Audited 31 Jul 2026·Last updated 28 Oct 2026·5 citations·Tier 1·0 uses

Enthalpy Calculator

Compute enthalpy H = U + pV from a system's internal energy in joules, its absolute pressure in pascals and its volume in cubic metres.

Enthalpy Calculator

J
Pa
m³
Enthalpy
7,026.5
Total enthalpy of the system in joules: internal energy plus pressure times volume, on the same reference as the U you entered. Divide by 1,000 for kilojoules.
Model scope
State-function identity for entered extensive U and V at pressure p; it does not infer properties, phase equilibrium, reaction enthalpy, reference state, flow-work sign convention, or changes between states.

Background.

Internal energy is the energy stored inside a system — the motion of its molecules and the bonds between them. But a system sitting in the open air also occupies space, and making that space meant pushing the surroundings out of the way. Enthalpy adds that cost to the stored energy: H = U + pV, where p is the pressure the system sits at and V the volume it takes up.

The pV term is not small for gases. Twenty litres of gas at one standard atmosphere, 101,325 pascals, has pushed back the atmosphere with 2,026.5 joules of work. For a solid or liquid the term is tiny: a mole of liquid water, about 18 millilitres, carries less than 2 joules of pV at the same pressure, so its enthalpy and internal energy are almost the same number.

Chemists and engineers prefer enthalpy because of what happens when a process runs at constant pressure, as most bench reactions and many flow processes do. The heat taken in or given out then equals the change in enthalpy, not the change in internal energy, because part of any heat supplied goes into expansion work against the surroundings. That is why tables list heats of reaction, combustion and vaporisation as ΔH values, and why steam tables list specific enthalpy for every state. Air-conditioning engineers lean on it too: the cooling a coil must deliver is the drop in enthalpy of the air passing over it, read off a psychrometric chart in kilojoules per kilogram of dry air, and a large share of that drop on a humid day is water vapour condensing rather than air getting colder.

Signs follow the system. A negative ΔH means the system handed heat to its surroundings, as every combustion does; a positive one means it absorbed heat, as melting ice or the ammonium nitrate dissolving in an instant cold pack does. Tables of standard enthalpies of formation rest on one more choice: each element in its most stable form at 1 bar is assigned zero, so a compound's value is the enthalpy change of making it from its elements, and a reaction's ΔH is the products' total minus the reactants'.

The arithmetic here is just that one identity, applied to a single state. Absolute internal energy cannot be measured, only changes in it, so U is whatever value you have relative to your chosen reference, and H inherits the same reference. To find an enthalpy change, work out H for the start and end states and subtract.

What is enthalpy calculator?

Enthalpy is a thermodynamic state function defined as internal energy plus the product of pressure and volume, H = U + pV. Like internal energy it is measured in joules and depends only on the current state of the system, not on how it got there. Its practical meaning comes from its change: for a process at constant pressure with only expansion work, ΔH equals the heat exchanged with the surroundings. Enthalpy is extensive: twice the material in the same state has twice the H. Tables therefore quote it per unit amount — kilojoules per mole in chemistry, kilojoules per kilogram in steam tables and engineering, and in older American references Btu per pound, where 1 Btu/lb is exactly 2.326 kJ/kg. For an ideal gas pV equals nRT, so H = U + nRT and both depend on temperature alone; for real gases, liquids and solids, pressure enters as well. Reaction enthalpies also shift with temperature, at a rate set by the difference in heat capacity between products and reactants, which is why tables state the temperature they apply to, usually 298.15 K.

How to use this calculator.

  1. Enter the internal energy U of the system in joules, measured from the same reference state as any table you are using. Steam tables, for instance, set the internal energy of liquid water at its triple point, 0.01 °C, to zero, so their values only mean something against each other. The page accepts only positive values, so a state below its reference — ice, on the steam-table scale — has to be added up by hand.
  2. Enter the absolute pressure in pascals. Multiply kilopascals by 1,000 and bar by 100,000; one standard atmosphere is 101,325 Pa. Gauge pressure readings need atmospheric pressure added first.
  3. Enter the volume the system occupies in cubic metres. Divide litres by 1,000, so 20 L is 0.02 m³.
  4. Read the enthalpy in joules. Because U and V here are totals for the whole system, so is H; for a per-kilogram or per-mole value, enter specific or molar U and V instead.

The formula.

H = U + pV

H = U + p × V. U is the internal energy in joules, p the absolute pressure in pascals and V the volume in cubic metres. The units fit without any conversion factor because one pascal is one newton per square metre, so pascals times cubic metres gives newton-metres, which are joules. The pV product is the work needed to create room for the system at pressure p against its surroundings. For gases at everyday pressures it can be a sizeable share of H; for liquids and solids it is usually negligible, and H is close to U. Doubling either the pressure or the volume doubles the pV contribution while leaving U unchanged. The page combines U, p and V for one state; it does not look up property data or compute a change between states.

Why the heat at constant pressure equals ΔH follows in two lines from the first law. With only expansion work, heat q added to a system raises its internal energy and pays for the expansion: q = ΔU + pΔV when p is constant. Because p does not change, pΔV is the same as Δ(pV), so q = Δ(U + pV) = ΔH. The definition holds for any substance; it is the shortcut that needs those conditions. If the pressure changes during the process, or the system does other work, such as a battery driving a current, the heat no longer equals ΔH. The same bookkeeping explains the gap between an ideal gas's two molar heat capacities: since H = U + nRT, warming a mole by one kelvin at constant pressure costs R, about 8.314 J, more than at constant volume.

