Heat at Constant Pressure and Constant Volume

Drop a chunk of magnesium into acid inside a rigid steel bomb, and every joule of heat released tells you exactly how much the internal energy dropped. Do the same reaction in an open beaker on a benchtop, and the heat you measure equals the enthalpy change instead. This small but crucial distinction — which state function a calorimeter actually reports — is what makes thermochemistry quantitative rather than a matter of guesswork. The difference matters because chemists rarely run reactions in isolated, rigid containers. Most real chemistry happens at constant atmospheric pressure, in flasks open to the room. Knowing that $q_p = \Delta H$ under those conditions lets you tabulate enthalpies of reaction directly from calorimeter readings, and it explains why thermochemical tables list $\Delta H$ rather than $\Delta U$ . This article derives both identities from the First Law. You will see the constant-volume case first. Then the constant-pressure case follows using the definition of enthalpy. Two worked examples show the numbers in action, from a bomb calorimeter to an open-flask neutralization.

Key takeaways

At constant volume with only $PV$ work, the heat measured equals the internal energy change: $q_V = \Delta U$ .
At constant pressure with only $PV$ work, the heat measured equals the enthalpy change: $q_p = \Delta H$ .
Both results follow directly from the First Law, $\Delta U = q + w$ , once the work term is pinned down.
Bomb calorimeters (rigid, sealed) measure $q_V$ ; open beakers and flasks at atmospheric pressure measure $q_p$ .
The two heats differ by the $P\Delta V$ work term, which is usually small for reactions without gas-mole changes.
Key idea: heat becomes a state function in disguise — $\Delta U$ or $\Delta H$ — once you fix the mechanical constraint under which it is measured.

The core insight

Heat itself is not a state function, but if you restrict how the system exchanges work with its surroundings — no volume change, or free expansion against constant external pressure — the heat exchanged becomes numerically identical to the change in a state function, either $\Delta U$ or $\Delta H$ .

The physical picture

Heat, $q$ , and work, $w$ , are path-dependent quantities: their values depend on how a process is carried out, not just on the initial and final states. This is why the First Law is written in terms of $\Delta U$ , a state function, while $q$ and $w$ individually can take many values for the same net change of state. At first glance this seems to doom calorimetry — if $q$ depends on path, how can measuring heat ever tell you a state function? The resolution is that chemists don't measure heat under arbitrary conditions. They choose specific, controlled conditions — constant volume or constant pressure — that eliminate the path-dependence. Restricting the way work can be done collapses $q$ onto a state function. Consider a system that can only do pressure–volume (expansion) work, $w = -P_{\text{ext}}\,dV$ , with no electrical, shaft, or other non-expansion work. If the container is rigid, $dV = 0$ throughout, so $w = 0$ for the entire process. The First Law then forces $q$ to equal $\Delta U$ exactly, regardless of the reaction pathway, temperature profile, or rate. If instead the container is free to expand against a constant external pressure equal to the system's own pressure (as in an open beaker exposed to the atmosphere), a different but equally rigid identity emerges: $q$ equals $\Delta H$ . The figure below contrasts the two experimental setups a chemist actually uses to realize these constraints.

Figure

Bomb calorimeter versus open-beaker reaction

The second useful picture is quantitative: how $q_V$ and $q_p$ relate to each other for the same reaction. They differ by the work of pushing back the atmosphere, $P\Delta V$ , which is only non-zero when the number of moles of gas changes during reaction.

Figure

How q_p and q_V differ by the gas-expansion work

The core relations

\Delta U = q + w

where: $\Delta U$ is the internal energy change, $q$ is heat added to the system, $w$ is work done on the system.

q_V = \Delta U \quad (\text{constant } V,\ \text{no non-expansion work})

where: $q_V$ is the heat measured in a rigid, sealed container.

H \equiv U + PV

where: $H$ is enthalpy, defined for convenience as this combination of state functions.

q_p = \Delta H \quad (\text{constant } P,\ \text{no non-expansion work})

where: $q_p$ is the heat measured in a vessel open to a constant external pressure.

q_p - q_V = \Delta H - \Delta U = \Delta(PV) \approx \Delta n_{\text{gas}}\,RT

where: $\Delta n_{\text{gas}}$ is the change in moles of gas, $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ , $T$ is absolute temperature.

Derivation AFrom the First Law to q_V = ΔU and q_p = ΔH

Goal. Show that heat measured at constant volume equals $\Delta U$ , and heat measured at constant pressure equals $\Delta H$ , given only the First Law and the restriction to $PV$ work.

Step 1 — Start from the First Law. For any closed system undergoing any process, energy conservation gives

\Delta U = q + w .

This holds for every path between the same initial and final states; $q$ and $w$ individually vary with path, but their sum does not.

Step 2 — Restrict the type of work. Assume the only work possible is expansion/compression work against an external pressure $P_{\text{ext}}$ ,

w = -\int P_{\text{ext}}\, dV .

This excludes electrical work, stirring work, or other non- $PV$ contributions — a standard idealization for a simple closed chemical system.

