What is an isochoric process?

An isochoric (or isovolumetric) process is one in which the volume of the system remains constant. Because no volume change occurs, the gas does no pressure-volume work (W = 0), and all heat added to or removed from the gas changes its internal energy.

Why is work always zero in an isochoric process?

Mechanical work in thermodynamics is defined as W = ∫P dV. When volume is constant, dV = 0 for every step of the process, so the integral is zero regardless of how pressure or temperature change.

What is the heat capacity at constant volume (Cv)?

Cv is the amount of heat needed to raise the temperature of one mole of gas by 1 K while keeping volume constant. For a monatomic ideal gas, Cv = (3/2)R ≈ 12.471 J/(mol·K). For a diatomic ideal gas, Cv = (5/2)R ≈ 20.785 J/(mol·K), where R = 8.314 J/(mol·K).

What is Gay-Lussac's Law?

Gay-Lussac's Law states that for a fixed amount of gas at constant volume, pressure is directly proportional to absolute temperature: P/T = constant, or P₁/T₁ = P₂/T₂. This is why a sealed container's pressure rises when heated.

How do I convert Celsius to Kelvin?

Add 273.15 to the Celsius value: K = °C + 273.15. All thermodynamic equations require absolute temperature in Kelvin. The tool accepts Celsius input and converts automatically.

What is the difference between isochoric and isobaric processes?

An isochoric process keeps volume constant (W = 0; Q = ΔU). An isobaric process keeps pressure constant (W = PΔV ≠ 0 in general; Q = nCpΔT). Because Cp > Cv for any ideal gas, more heat is required to achieve the same temperature rise at constant pressure than at constant volume.

Isochoric Process Calculator

Enter moles, temperatures, initial pressure, and gas type to find heat transferred, internal energy change, and final pressure for a constant-volume process.

Process Inputs

Gas Type

Cv = 12.471 J/(mol·K) — 3R/2

Amount of Gas (n) mol

Temperature Unit

Initial Temperature (T₁) K

Final Temperature (T₂) K

Initial Pressure (P₁)

Enter values on the left to see results.

Summary

An isochoric process (also called isovolumetric or isometric) is a thermodynamic change that occurs at constant volume. Because dV = 0, no pressure-volume work is done (W = 0). All heat exchanged goes directly into changing the internal energy of the gas: Q = ΔU = nCvΔT. The pressure varies in direct proportion to the absolute temperature (Gay-Lussac's Law): P₁/T₁ = P₂/T₂.

How it works

Enter the amount of gas in moles (n) and select the gas type (monatomic or diatomic).
Enter the initial temperature T₁ and final temperature T₂. Switch the unit toggle to use Celsius — values are converted to Kelvin internally.
Enter the initial pressure P₁. Switch the unit toggle between Pa and kPa.
Click Calculate. The tool applies ΔU = nCvΔT (Cv = 3R/2 for monatomic, 5R/2 for diatomic) and Gay-Lussac's Law P₂ = P₁T₂/T₁.
Work done W is always 0 J for a constant-volume process — no volume change means no PdV work.
Results update instantly whenever any input changes.

Use cases

Thermodynamics homework involving constant-volume heating or cooling of an ideal gas.
Calculating the pressure rise inside a sealed rigid container when heated.
Verifying Gay-Lussac's Law relationships between pressure and temperature.
Engine and combustion analysis where the combustion stroke is approximated as isochoric.
Teaching the first law of thermodynamics for processes with zero work output.
Comparing isochoric vs. isobaric heat capacity differences (Cv vs. Cp).

Frequently Asked Questions

Last updated: 2026-06-13 · Reviewed by Nham Vu