Constant pressure is a condition where the total pressure of a system stays fixed during a process, so volume is free to change. In AP Chem, it matters most in Topic 7.9, where adding an inert gas at constant pressure expands the container and shifts a gas-phase equilibrium toward the side with more moles of gas.
Constant pressure describes a system where the total pressure doesn't change while a process happens. Picture a container with a movable piston open to the atmosphere. If you add gas or heat the system, the piston slides and the volume changes so the pressure stays the same. That trade-off is the whole point. At constant pressure, volume is the variable doing the adjusting.
In Unit 7, this condition shows up in Le Châtelier's principle problems (Topic 7.9). The classic setup is adding an inert gas like argon to an equilibrium mixture. At constant volume, nothing happens, because the partial pressures of the reacting gases don't change. But at constant pressure, the container must expand to make room for the argon. That expansion lowers every reacting gas's partial pressure, which is the same stress as increasing the volume. The equilibrium responds by shifting toward the side with more moles of gas.
This term lives in Unit 7 (Equilibrium), Topic 7.9, supporting learning objective 7.9.A: identify the response of a system at equilibrium to an external stress using Le Châtelier's principle. Essential knowledge 7.9.A.1 specifically lists "change in volume/pressure of a gas-phase system" as one of the stresses you have to handle. The constant-pressure condition is the detail that decides whether a stress actually changes anything. Adding inert gas at constant volume is a non-stress, while adding it at constant pressure is a real volume change in disguise. Recognizing which scenario you're in is exactly the kind of careful reading the exam rewards.
Keep studying AP Chemistry Unit 7
Le Châtelier's Principle (Unit 7)
Constant pressure is one of the conditions you check before predicting a shift. Adding an inert gas at constant pressure forces the volume up, dilutes the partial pressures, and pushes the equilibrium toward the side with more moles of gas. It's a volume increase wearing a disguise.
Moles of Gas (Unit 7)
You can't predict a constant-pressure shift without counting moles of gas on each side of the equation. For PCl5(g) ⇌ PCl3(g) + Cl2(g), products have 2 moles versus 1 for reactants, so expansion at constant pressure shifts the system right.
Boyle's Law (Unit 3)
Boyle's Law says pressure and volume are inversely related at fixed temperature. That's why holding pressure constant means volume must change instead. The gas-law logic from Unit 3 is what makes the Unit 7 inert-gas trick work.
ΔH° and Enthalpy (Unit 6)
Enthalpy is defined as heat flow at constant pressure, which is why coffee-cup calorimetry (open to the atmosphere) measures ΔH directly. Same condition, different unit.
Multiple-choice questions love the inert-gas setup. A typical stem gives you a gas-phase equilibrium like PCl5(g) ⇌ PCl3(g) + Cl2(g), adds argon "while maintaining constant total pressure," and asks what you need to know or which way the system shifts. The move is always the same. First, recognize that constant pressure plus added gas means the volume increased. Second, count moles of gas on each side. Third, shift toward the side with more moles. The trap answer assumes nothing happens, which is only true at constant volume. No released FRQ has used this exact phrase, but justifying equilibrium shifts with Le Châtelier reasoning is a standard FRQ task, and you'd cite the change in partial pressures, not just say "Le Châtelier" and stop.
These two conditions give opposite answers to the same question. Add an inert gas at constant volume and the partial pressures of the reacting gases don't change, so the equilibrium doesn't shift at all. Add it at constant pressure and the container expands, every reacting gas's partial pressure drops, and the equilibrium shifts toward the side with more moles of gas. Always check which condition the problem states before answering.
Constant pressure means the total pressure stays fixed, so the volume of the system changes instead.
Adding an inert gas at constant pressure increases the volume and lowers the partial pressures of all reacting gases, shifting the equilibrium toward the side with more moles of gas.
Adding an inert gas at constant volume does nothing to the equilibrium, because the partial pressures of the reacting species don't change.
To predict a constant-pressure shift, you must compare the total moles of gas on the reactant and product sides of the balanced equation.
If both sides have the same number of moles of gas, a constant-pressure volume change causes no shift at all.
This condition supports LO 7.9.A, since changes in volume/pressure of a gas-phase system are one of the stresses listed in essential knowledge 7.9.A.1.
It means the total pressure of the system stays the same during a process, usually because the container can expand or contract (think movable piston or open beaker). When pressure is held constant, volume is the variable that changes.
It depends on the condition, and this is the classic trap. At constant volume, no shift happens because the reacting gases' partial pressures don't change. At constant pressure, the volume expands, partial pressures drop, and the equilibrium shifts toward the side with more moles of gas.
At constant volume, adding a non-reacting gas changes nothing. At constant pressure, adding any gas forces the container to expand, which acts like a volume increase and shifts the equilibrium toward the side with more moles of gas. Same argon, opposite outcomes.
Because a constant-pressure expansion lowers all partial pressures, the system relieves that stress by making more gas particles. For PCl5(g) ⇌ PCl3(g) + Cl2(g), the product side has 2 moles of gas versus 1 on the reactant side, so the equilibrium shifts right.
Yes. Enthalpy change is defined as the heat transferred at constant pressure, which is why open coffee-cup calorimetry experiments in Unit 6 measure ΔH directly. In Unit 7, the same condition shows up in Le Châtelier problems instead.