Heat at Constant Pressure and Volume

Introduction

The heat measured for a reaction depends on experimental constraints. In a rigid constant-volume vessel, no pressure-volume work occurs, so heat can equal the internal-energy change. At constant pressure, expansion or compression work may occur, and heat can equal the enthalpy change. Both equalities require that other work modes and material transfers be handled correctly.

Core explanation

For a closed system under the chemistry convention, ΔU = q + w. If the volume is fixed, ΔV = 0, so pressure-volume work is zero. If no other work occurs, ΔU = q V. A bomb calorimeter approximates this situation during a reaction in a rigid sealed vessel. The actual calorimeter measurement must still account for heat absorbed by the apparatus and surrounding water rather than assuming the thermometer directly reads ΔU.

At constant pressure, with only pressure-volume work, w = −PΔV. Then ΔU = q p − PΔV. Because H = U + PV and pressure is constant, ΔH = ΔU + PΔV = q p. This is the basis for many coffee-cup calorimetry interpretations. The ambient pressure is approximately constant, and the reaction heat is inferred from the temperature change of solution and calorimeter.

The two heat amounts can differ for gas reactions. If a reaction creates more moles of gas at constant pressure, the system expands and does work on surroundings. Some energy change appears as work, so q p = ΔH need not equal q V = ΔU. For an ideal-gas reaction at one temperature, the difference can often be expressed as ΔH − ΔU = Δn gRT, where Δn g counts product gas moles minus reactant gas moles in the balanced reaction. Solids and liquids do not enter that simple gas-mole count.

If a reaction involves substantial electrical work, then at fixed volume q V is not equal to ΔU by itself, because w electrical also contributes. Likewise, at fixed pressure q p need not equal ΔH if non-P–V work is exchanged. Material flow creates additional complications. The equations are powerful, but their assumptions should be stated whenever a problem is not a simple sealed reaction.

Physical states and reaction extent matter. Water formed as liquid versus vapor changes enthalpy, and doubling a balanced equation doubles the reaction energy. A pressure condition does not erase the need to define initial and final states. Nor does “constant pressure” necessarily mean constant temperature; an exothermic reaction may warm its surroundings.

Measurement signs require care. If an exothermic reaction warms the calorimeter, the reaction system has negative heat while the calorimeter absorbs positive heat. The observed temperature rise is used with heat capacity to estimate q cal, then q rxn ≈ −q cal if losses are negligible. The correct q can then be associated with ΔH or ΔU according to the constraint.

Step-by-step reasoning

1. Identify whether the vessel is rigid or pressure-held. 2. Decide whether matter and non-P–V work cross the boundary. 3. For rigid P–V-only conditions, set w PV = 0 and q V = ΔU. 4. For pressure-held P–V-only conditions, use q p = ΔH. 5. Convert the measured calorimeter heat to the reaction-system sign.

Visual explanation

Draw a sealed rigid bomb and a movable atmospheric-pressure piston. Put ΔV = 0 beside the bomb and an expansion arrow beside the piston. Under the first write q V = ΔU; under the second write q p = ΔH, with a note “only P–V work and suitable closed-system conditions.”

Real-world analogy

If a machine is bolted in place, supplied energy cannot become boundary-motion work; if it can move a weight, some energy may leave mechanically. Rigid and movable calorimetric constraints similarly partition energy differently, though chemical heat also depends on the stated states.

Real-world example

Combustion in a bomb calorimeter is measured at nearly fixed volume and gives a quantity related to ΔU. Neutralisation in an open insulated cup is near atmospheric pressure and is commonly interpreted in terms of ΔH. The difference is experimental condition, not a disagreement about energy conservation.

Why?

Why does a rigid container remove P–V work even if pressure rises? Work requires boundary displacement; with ΔV = 0 there is no pressure-volume displacement, although the gas can gain internal energy and pressure.

Common misconception

“Any heat measured in any vessel is ΔH.” It equals ΔH only under suitable constant-pressure conditions with appropriate work and boundary assumptions. A rigid sealed vessel points instead to ΔU for P–V-only work.

Worked example

A sealed rigid reaction releases 2.0 kJ to its calorimeter. For the reaction system q V = −2.0 kJ, so ΔU = −2.0 kJ if no other work occurs. If the same specified reaction at constant pressure has PΔV = +0.3 kJ, then ΔH = ΔU + PΔV = −1.7 kJ, and q p = −1.7 kJ. The two heats differ because the pressure-held reaction also does expansion work.

Quick check

1. Under P–V-only conditions, which state-function change equals heat in a rigid sealed vessel? Answer: ΔU, because P–V work is zero at constant volume.

Exam focus

Always state the constraint and work assumptions before equating q with ΔH or ΔU. For calorimetry, reverse the sign between reaction and calorimeter heat when losses are negligible.

Advanced insight

The distinction between q p and q V is often small for condensed-phase reactions but can be significant when the number of gas moles changes. The ideal-gas Δn gRT correction quantifies the difference under a stated temperature and reaction extent.

Summary

At fixed volume with no other work, q V = ΔU. At constant pressure with only P–V work, q p = ΔH. Gas expansion and compression make these quantities differ, and calorimeter signs must be assigned from the chosen system boundary.

Practice questions

1. Why does a rigid bomb reaction have zero P–V work? Answer: Its boundary does not change volume, so ΔV = 0. 2. A calorimeter gains 500 J from a reaction. What is reaction-system q if losses are negligible? Answer: −500 J, because the reaction loses the heat the calorimeter gains. 3. If electrical work enters a fixed-volume cell, does q V alone equal ΔU? Answer: No. ΔU = q V + w electrical when that work mode is present.