Combined Gas Law Calculator

Isochoric Process Calculator

Solve P₁/T₁ = P₂/T₂ for a gas sealed in a fixed volume. Leave one field blank, enter the other three, and get an instant result with full step-by-step working.

Isochoric Process Solver

The Thermodynamics of an Isochoric Process

An isochoric process fixes volume, which happens whenever a gas is sealed inside a rigid container. A steel cylinder, a closed can, a bomb calorimeter. With the boundary unable to move, the gas can't push against anything, so it can't do mechanical work no matter how much its pressure rises.

That makes the energy bookkeeping unusually clean: since W = 0, the first law of thermodynamics reduces to ΔU = Q, every joule of heat supplied goes directly into internal energy (and temperature), none of it diverted into expansion work. This is why bomb calorimeters, used to measure the heat released by combustion reactions precisely, are built as rigid, constant-volume vessels.

On a pressure-volume diagram, an isochoric process is a vertical line. Volume never changes while pressure climbs or falls with temperature.

Constant Volume, Changing Pressure 1 2 Volume (V) Pressure (P)

A vertical line on the P–V diagram. Volume never changes, only pressure.

Isochoric vs. the Other Basic Processes

ProcessHeld constantRelationshipCalculator
IsochoricVolumeP₁/T₁ = P₂/T₂this page
IsothermalTemperatureP₁V₁ = P₂V₂Isothermal process
IsobaricPressureV₁/T₁ = V₂/T₂Isobaric process
AdiabaticHeat exchange (Q = 0)P₁V₁^γ = P₂V₂^γAdiabatic process

Worked Example: Pressure Rise in a Sealed Tank

Problem: A sealed rigid tank holds gas at 3.00 atm and 20°C. If it's heated to 150°C, what is the new pressure?

Given: P₁ = 3.00 atm, T₁ = 20°C = 293.15 K, T₂ = 150°C = 423.15 K. Gay-Lussac's Law: P₁/T₁ = P₂/T₂ → P₂ = P₁T₂/T₁ P₂ = 3.00 × 423.15 / 293.15 P₂ ≈ 4.33 atm

A pressure increase of nearly 45% from a 130°C temperature rise illustrates exactly why sealed gas containers, propane tanks, aerosol cans, compressed gas cylinders, carry explicit warnings against exposure to heat: since volume can't give, all of that thermal energy shows up as rising internal pressure, which can eventually exceed a container's rated limit.

Common Mistakes When Analyzing an Isochoric Process

A frequent mistake, shared with the standalone Gay-Lussac's Law calculation this process describes, is forgetting to convert temperature to Kelvin. P₁/T₁ = P₂/T₂ is a direct proportion and gives a badly wrong pressure ratio if fed an unconverted Celsius value.

A second mistake is assuming "sealed container" automatically means "isochoric." A sealed but flexible container, a balloon, a plastic bottle, will bulge slightly as internal pressure rises, meaning volume is not perfectly constant even though no gas escapes. True isochoric behavior requires a genuinely rigid container, like a steel cylinder or a bomb calorimeter.

A third pitfall is assuming work is done during an isochoric process because pressure is clearly changing. Work depends on volume changing (W = ∫P dV), not pressure. Since volume is fixed by definition, W = 0 always, regardless of how dramatically pressure rises or falls.

Finally, be careful not to apply Gay-Lussac's Law-style reasoning to a container that is simultaneously being heated and vented or filled. If the amount of gas is changing as well as its temperature, the simple P₁/T₁ = P₂/T₂ relationship no longer holds, and a more general form of the ideal gas law is needed instead.

Isochoric Process FAQ

What is an isochoric process?
An isochoric process (also called isometric or constant-volume) is one in which volume stays fixed throughout, from the Greek 'iso' (equal) and 'khora' (space/volume). It describes gas sealed inside a rigid container. The resulting pressure-temperature relationship is P₁/T₁ = P₂/T₂, Gay-Lussac's Law.
How is an isochoric process different from Gay-Lussac's Law?
They describe the same relationship, P₁/T₁ = P₂/T₂. 'Gay-Lussac's Law' is the historical empirical name; 'isochoric process' is the thermodynamic classification, useful because it tells you immediately that no work is done (since work requires a volume change) and all heat added goes directly into raising internal energy.
How much work is done during an isochoric process?
None. Work done by a gas is W = ∫P dV, and since volume never changes in an isochoric process, dV = 0 throughout, so W = 0 always. By the first law of thermodynamics, ΔU = Q − W = Q, meaning every joule of heat added goes directly into raising the gas's internal energy (and hence its temperature). None is 'spent' pushing a boundary outward.
Why must temperature be in Kelvin for this calculator?
P₁/T₁ = P₂/T₂ is a direct proportion, which is only valid on an absolute temperature scale where zero truly means zero. This calculator converts any °C, °F, or °R value you enter into Kelvin internally before solving, so you never need to convert by hand.
What is a real-world example of an isochoric process?
A sealed aerosol can, a rigid gas cylinder, or a pressure cooker with its valve closed are all approximately isochoric systems when heated. Their rigid walls hold volume fixed, so added heat shows up entirely as rising pressure and temperature, which is why these containers carry warnings about overheating.
What units does this calculator support?
Pressure can be entered in Pa, kPa, MPa, bar, mbar, atm, mmHg, torr, psi, or inHg. Temperature can be entered in Kelvin, Celsius, Fahrenheit, or Rankine. Mix units freely. The calculator converts internally before solving.
How is an isochoric process used to measure heat of combustion?
A bomb calorimeter burns a sample inside a small, rigid, sealed steel vessel submerged in a water bath. Because the vessel's volume cannot change, all the chemical energy released by combustion converts into internal energy (temperature rise) with none diverted into expansion work. Letting chemists calculate the exact heat of combustion from the measured temperature rise of the surrounding water, with no work-done correction needed.
What is the constant-volume specific heat capacity, Cv?
Cv is the amount of heat needed to raise one mole (or one gram) of a substance by one degree at constant volume. Because no work is done in an isochoric process, all of that heat goes directly into internal energy, making Cv a direct measure of how a gas's internal energy responds to temperature. It is always smaller than the constant-pressure specific heat, Cp, for exactly the reason described in our Isobaric Process page.
How does an isochoric process appear on a pressure-volume graph?
As a perfectly vertical line. Volume never changes while pressure rises or falls along that single volume value. This is visually the most distinctive of the four basic process types on a P-V diagram, since isothermal and adiabatic curves both slope downward and isobaric processes appear as horizontal lines; only the isochoric process produces a vertical line.

Related Calculators

⇄ Combined Gas Law P/T Gay-Lussac's Law ΔP=0 Isobaric Process Q=0 Adiabatic Process