What is an isochoric process?

A thermodynamic process at constant volume. Gay-Lussac's law (P₁/T₁ = P₂/T₂) describes the pressure-temperature relation.

Is any work done in an isochoric process?

No. Because ΔV = 0, the work W = PΔV is zero. All heat input goes into changing internal energy: Q = ΔU = nCvΔT.

Cv is the molar heat capacity at constant volume. For an ideal monatomic gas, Cv ≈ 12.47 J/(mol·K); for diatomic gases, ≈20.8 J/(mol·K).

How is it different from isobaric?

Isochoric: V fixed, no work, heat raises pressure and T. Isobaric: P fixed, work is done as V expands, more heat needed for the same ΔT.

A real-world example?

Heating a sealed rigid pressure cooker or a closed gas cylinder raises pressure with temperature, because the volume cannot change.

Can pressure keep rising forever in an isochoric process?

No. A real rigid container has a burst pressure limit. This is why sealed cans and rigid cylinders carry warnings against heating: Gay-Lussac's law predicts pressure rising in direct proportion to absolute temperature, with no upper bound built into the equation itself.

Isochoric Process Calculator

Compute T₂ = T₁·P₂/P₁ and Q = nCvΔT for a sealed rigid container.

Isochoric Process Calculator

Constant V. T₂ = T₁·P₂/P₁, Q = nCvΔT (Cv ≈ 20.8 J/mol·K).

P₁

T₁

P₂

n (mol)

T₂

600 K

T₂ = T₁·P₂/P₁

Heat (Q)

6240 J

Q = nCvΔT

Isochoric Process and Gay-Lussac's Law

An isochoric (constant-volume) process has the gas trapped in a rigid container. Heating raises pressure and temperature together according to Gay-Lussac's law: P₁/T₁ = P₂/T₂. Because the gas cannot expand, no work is done. All heat input goes into internal energy.

Heat Transfer Equation

Q = ΔU = n × Cv × ΔT. The heat capacity at constant volume Cv is smaller than Cp by exactly R (for an ideal gas), reflecting the fact that no work-against-surroundings has to be done in an isochoric process.

Worked Example: Heating a Sealed Gas Cylinder

1 mol of a monatomic ideal gas (Cv = 12.47 J/(mol·K)) is sealed in a rigid cylinder at P₁ = 100 kPa and T₁ = 300 K. Sunlight warms the cylinder until the pressure reaches P₂ = 150 kPa. By Gay-Lussac's law, T₂ = T₁ × P₂/P₁ = 300 × 150/100 = 450 K, a rise of ΔT = 150 K. Because the volume cannot change, all the heat goes into internal energy: Q = ΔU = n × Cv × ΔT = 1 × 12.47 × 150 = 1871 J. No work is done on or by the gas in this process, since W = P × ΔV and ΔV = 0.

Examples

Pressure cooker: sealed lid prevents volume change; pressure climbs as temperature rises.
Closed steel cylinder of gas: warming a gas bottle in sunlight raises internal pressure, a real safety concern for storage.
Constant-volume calorimetry: bomb calorimeters measure heat of reaction in a rigid vessel, exploiting Q = ΔU directly.