Charles Law Calculator

Isobaric Process Calculator

Solve V₁/T₁ = V₂/T₂ for a process at constant pressure, then add pressure and moles to see the work done, the heat added, and the change in internal energy.

Solved

Computed from the other three values.

Final Temperature

T₂ = 375 K

T₂ = T₁ × V₂ / V₁ = 300 × 2.5 / 2 = 375 K

What Is an Isobaric Process?

An isobaric process is any change in a gas that happens at constant pressure — the "iso" means "same" and "baric" refers to pressure. A gas heated inside a cylinder with a freely moving piston, or a parcel of air warmed by the sun while surrounded by the constant pressure of the atmosphere, are both isobaric: pressure stays fixed while volume and temperature move together.

Definition

An isobaric process holds the pressure of a fixed mass of gas constant while its volume and temperature change in direct proportion to each other: heat it and it expands; cool it and it contracts, with volume divided by temperature always landing on the same number.

That constant ratio is not a coincidence — it's the same relationship stated by Charles’ law, and this is the key thing to understand about this page: an isobaric process and Charles’ law are the same physics described from two different angles. This is exactly what the Charles law calculator itself solves for V and T — this page adds the work and heat on top, using the same V₁/T₁ = V₂/T₂ solve underneath its calculator.

Because of that, the four core fields above behave identically to the homepage's calculator: enter any three of V₁, T₁, V₂ and T₂ and the fourth fills in on its own. The new part is the "Advanced: work and heat" section, which turns that same solved state into an energy account — how much work the gas did, how much heat it took in, and how its internal energy changed.

Work, Heat, and Internal Energy at Constant Pressure

When a gas expands against a constant external pressure, it pushes its surroundings out of the way, and pushing something requires work. That work done by the gas is W = P × (V₂ − V₁), with pressure in pascals and volume in cubic metres so the answer comes out in joules. If the gas expands, V₂ is larger than V₁ and W is positive; if it’s compressed instead, W comes out negative — the surroundings did the work on the gas, not the other way around.

Heat behaves differently at constant pressure than at constant volume, because some of the heat you add goes into that expansion work rather than into raising the temperature. The heat added is Q = n × Cp × (T₂ − T₁), where n is the amount of gas in moles, temperatures are in kelvin, and Cp is the molar heat capacity at constant pressure. This calculator assumes a diatomic gas such as air or nitrogen, for which Cp = (7/2)R — an assumption that ties back to the ideal gas model, since Cp and its constant-volume counterpart Cv are only fixed numbers like this because the gas is treated as ideal in the first place.

Put the two together and you get the first law of thermodynamics: ΔU = Q − W. Whatever heat goes in either does work pushing the surroundings back, or stays behind as a change in internal energy — there's nowhere else for it to go. If your readings came from a thermometer in Celsius or Fahrenheit rather than kelvin, run them through the temperature converter first, or simply pick the right unit from each temperature field's own dropdown above.

Isobaric Process Examples: Step by Step

Example 1

A Heated Piston

A piston holds V₁ = 2 L of air at T₁ = 300 K under a constant pressure of P = 1 atm (101,325 Pa). It's heated until the volume reaches V₂ = 2.5 L, with n = 0.08 mol of air inside. Find T₂, the work done, the heat added, and the change in internal energy.

  • Charles’ law: T₂ = T₁ × V₂ / V₁ = 300 × 2.5 / 2 = 375 K
  • Work: W = P × (V₂ − V₁) ≈ 50.66 J
  • Heat: Q = n × Cp × (T₂ − T₁) ≈ 174.6 J
  • Internal energy: ΔU = Q − W ≈ 123.9 J

About 71% of the heat stayed behind as a rise in internal energy; the rest pushed the piston outward.

Example 2

Atmospheric Heating

A parcel of air with V₁ = 1 m³ at T₁ = 288 K expands to V₂ = 1.05 m³ at a constant atmospheric pressure of P = 100 kPa. Find the new temperature and the work done, with the amount of gas left unspecified.

  • Charles’ law: T₂ = T₁ × V₂ / V₁ = 288 × 1.05 / 1 = 302.4 K
  • Work: W = P × (V₂ − V₁) = 100,000 × 0.05 = 5,000 J

Without a moles value there's nothing to multiply Cp by, so Q and ΔU stay hidden — the same way the calculator above behaves when that field is empty.

Real Examples of Isobaric Processes

A piston free to slide inside a cylinder is the textbook picture of an isobaric process: as long as it moves without friction, the gas inside always pushes back with just enough pressure to balance whatever presses on it from outside. Heat the gas and it expands at that same constant pressure; cool it and it contracts the same way — the same beach-ball and balloon pictures Charles’ law uses on the homepage, just with a piston standing in for the balloon's stretchy skin.

The atmosphere provides the other everyday example. A parcel of air near the ground sits under roughly constant pressure at any given altitude, so when the sun warms it, it expands isobarically rather than building up pressure the way a sealed container would. That's part of why hot air rises: warmed air becomes less dense at the same pressure, exactly as an isobaric expansion predicts.

Not every real process holds pressure this cleanly constant, though. A sealed rigid tank being heated, or a weather system where pressure and temperature both shift, isn't isobaric at all — for those, the combined gas law calculator is the right tool, since it lets pressure, volume, and temperature move all at once.

Frequently Asked Questions About Isobaric Processes

  • What is an isobaric process?

    An isobaric process is any change in a gas that happens at constant pressure. Volume and temperature are free to change, but their ratio stays fixed — which is exactly the relationship Charles’ law describes.

  • Is an isobaric process the same thing as Charles’ law?

    Yes, physically they are the same law. "Isobaric process" is the thermodynamics name for a constant-pressure change; "Charles’ law" is the name most people learn it under in chemistry class. Both reduce to V₁/T₁ = V₂/T₂.

  • How do you calculate work done in an isobaric process?

    Work done by the gas is W = P × (V₂ − V₁), with pressure in pascals and volume in cubic metres, so the result comes out in joules. If the gas expands, V₂ > V₁ and the gas does positive work on its surroundings; if it’s compressed, the work is negative.

  • What formula gives the heat added at constant pressure?

    Q = n × Cp × (T₂ − T₁), where n is the amount of gas in moles, Cp is the molar heat capacity at constant pressure, and the temperatures are in kelvin. This calculator assumes a diatomic gas, for which Cp = (7/2)R.

  • What is the first law of thermodynamics in this context?

    ΔU = Q − W: the change in a gas’s internal energy equals the heat added to it minus the work it does on its surroundings. At constant pressure, all three quantities can be found from the same V₁, T₁, V₂, T₂ state used to solve Charles’ law.

  • Why does this calculator assume a diatomic gas?

    Cp depends on how a gas molecule stores energy. Diatomic gases such as nitrogen, oxygen, and ordinary air store energy in rotation as well as translation, giving Cp = (7/2)R — a good approximation for most everyday air-based examples. A monatomic gas like helium or argon has a different Cp, (5/2)R, and would need a different constant.

  • Does real gas behaviour match the isobaric process exactly?

    Closely, for air and similar gases at ordinary pressures and temperatures. It is an ideal-gas model, so it becomes less exact near a phase change or at very high pressure — the same caveat that applies to Charles’ law and the ideal gas law generally.

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