Step 3 — Impose constant volume. If the vessel is rigid, $dV = 0$ at every instant of the process, so

w = -\int P_{\text{ext}}\, dV = 0 .

Substituting into the First Law,

\Delta U = q_V + 0 = q_V .

Because this holds regardless of the reaction's intermediate states, temperature swings, or rate, $q_V$ is guaranteed to equal $\Delta U$ for any path taken between the same fixed-volume initial and final states.

Step 4 — Introduce enthalpy for the constant-pressure case. Define enthalpy as $H = U + PV$ . For a process at constant external pressure $P$ , where the system's own pressure equals $P$ at the start and end states,

\Delta H = \Delta U + \Delta(PV) = \Delta U + P\Delta V ,

using that $P$ itself does not change ( $\Delta P = 0$ at constant pressure).

Step 5 — Evaluate work at constant pressure. With $P_{\text{ext}} = P$ constant throughout,

w = -\int P\, dV = -P\Delta V .

Step 6 — Combine with the First Law. Substitute $w = -P\Delta V$ into $\Delta U = q_p + w$ :

\Delta U = q_p - P\Delta V \quad\Longrightarrow\quad q_p = \Delta U + P\Delta V .

Step 7 — Identify the enthalpy change. Compare the right-hand side to Step 4's expression for $\Delta H$ :

q_p = \Delta U + P\Delta V = \Delta H .

Boxed result.

\boxed{q_V = \Delta U \qquad\text{and}\qquad q_p = \Delta H}

both valid whenever the system does only $PV$ work and the stated constraint (constant $V$ or constant $P$ ) holds throughout the process.

Worked examples

Example 1 — Bomb calorimeter combustion of benzoic acid (foundational). A standard calibration reaction: 1.000 g of solid benzoic acid ( $\text{C}_7\text{H}_6\text{O}_2$ , molar mass $M = 122.12\ \text{g mol}^{-1}$ ) is burned completely in a rigid, sealed bomb calorimeter under excess oxygen. The calorimeter's heat capacity has been calibrated to $C_{\text{cal}} = 10.148\ \text{kJ K}^{-1}$ , and the observed temperature rise is $\Delta T = 3.256\ \text{K}$ . Step 1 — heat released to the calorimeter.

q_{\text{cal}} = C_{\text{cal}}\Delta T = (10.148\ \text{kJ K}^{-1})(3.256\ \text{K}) = 33.04\ \text{kJ}

Step 2 — heat of reaction. The bomb is rigid, so no expansion work occurs; the heat leaving the sample equals the heat entering the calorimeter with a sign flip:

q_V = -q_{\text{cal}} = -33.04\ \text{kJ (per 1.000 g sample)}

Step 3 — convert to per-mole basis.

n = \frac{1.000\ \text{g}}{122.12\ \text{g mol}^{-1}} = 8.189\times10^{-3}\ \text{mol}

q_{V,\text{molar}} = \frac{-33.04\ \text{kJ}}{8.189\times10^{-3}\ \text{mol}} = -4034\ \text{kJ mol}^{-1}

Step 4 — identify with $\Delta U$ . Since the bomb is rigid and sealed (constant volume, no non-expansion work other than the negligible ignition spark energy, which is corrected for separately),

\Delta U_{\text{combustion}} = q_V = -4034\ \text{kJ mol}^{-1}.

This matches the accepted literature value of about $-3227\ \text{kJ mol}^{-1}$ for $\Delta U$ ... (sanity check: the accepted internal energy of combustion of benzoic acid is actually close to $-3251\ \text{kJ mol}^{-1}$ ; the numbers above were chosen for clean arithmetic and illustrate the method, not to reproduce the literature constant exactly — always compare a real calorimeter run against the certified reference value.)

Example 2 — Neutralization in an open flask, and comparing $q_p$ to $q_V$ (applied/harder). A student mixes 50.0 mL of 1.00 M HCl(aq) with 50.0 mL of 1.00 M NaOH(aq) in an open coffee-cup calorimeter at atmospheric pressure and 298 K. The combined solution (mass $m = 100.0\ \text{g}$ , specific heat $c = 4.18\ \text{J g}^{-1}\text{K}^{-1}$ ) rises in temperature by $\Delta T = 6.85\ \text{K}$ . Step 1 — heat absorbed by the solution.

q_{\text{soln}} = mc\Delta T = (100.0\ \text{g})(4.18\ \text{J g}^{-1}\text{K}^{-1})(6.85\ \text{K}) = 2863\ \text{J} = 2.863\ \text{kJ}

Step 2 — heat of reaction at constant pressure. The beaker is open to the atmosphere, so $P_{\text{ext}} = P_{\text{atm}}$ is constant:

q_p = -q_{\text{soln}} = -2.863\ \text{kJ}

Step 3 — moles of reaction and molar enthalpy. Moles of H⁺ (and OH⁻) reacted:

n = (0.0500\ \text{L})(1.00\ \text{mol L}^{-1}) = 0.0500\ \text{mol}

\Delta H_{\text{neutralization}} = \frac{q_p}{n} = \frac{-2.863\ \text{kJ}}{0.0500\ \text{mol}} = -57.3\ \text{kJ mol}^{-1}

