Isothermal Process Calculator
Enter initial pressure/volume and final volume to compute the work, heat and internal-energy change of an ideal-gas isothermal process.
Input Data
Results
At a glance:An isothermal process keeps temperature constant. For an ideal gas, P·V = n·R·T = const (Boyle's law), so when volume changes from V1 to V2 the pressure becomes P2 = P1·V1/V2. The work done by the gas is W = ∫P dV = n·R·T·ln(V2/V1) = P1·V1·ln(V2/V1). Because ideal-gas internal energy depends only on T, ΔU = 0, so by the first law ΔU = Q − W the heat transferred equals the work: Q = W (heat flows in during expansion, out during compression). This tool computes P2, W, Q and ΔU from P1, V1, V2 and (n, T for the work formula).
Formula
Boyle: P1·V1 = P2·V2
Work: W = n·R·T·ln(V2/V1) = P1·V1·ln(V2/V1)
ΔU = 0, Q = W
$$P_1 V_1 = P_2 V_2, \quad W = nRT \ln\frac{V_2}{V_1}, \quad \Delta U = 0, \quad Q = W$$How to Use
- Enter initial pressure P1 and volume V1, final volume V2.
- Enter n and T to evaluate the work.
- The calculator returns P2, W, Q and ΔU.
Case Studies
Slow expansion
P1V1 = 1000 J equivalent, V2/V1 = 2, nT gives W = P1V1·ln2 ≈ 693 J.
Heat in = 693 J, ΔU = 0.
Typical Carnot isothermal branch.
FAQ
Why is ΔU zero in an isothermal ideal gas?
Internal energy of an ideal gas depends only on temperature; if T is constant, U is constant, so ΔU = 0 and all the heat added becomes work (or all work done on it is rejected as heat).
Work is positive for expansion?
By the convention W = work done by the gas, expansion (V2>V1) gives W>0 (gas does work, absorbs heat); compression gives W<0 (work done on gas, heat leaves).
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References
Content review: Calculatorism Science Team. Results are for reference only; please refer to the relevant authorities for the official figures.