Charles Law Calculator

Isochoric Process Calculator

Solve P₁/T₁ = P₂/T₂ at constant volume, then add the amount of gas to see how much heat it takes and where that energy goes.

Solved

Computed from the other three values.

Final Pressure

P₂ = 1.273 atm

P₂ = P₁ × T₂ / T₁ = 1 × 373.15 / 293.15 = 1.273 atm

What Is an Isochoric Process?

An isochoric process — also called isovolumetric or isometric — is a change in a gas's pressure and temperature that happens while its volume stays fixed. Picture a sealed, rigid steel tank: no matter how hard you heat it or cool it, the walls don't move, so the volume trapped inside never changes. Only pressure and temperature are free to respond.

Definition

An isochoric process holds the volume of a fixed mass of gas constant while its pressure and temperature change together, in direct proportion to one another.

Here's the key thing to notice: that direct proportion is exactly Gay-Lussac's law, P/T = constant, written for two states as P₁/T₁ = P₂/T₂. In other words, an isochoric process is the physical situation Gay-Lussac's law describes — the same equation, the same physics. If all you need is the plain P₁/T₁ = P₂/T₂ solve without any heat bookkeeping, the Gay-Lussac's law calculator handles exactly that. This page builds on top of it: once the core four values are resolved, it also works out how much heat had to be added or removed to get there, and what that heat did to the gas's internal energy.

If your numbers arrived in Celsius or Fahrenheit, convert them to kelvin first with the temperature converter — the calculator above does this automatically as you type, but it's worth knowing the absolute scale is what the underlying ratio actually depends on. And if you suspect the container isn't perfectly rigid, or that some gas might escape as it heats, the process may not be truly isochoric — the combined gas law calculator is the right tool once volume is also allowed to change.

Heat Transfer at Constant Volume

Solving P₁/T₁ = P₂/T₂ tells you the new pressure or temperature, but it doesn't say anything about the energy involved in getting there. That's what the "Advanced: heat transfer" section above adds. Once the core four values resolve and you enter the amount of gas n in moles, the calculator applies:

Formula

Q = n × Cv × (T₂ − T₁)

with Cv = (5/2)R ≈ 20.79 J/(mol·K), assuming a diatomic ideal gas such as air or nitrogen — the standard textbook assumption for this kind of problem, and the one used throughout this calculator and its worked examples below.

The reason the calculator can stop there, without a separate work term, comes from how work is defined for an expanding or compressing gas: W = P ΔV. At constant volume, ΔV = 0, so W = 0 — no matter how large the pressure or temperature swing is, the gas does no mechanical work and none is done on it. The first law of thermodynamics, ΔU = Q − W, then collapses to something very clean:

Key result

ΔU = Q — every joule of heat added at constant volume goes directly into the gas's internal energy. None of it is "spent" pushing back a piston or expanding a boundary, because there's no boundary movement to spend it on.

It's worth contrasting this with the isobaric case, where pressure — not volume — is held constant. There, a heated gas is free to expand, so some of the heat you add does go into work as the gas pushes its surroundings back; only the remainder raises the internal energy. The isochoric case is the tidy opposite: with the "escape valve" of expansion removed, heat has nowhere to go but into internal energy, exactly as it does on the isothermal page's own clean fact that ΔU = 0 forces Q = W. Isochoric and isothermal end up being mirror-image "one variable does all the work" cases of the first law.

Isochoric Process Formula and Symbols

Rearranged for whichever core value is unknown: P₂ = P₁ × T₂ / T₁, P₁ = P₂ × T₁ / T₂, T₂ = T₁ × P₂ / P₁, and T₁ = T₂ × P₁ / P₂. The calculator above picks the right rearrangement automatically based on which field you leave blank, and temperatures are always converted to kelvin internally before the ratio is taken.

The symbols in the isochoric process formula and their units
Symbol Meaning Unit
P₁ Initial pressure kPa, atm, mmHg, psi
T₁ Initial temperature K, °C, °F
P₂ Final pressure kPa, atm, mmHg, psi
T₂ Final temperature K, °C, °F
n Amount of gas (advanced) mol

Isochoric Process Examples: Step by Step

Example 1

Heating a Sealed, Rigid Container

A rigid, sealed container holds gas at P₁ = 1 atm and T₁ = 293.15 K (20°C). It's heated at constant volume until T₂ = 373.15 K (100°C). Find the new pressure, then — with n = 0.5 mol of gas inside — find the heat added and the change in internal energy.