A worked example.

Example

A gas sample has an internal energy of 5,000 J and occupies 0.02 m³ (20 litres) at 101,325 Pa, one standard atmosphere. The pressure-volume term is 101,325 × 0.02 = 2,026.5 J. Adding it to the internal energy gives H = 5,000 + 2,026.5 = 7,026.5 J. The pV share is about 29% of the total, a reminder of how much room a gas takes up. If the gas behaved ideally at 25 °C, those 20 litres at one atmosphere would hold about 0.8175 mol, since pV = nRT. Swap in 18 millilitres of liquid water instead and the pV term at the same pressure shrinks to about 1.824 J, which is why enthalpy and internal energy are treated as nearly equal for liquids. Warm the same gas at constant pressure until it fills 25 litres and the pV term grows by 101,325 × 0.005 = 506.625 J. That much of the heat supplied went into pushing back the atmosphere rather than into the gas itself, which is exactly the difference between ΔH and ΔU for the warming.

pressure Pa101,325
volume M30.02
internal Energy J5,000

Frequently asked questions.

What is the difference between enthalpy and internal energy?
Internal energy counts the energy stored within the system. Enthalpy adds pV, the work done to make room for the system at its pressure. At constant volume the heat exchanged equals the change in internal energy; at constant pressure it equals the change in enthalpy, which is why open-vessel chemistry uses ΔH.
How do I get an enthalpy change rather than an absolute enthalpy?
Calculate H for the initial state and for the final state, each with its own U, p and V, then subtract. At constant pressure this reduces to ΔH = ΔU + pΔV. Because absolute U is never known, only differences between states carry physical meaning, and both states must use the same reference.
Why does the pV term matter so little for liquids and solids?
Condensed phases occupy very little volume per mole. A mole of liquid water fills about 18 millilitres, so at one atmosphere its pV is around 1.824 J, small beside the energy involved in melting, boiling or reacting it, which runs to kilojoules per mole. At 25 °C a mole of gas at the same pressure fills over a thousand times as much space.
Should I use gauge or absolute pressure?
Absolute pressure. The pV term represents work against the actual pressure acting on the system, which includes the atmosphere. A tyre gauge reading of 200 kPa corresponds to roughly 300 kPa absolute at sea level, and entering the gauge value would understate the pV contribution by a third. Vacuum work makes the point sharper: a gauge reading of −80 kPa is about 21.3 kPa absolute, and that absolute figure is the one that belongs in the pV term.
How do litre-atmospheres convert to joules?
One litre-atmosphere is 0.001 m³ × 101,325 Pa = 101.325 J. Older chemistry texts often express pV work in litre-atmospheres, so multiply by 101.325 to get joules before comparing with energy values. Twenty litre-atmospheres, as in the default example, is 2,026.5 J.
How do I turn a bomb-calorimeter ΔU into ΔH?
A bomb calorimeter is sealed, so the heat it measures is released at constant volume and equals ΔU. For a reaction whose gases behave ideally at temperature T, ΔH = ΔU + Δn_gas × R × T, where Δn_gas is the moles of gaseous products minus the moles of gaseous reactants. Burning methane to liquid water, CH₄ + 2O₂ → CO₂ + 2H₂O, has Δn_gas = −2, so at 298.15 K ΔH is about 4.958 kJ per mole of methane more negative than ΔU.
Why do turbines, boilers and throttling valves work in enthalpy rather than internal energy?
Fluid flowing through a device is pushed in by the fluid behind it and pushes out the fluid ahead, and that flow work amounts to p × v per kilogram. Enthalpy already includes it, so a steady-flow energy balance reduces to heat added minus shaft work, per kilogram, equals the change in h, ignoring changes in speed and height. An insulated turbine's work per kilogram is simply the enthalpy drop across it, and a throttling valve, with no heat or work, leaves enthalpy unchanged.
Does raising the pressure always raise the enthalpy?
Not necessarily. Compress an ideal gas while holding its temperature constant and the volume halves as the pressure doubles, so pV stays put; U, which for an ideal gas depends only on temperature, stays put too, and H is unchanged: 202,650 Pa × 0.01 m³ is still 2,026.5 J. Doubling the pressure on the page while leaving U and V alone describes a different state, not a compression of the same gas.
How much of the heat that boils water goes into pushing back the atmosphere?
A small but real share. Water's enthalpy of vaporisation at 100 °C and one atmosphere is about 40.65 kJ per mole. Treating the steam as an ideal gas and neglecting the liquid's own small volume, the expansion work pΔV is close to RT, 8.314 × 373.15 ≈ 3.103 kJ per mole. So about 37.55 kJ of each mole's 40.65 goes into pulling the molecules apart, and roughly 7.6% into making room for the vapour.
Does a negative ΔH mean a reaction will happen by itself?
No. Whether a change runs on its own at constant temperature and pressure is decided by the Gibbs energy change, ΔG = ΔH − TΔS, which weighs the heat released against the change in entropy. Ice melting in a 10 °C room absorbs heat, so its ΔH is positive, yet it happens unaided because the entropy gain outweighs it; an instant cold pack works the same way. Enthalpy tells you how much heat to expect, not whether the change will go.

How this page was produced

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Method
H = U + pV
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