Isothermal Process Calculator
Solve P₁V₁ = P₂V₂ for a gas held at constant temperature. Leave one field blank, enter the other three, and get an instant result with full step-by-step working.
Isothermal Process Solver
The Thermodynamics of an Isothermal Process
In thermodynamics, a process is classified by which variable is held fixed while the others change. An isothermal process holds temperature fixed, which for an ideal gas immediately implies that its internal energy does not change either, since internal energy depends only on temperature for an ideal gas.
That has a striking consequence for energy bookkeeping: by the first law of thermodynamics, ΔU = Q − W, and since ΔU = 0 for an isothermal process, any work the gas does on its surroundings (W) must be exactly balanced by heat flowing in from the surroundings (Q). An isothermally expanding gas is, in effect, converting absorbed heat directly into work, with none left over to raise its own temperature.
On a pressure-volume diagram, an isothermal process traces the same hyperbolic curve used to illustrate Boyle's Law, the two describe identical physics, just from different disciplinary traditions (chemistry's empirical gas laws versus physics' thermodynamic process classification).
The shaded area under the P–V curve equals the work done, W = nRT·ln(V₂/V₁).
Isothermal vs. the Other Basic Processes
| Process | Held constant | Relationship | Calculator |
|---|---|---|---|
| Isothermal | Temperature | P₁V₁ = P₂V₂ | this page |
| Isobaric | Pressure | V₁/T₁ = V₂/T₂ | Isobaric 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 Isothermal Expansion
Problem: 1.00 mol of gas at 300 K expands isothermally from 5.00 L to 15.0 L. How much work does it do?
Since temperature never changes, all 2,740 joules of work the gas performs on its surroundings must be matched by exactly 2,740 joules of heat flowing in from outside. Internal energy for an ideal gas depends only on temperature, and here temperature is fixed. Compare this against the adiabatic process, where no heat can flow in and the gas would cool instead.
Common Mistakes When Analyzing an Isothermal Process
A frequent mistake is assuming that "no temperature change" automatically means "no energy changes hands." An isothermal process still involves heat flowing in or out and work being done. It's specifically the internal energy that stays constant for an ideal gas, not the energy flows themselves. Students sometimes incorrectly conclude that Q = 0 during an isothermal process, when in fact Q = W (they're equal, not zero).
A second mistake is confusing an isothermal process with an adiabatic one, since both can describe gas expansion or compression. The distinguishing question is whether temperature changes: isothermal means it doesn't (heat is exchanged to prevent it from changing); adiabatic means no heat is exchanged at all (so temperature necessarily does change).
A third pitfall is forgetting that the work formula W = nRT·ln(V₂/V₁) requires temperature in Kelvin and gives work in joules only when R, V, and T are all in consistent SI-compatible units. Mixing units here (like liters with the SI value of R) produces an answer off by a large factor, since the natural log itself is unitless but the RT prefactor is not.
Finally, remember that a genuinely isothermal process requires either very slow heat exchange with a large thermal reservoir, or deliberate temperature control. Most fast, real-world gas expansions and compressions are closer to adiabatic than isothermal, so it's worth double-checking which idealization actually fits the physical situation you're modeling.