process.
The correction is small for anything dense. A mole of liquid water occupies about 18 cubic centimetres, so its PV term at atmospheric pressure is negligible next to its internal energy, which is why solution calorimetry often treats ΔU and ΔH as the same figure. Gases are a different story: a mole of steam at 100 degrees Celsius occupies roughly 1,600 times the volume of the liquid it condensed from, so its PV term is not a rounding error. That is exactly why power-plant engineers reading steam tables track enthalpy rather than internal energy — a turbine is a flow process, and the PV term is the flow work the steam performs just by crossing the boundary, before it turns a single blade.
- Enter the system's internal energy in the Internal energy field — kilojoules by default, with plain joules also on the unit menu.
- Enter the surrounding absolute pressure in the Pressure field, choosing kilopascals, pascals, or atmospheres.
- Enter the volume the system occupies in the Volume field, in litres, millilitres, or cubic metres.
- Read the result: the Enthalpy field returns H = U + PV instantly, shown in joules or kilojoules.
Worked example — 500 kJ system at atmospheric pressure
Take a closed system holding 500,000 J of internal energy, sitting at standard atmospheric pressure — 101,325 Pa — while occupying 0.01 cubic metres, or 10 litres. The pressure–volume product is PV = 101,325 × 0.01 = 1,013.25 J. Adding that to the internal energy gives H = 500,000 + 1,013.25 = 501,013.25 J, or 501.01 kJ — the exact figure this instrument returns for those three inputs.
The PV term here is only about 0.2 percent of the internal energy, typical for a system that is mostly liquid or solid, since dense phases occupy so little volume that pushing back the atmosphere costs almost nothing. That is why constant-pressure calorimetry can read the heat flow straight off as ΔH without separately tracking expansion work — the correction only grows significant once a reaction changes the number of moles of gas present, or the system is a gas to begin with.
Questions
Why does enthalpy add PV instead of just using internal energy?
Because at constant pressure, the work a system does expanding against its surroundings is already built into PV, so adding it to U turns an awkward, process-dependent heat exchange into ΔH, a state-function difference. That lets you subtract two enthalpy values and get the exact heat of a constant-pressure process without redoing a work calculation each time.
Why do bomb calorimeters report a different number than the enthalpy of reaction?
A bomb calorimeter is a sealed, rigid steel vessel, so its volume cannot change and no PV work happens — the heat it measures is ΔU directly. Enthalpy of reaction, ΔH, comes from adding the PΔV correction for the same reaction run in open air instead; for reactions that change the moles of gas present, that correction can reach several kilojoules per mole, not a rounding error.
Should the Pressure field use absolute or gauge pressure?
Absolute. The PV term represents real work done against the total surrounding pressure, so a gauge reading, which already has atmospheric pressure subtracted out, would silently drop about 101,325 pascals from the calculation. If your instrument reads gauge pressure, add atmospheric pressure to it before entering the value here.
Why do engineers track enthalpy instead of internal energy for turbines and pumps?
Because those are flow processes: fluid crosses a boundary carrying its own PV as flow work, on top of whatever internal energy it holds. Steady-flow energy balances for turbines, compressors, and heat exchangers are written directly in terms of enthalpy for this reason — steam tables list H, not U, because H is the quantity that actually balances at the inlet and outlet.
Can the enthalpy result come out negative?
Yes, if the internal energy entered is negative, or small enough that it dominates over the positive PV term. Enthalpy has no absolute zero — only differences between two states are physically meaningful — so the sign depends entirely on whatever reference point the internal energy figure was measured against, not on anything this calculator assumes.
What if pressure or volume changes during a process?
This formula gives enthalpy at one specific state, defined by the U, P, and V entered for that instant; it is not a path integral. If pressure varies during a process, compute H at the start and end states separately and take the difference, ΔH = H₂ − H₁, rather than folding a varying pressure into a single calculation here.