Keep a gas at constant temperature while it does work

Follow the energy while temperature stays fixed

Imagine one mole of gas moving a frictionless piston against a load that is adjusted in tiny steps. A heat bath holds the gas at 300 K. Predict the energy source before moving the final-volume control from 20 L to 40 L.

The solid curve is isothermal, meaning constant temperature throughout. For the modeled ideal gas, internal energy depends only on temperature, so its change ΔU is zero. The first law, ΔU = Q − W, then requires Q = W. Here Q is heat into the gas and W is work by the gas. The gas can transfer energy as work without cooling because the bath supplies the same energy as heat.

This differs from the dashed, insulated process. Both start at the same state and finish at your selected volume, but their final pressures and temperatures generally differ. Read the full endpoint rows before comparing energy changes.

Read pressure and area together

At 300 K, one mole has nRT = 1 × 8.314 × 300 = 2494.2 L kPa. Here n is amount in moles, R is the gas constant, and T is absolute temperature in kelvin. Since P = nRT/V, initial pressure is 2494.2/20 = 124.710 kPa, and final pressure at 40 L is 62.355 kPa. These are absolute pressures.

The pressure is falling during expansion, so multiplying either endpoint pressure by the whole volume change gives the wrong work. With load matching in the reversible limit, each small slice contributes pressure times its small volume change. The sum is the area under the curve, taken positive when volume increases and negative when it decreases. A litre-kilopascal is one joule.

Use a volume ratio

For this isothermal ideal-gas path, W = nRT ln(Vf/Vi). Vi and Vf are initial and final volumes in the same units. The natural logarithm ln acts on their dimensionless ratio. It measures the accumulated effect of the decreasing pressure; it is not multiplication by the volume ratio. You can use the calculated ledger without deriving logarithms.

At 300 K, doubling volume gives ln(2) approximately 0.693147. Thus W = 2494.2 × 0.693147 = 1728.848 J, or 1.729 kJ. The same heat enters from the bath. Halving volume gives ln(0.5) = −0.693147, so work by the gas is −1728.848 J. The surroundings do positive work on the gas and the bath removes heat to maintain 300 K.

Capture both settings. The magnitudes agree because the volume ratios are reciprocals, although the final pressures differ. Set 20 L: ln(1) = 0, so this prescribed path has no work or heat transfer. Returning to the initial state through a different cycle need not have zero net work.

Optional: why the logarithm appears

The reversible work is the integral of P dV. Substituting P = nRT/V with constant n and T gives nRT times the integral of dV/V. Between Vi and Vf, that integral is ln(Vf/Vi). No logarithm of a dimensional volume is needed.

Transfer to a different expansion

For one mole at 250 K, compare isothermal expansion from 20 L to 30 L with the same volume increase from 30 L to 40 L. Which does more work? Use ln(1.5) approximately 0.405465 and ln(4/3) approximately 0.287682.

Check the isothermal transfer

The first expansion gives 8.314 × 250 × 0.405465 ≈ 842.759 J; the second gives about 597.947 J. The first starts at higher pressure and has greater area over the same 10 L width. Heat into the gas equals work in each case; internal-energy change is zero.

The bath and load control are ideal assumptions, not an apparatus design or process-time prediction. For a non-quasistatic expansion, do not automatically integrate equilibrium gas pressure to obtain external work. Consult the OpenStax thermodynamic-process reference for process definitions.