101,325
Pa
50,662.5
Pa
2
0.693147
702.33
J
702.33
J
0
J
101,325
Pa
50,662.5
Pa
2
0.693147
702.33
J
702.33
J
0
J
The Isothermal Process Calculator analyzes thermodynamic processes that occur at constant temperature. In an isothermal process, a system exchanges heat with its surroundings slowly enough that thermal equilibrium is maintained throughout, keeping the temperature fixed while pressure and volume change according to Boyle's law: $$P_1 V_1 = P_2 V_2$$.
Isothermal processes are fundamental to understanding heat engines, gas compression, and biological systems. When an ideal gas expands isothermally, it does work on the surroundings while absorbing an equal amount of heat from the reservoir. The internal energy of an ideal gas depends only on temperature, so $$\Delta U = 0$$ for any isothermal change. This elegant relationship makes isothermal processes one of the most tractable cases in thermodynamics.
The work done by the gas during an isothermal expansion or compression is given by $$W = nRT\ln\left(\frac{V_2}{V_1}\right)$$, where $$n$$ is the number of moles, $$R = 8.314\;\text{J/(mol·K)}$$ is the universal gas constant, and $$T$$ is the absolute temperature. Because $$\Delta U = 0$$, the first law of thermodynamics requires $$Q = W$$, meaning all absorbed heat converts entirely into work. This calculator computes the final pressure, work, and heat transfer for any isothermal process involving an ideal gas.
Engineers use isothermal analysis for slow compression in reciprocating compressors, idealized stages in multi-stage compression with intercooling, and as a reference process in Carnot cycle analysis. Physics and chemistry students encounter isothermal processes when studying gas laws, thermodynamic cycles, and equilibrium phenomena.
The calculator applies the ideal gas law under constant-temperature conditions:
Boyle's Law: $$P_1 V_1 = P_2 V_2 \implies P_2 = \frac{P_1 V_1}{V_2}$$
Work done by the gas: $$W = \int_{V_1}^{V_2} P\,dV = nRT\ln\left(\frac{V_2}{V_1}\right)$$
First Law of Thermodynamics: $$\Delta U = Q - W$$
Since the internal energy of an ideal gas depends only on temperature and $$T$$ is constant: $$\Delta U = 0 \implies Q = W$$
Positive work means the gas expands and does work on the surroundings. Negative work means the gas is compressed and work is done on the gas. The heat $$Q$$ has the same sign: positive means heat flows into the system, negative means heat flows out.
A positive work value indicates isothermal expansion — the gas pushes outward, doing work while absorbing heat from the thermal reservoir to maintain constant temperature. A negative work value indicates isothermal compression — external work is done on the gas and an equal amount of heat is rejected to the reservoir. The internal energy change is always exactly zero for an ideal gas isothermal process, confirming energy conservation via $$Q = W$$. Compare the volume ratio to understand the magnitude: doubling the volume yields $$W = nRT\ln 2 \approx 0.693\,nRT$$.
Inputs
Results
1 mol of ideal gas at 300 K doubles its volume. The pressure halves to ~50,663 Pa and the gas does 1728.85 J of work, absorbing an equal amount of heat.
Inputs
Results
2 mol at 400 K compressed to half volume. Pressure doubles to 400 kPa. Work is negative (−4609 J), meaning 4609 J of work was done on the gas and the same heat was rejected.
An isothermal process is a thermodynamic process in which the temperature of the system remains constant throughout. For an ideal gas, this means the product $$PV$$ stays constant (Boyle's law). The system must exchange heat with a thermal reservoir to maintain its temperature as it does work or has work done on it.
For an ideal gas, internal energy depends solely on temperature: $$U = \frac{f}{2}nRT$$, where $$f$$ is the degrees of freedom. Since temperature is constant in an isothermal process, $$\Delta U = 0$$. This is a property specific to ideal gases; real gases may have small internal energy changes due to intermolecular forces.
Isothermal work is calculated using $$W = nRT\ln(V_2/V_1)$$, which accounts for the changing pressure as volume changes. Isobaric work uses the simpler formula $$W = P\Delta V$$ because pressure is constant. For the same volume change, isothermal expansion from a high pressure does more work than isobaric expansion at the lower final pressure.
A perfectly isothermal process requires infinitely slow (quasi-static) operation so the system always remains in thermal equilibrium with the reservoir. In practice, slow processes with good thermal contact approximate isothermal conditions. Multi-stage compression with intercooling is an engineering approach to approximate isothermal compression.
An isothermal process appears as a hyperbola on a PV diagram because $$PV = \text{constant}$$. The curve is called an isotherm. Higher temperatures correspond to isotherms farther from the origin. The area under the curve between $$V_1$$ and $$V_2$$ equals the work done.
The Carnot cycle consists of two isothermal processes (expansion at $$T_H$$ and compression at $$T_C$$) and two adiabatic processes connecting them. The isothermal expansion absorbs heat from the hot reservoir while doing work, and the isothermal compression rejects heat to the cold reservoir. The Carnot efficiency $$\eta = 1 - T_C/T_H$$ is the maximum possible for any heat engine.
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