Isobaric Process Calculator
Solve V₁/T₁ = V₂/T₂ for a gas held at constant pressure. Leave one field blank, enter the other three, and get an instant result with full step-by-step working.
Isobaric Process Solver
The Thermodynamics of an Isobaric Process
An isobaric process fixes pressure and lets volume respond to changing temperature. Physically, this means the boundary containing the gas (a piston, a balloon skin) is free to move, always settling at whatever volume keeps internal pressure matched to a constant external pressure.
Because pressure is constant, calculating work is unusually simple compared to the other basic processes: W = P(V₂ − V₁). Contrast that with an isothermal process, where work requires a logarithm, or an adiabatic process, where it requires the heat capacity ratio γ.
On a pressure-volume diagram, an isobaric process appears as a flat horizontal line: pressure never changes, while volume slides left or right along it as temperature drives expansion or compression.
A flat, horizontal line on the P–V diagram. Pressure never changes, only volume.
Isobaric vs. the Other Basic Processes
| Process | Held constant | Relationship | Calculator |
|---|---|---|---|
| Isobaric | Pressure | V₁/T₁ = V₂/T₂ | this page |
| Isothermal | Temperature | P₁V₁ = P₂V₂ | Isothermal process |
| Isochoric | Volume | P₁/T₁ = P₂/T₂ | Isochoric process |
| Adiabatic | Heat exchange (Q = 0) | P₁V₁^γ = P₂V₂^γ | Adiabatic process |
Worked Example: Work Done in an Isobaric Expansion
Problem: A gas at 1.50 atm expands from 3.00 L to 7.00 L at constant pressure. How much work does it do, in joules?
Note the unit conversion step: pressure and volume both had to be converted to SI units (pascals and cubic meters) before multiplying, since 1 Pa × 1 m³ = 1 joule exactly, while atm × L does not directly equal joules. This is the simplest work formula among the four basic thermodynamic processes precisely because pressure factors straight out of the W = ∫P dV integral.
Common Mistakes When Analyzing an Isobaric Process
A frequent mistake is forgetting to convert temperature to Kelvin in V₁/T₁ = V₂/T₂, exactly as with the standalone Charles' Law formula this process describes. Since it's a direct proportion, an unconverted Celsius or Fahrenheit value produces a badly wrong volume ratio.
A second mistake is computing work done, W = P(V₂ − V₁), with pressure and volume left in everyday units like atm and liters rather than converting to pascals and cubic meters first. Since 1 Pa × 1 m³ = 1 joule exactly, but 1 atm × 1 L does not, skipping the SI conversion gives a numerically wrong answer in joules even though the physics is done correctly.
A third pitfall is assuming a process is isobaric just because it happens in the open air, while atmospheric pressure is close to constant on human timescales, a rapidly moving piston or a fast chemical reaction can create brief local pressure differences that make the "constant pressure" assumption less accurate than it first appears.
Finally, don't confuse isobaric work with isothermal work. The two use entirely different formulas (W = PΔV versus W = nRT·ln(V₂/V₁)), and mixing them up is a common exam mistake since both processes can superficially look like "a gas expanding."