Isochoric Process Calculator

Calculator Inputs

This page uses a single-column content flow, while the calculator fields switch between three, two, and one columns by screen size.

Thermal Basis

Gas Preset

Amount Basis

Amount, n (mol)

Mass (kg)

Constant Volume (m³)

Custom Molar Cv (kJ/mol·K)

Specific cv (kJ/kg·K)

Molar Mass (g/mol)

Useful when converting between molar and specific heat capacities.

Initial Pressure, P₁ (kPa)

Initial Temperature, T₁ (K)

Final Pressure, P₂ (kPa)

Optional if final temperature is known.

Final Temperature, T₂ (K)

Optional if final pressure is known.

Formula Used

Core Relations

Isochoric pressure-temperature law:
P₁ / T₁ = P₂ / T₂

Boundary work:
W = ∫P dV = 0

First law at constant volume:
Q = ΔU + W, so Q = ΔU

Energy and Entropy

Internal energy change:
ΔU = nCvΔT or ΔU = mcvΔT

Entropy change:
ΔS = nCv ln(T₂/T₁) or ΔS = mcv ln(T₂/T₁)

Temperature change:
ΔT = T₂ − T₁

Absolute temperature in kelvin is required because proportional pressure-temperature behavior only holds on an absolute scale.

How to Use This Calculator

Select a thermal basis. Choose a preset gas or enter a custom heat capacity.
Select how the gas amount is known. Use moles, mass, or derive moles from the initial PVT state.
Enter the initial pressure and temperature in kPa and kelvin.
Provide either the final pressure or final temperature. You may enter both for validation.
Press the calculate button to view the result block, export files, and inspect the pressure-temperature graph.

Example Data Table

Case	Basis	Inputs	Final State	Heat / ΔU
Example 1	2 mol, diatomic	P₁ = 100 kPa, T₁ = 300 K, T₂ = 450 K	P₂ = 150.0000 kPa	6.2359 kJ
Example 2	1 mol, monatomic	P₁ = 150 kPa, T₁ = 400 K, P₂ = 225 kPa	T₂ = 600.0000 K	2.4943 kJ
Example 3	1.2 kg, cv = 0.718	P₁ = 95 kPa, T₁ = 290 K, T₂ = 330 K	P₂ = 108.1034 kPa	34.4640 kJ

Frequently Asked Questions

1) What remains constant during an isochoric process?

Volume remains constant throughout the process. Pressure and temperature may change, but the gas does not expand or compress.

2) Why is the boundary work zero?

Boundary work equals the integral of pressure with respect to volume. Because volume does not change, that integral becomes zero.

3) Can pressure decrease in an isochoric process?

Yes. If the gas cools while volume stays fixed, pressure decreases in direct proportion to absolute temperature.

4) Why must temperature be entered in kelvin?

Kelvin is an absolute temperature scale. The relation P₁/T₁ = P₂/T₂ is valid only when absolute temperatures are used.

5) When should I use the mass basis?

Use mass basis when the gas amount is known in kilograms or when your heat capacity data is given as kJ per kilogram-kelvin.

6) What does the derived moles mode do?

It calculates the number of moles from the initial pressure, constant volume, and initial temperature using the ideal gas law.

7) Is this suitable for real gases?

It is best for ideal-gas style analysis. Real-gas deviations can matter at very high pressures or near phase-change conditions.

8) Can heat transfer be negative here?

Yes. During cooling, ΔT becomes negative, so ΔU and Q also become negative while boundary work remains zero.