  • Apply: P₂ = P₁ × T₂ / T₁ = 1 × 373.15 / 293.15 = 1.273 atm
  • Heat: Q = 0.5 × 20.79 × (373.15 − 293.15) = 831.4 J
  • Answer: P₂ = 1.273 atm, Q = ΔU = 831.4 J

All 831.4 J of heat becomes internal energy — none of it does mechanical work, since the container's rigid walls never move.

Example 2

A Pressure Cooker Building Pressure

A sealed, rigid pressure cooker holds gas at P₁ = 100 kPa and T₁ = 298.15 K (25°C). It heats up, at constant volume, until the pressure reaches P₂ = 250 kPa, just before the release valve would open. Find the temperature at that point.

  • Apply: T₂ = T₁ × P₂ / P₁ = 298.15 × 250,000 / 100,000 = 745.4 K
  • Answer: T₂ = 745.4 K (about 472°C)

That's well above the temperature at which the safety valve is designed to release — a reminder of just how fast pressure climbs when volume can't give at all.

Real Examples of Isochoric Processes

A sealed, rigid container being heated. Whether it's a steel cylinder, a can, or a laboratory bomb calorimeter, a fixed volume of gas trapped inside rigid walls and placed over a heat source is the textbook isochoric setup — pressure climbs in direct proportion to absolute temperature until something gives.

A pressure cooker before the valve releases. While the valve stays shut, the cooking chamber's volume is fixed, so the steam and air inside heat up isochorically, building pressure well above atmospheric until the valve finally lets some gas escape and the process stops being isochoric.

An aerosol can left in a hot car. The can's metal walls don't expand meaningfully as it warms up, so the propellant gas inside undergoes an isochoric temperature-pressure rise — which is exactly why aerosol cans carry warnings against leaving them somewhere hot.

For a broader look at the isochoric page's counterpart on the primary gas law calculator, which covers the isobaric case where pressure — not volume — is what's held fixed, see how that page's V/T solver compares to this one's P/T solver.

Frequently Asked Questions About Isochoric Processes

  • What is an isochoric process?

    An isochoric process (also called isovolumetric or isometric) is a change in a gas’s state where the volume stays fixed while pressure and temperature change. Because volume never changes, no work is done on or by the gas.

  • Is an isochoric process the same as Gay-Lussac’s law?

    Yes — Gay-Lussac’s law, P₁/T₁ = P₂/T₂, is exactly the isochoric relationship between pressure and temperature. This page uses the same solver as the Gay-Lussac’s law calculator, plus an extra step for heat transferred and internal-energy change.

  • What is the formula for heat in an isochoric process?

    Q = n Cv (T₂ − T₁), where n is the amount of gas in moles, Cv is the molar heat capacity at constant volume, and T₁, T₂ are the initial and final temperatures in kelvin. For a diatomic ideal gas such as air or nitrogen, Cv = (5/2)R ≈ 20.79 J/(mol·K).

  • Why is no work done in an isochoric process?

    Work done by an expanding or compressing gas is W = PΔV. Since volume is constant, ΔV = 0, so W = 0 regardless of how much the pressure or temperature changes.

  • Why does all the heat become internal energy?

    The first law of thermodynamics says ΔU = Q − W. With W = 0 at constant volume, that reduces to ΔU = Q — every joule of heat added raises the gas’s internal energy directly, with none of it "spent" doing mechanical work.

  • What are some real-world examples of isochoric processes?

    A sealed, rigid container being heated on a stove, a pressure cooker before its valve releases, and an aerosol can left in a hot car are all approximately isochoric — their volume is fixed by a rigid wall, so any heating shows up as a pressure rise.

  • What if the container isn’t perfectly rigid, or gas can escape?

    If volume actually changes during the process, it isn’t isochoric — use the combined gas law calculator instead, which lets pressure, volume, and temperature all vary at once.

Need a different variable held constant, or want to convert your inputs first? These companion tools cover the rest of the gas law family.