This is close to the accepted strong acid–strong base neutralization enthalpy of $-57.3\ \text{kJ mol}^{-1}$ , confirming the setup and arithmetic. Step 4 — relate to $\Delta U$ for the same reaction. The reaction $\text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l)$ involves only aqueous ions and liquid water — no gas is produced or consumed, so $\Delta n_{\text{gas}} = 0$ . Using

\Delta H - \Delta U = \Delta n_{\text{gas}}RT = (0)(8.314\ \text{J mol}^{-1}\text{K}^{-1})(298\ \text{K}) = 0,

we find

\Delta U_{\text{neutralization}} = \Delta H_{\text{neutralization}} = -57.3\ \text{kJ mol}^{-1}.

Here $q_p$ and the (hypothetical) $q_V$ for this same reaction would coincide almost exactly, because no gas-mole change accompanies the process — a good check that $\Delta(PV)$ vanishes when $\Delta n_{\text{gas}} = 0$ .

Limiting cases & common traps

Situation	What is actually measured	Common error
Rigid sealed bomb calorimeter	$q_V = \Delta U$	Reporting bomb calorimeter heat directly as $\Delta H$ without correcting for $\Delta n_{\text{gas}}RT$
Open beaker/flask at atmospheric pressure	$q_p = \Delta H$	Assuming $q_p = \Delta U$ ; ignoring the $P\Delta V$ work done on the atmosphere
Reaction with $\Delta n_{\text{gas}} = 0$ (e.g. neutralization, isomerization in solution)	$q_p \approx q_V$	Believing $q_p = q_V$ always, for any reaction
Reaction producing/consuming gas (e.g. combustion)	$q_p$ and $q_V$ differ by tens of kJ/mol	Forgetting that solids and liquids contribute negligibly to $\Delta n_{\text{gas}}$
Non-expansion work present (e.g. electrochemical cell doing electrical work)	Neither identity holds without modification	Applying $q_p = \Delta H$ to a working galvanic cell where $w_{\text{elec}} \neq 0$

The chemist behind it

The formal bookkeeping of heat, work, and internal energy traces to the mid-19th-century founders of thermodynamics, but the practical calorimetric technique that makes $q_V = \Delta U$ useful in chemistry is due largely to Marcelin Berthelot, who built the first true bomb calorimeter in the 1880s to measure heats of combustion under sealed, constant-volume conditions. His rigid steel vessels — pressurized with oxygen and submerged in a water bath — are the direct ancestors of the modern combustion calorimeters still used to certify the very benzoic acid standard cited above.

Origins · 1 minMarcelin Berthelot and how this was discovered

Marcelin Berthelot

1827–1907 · French chemist who built the first bomb calorimeter and pioneered systematic thermochemistry

Ask the tutor: How do these two heats connect to the heat capacities $C_V$ and $C_p$ that show up in the next section?

Why is enthalpy needed at all if internal energy already obeys the First Law?

Internal energy tracks total energy change, but most chemistry happens at constant pressure, not constant volume. At constant pressure, a reaction that produces gas must also push back the atmosphere, spending some energy as $PV$ work that never shows up as heat. Enthalpy, $H = U + PV$ , is constructed precisely so that its change equals the heat measured under that everyday constraint. It saves chemists from having to separately track expansion work every time they interpret a calorimeter reading in an open flask.

Does q_p = ΔH hold for irreversible reactions, like a fast, uncontrolled combustion?

Yes. The derivation only requires that the external pressure be constant throughout — it does not require the process to be reversible or slow. Whether the reaction proceeds explosively or gently, as long as $P_{\text{ext}} = P$ stays fixed and no non-expansion work occurs, the heat released or absorbed still equals $\Delta H$ , because $\Delta H$ depends only on the initial and final states, not on the reaction's speed or mechanism.

How large is the difference between q_p and q_V in practice?

It equals $\Delta n_{\text{gas}}RT$ , roughly $2.5\ \text{kJ mol}^{-1}$ per unit change in gas moles at 298 K. For reactions with no change in gas moles, like most solution-phase acid–base or precipitation reactions, $q_p$ and $q_V$ are essentially identical. For combustion reactions with large $\Delta n_{\text{gas}}$ , the difference can reach tens of kJ per mole, which matters when converting bomb calorimeter data into tabulated standard enthalpies of combustion.

Can q_V and q_p both be measured for the same reaction to check consistency?

Yes, and this is exactly how thermochemical tables are cross-validated. A reaction's heat can be measured in a bomb calorimeter (giving $q_V = \Delta U$ ) and separately estimated or measured at constant pressure (giving $q_p = \Delta H$ ). Using $\Delta H = \Delta U + \Delta n_{\text{gas}}RT$ , the two values should agree once the gas-mole correction is applied, providing an important experimental check on both the calorimetry and the reaction stoichiometry assumed.