The Isothermal Process: Work as Area

Hold a gas at one temperature and let it change volume, and it traces the curve P = nRT/V. The blue line is that isotherm; the gold region under it, between V1 and V2, is the work — and the card beside it is the same area written as a logarithm. Press Free expansion instead and the curve and the shading disappear: a gas rushing into a vacuum is not in equilibrium on the way, so there is no path to draw and no area to measure, which is exactly why the work is zero. Both states are drawn against one pair of axes. The two pressures come from PV = nRT, so they are the curve restated, not a check on it.

Area under the curve, by numerical integration1728.8 J
Pressure before P1249.42 kPa
Pressure after P2124.71 kPa
Volume ratio V2/V12.000
Pressure ratio P1/P22.000
Temperature in Celsius26.85 °C

What is happeningThe gas expands and does 1728.8 J of work on its surroundings, drawing the same 1728.8 J in as heat.

Work done by the gas  W = nRT ln(V2/V1)
1728.8 J
Heat absorbed  Q = W, because ΔU = 0
1728.8 J
Change in internal energy  ΔU — exactly zero for an ideal gas
0.0 J
Entropy change  ΔS = nR ln(V2/V1)
5.763 J/K
Amount of gas n1.0 mol
Temperature T (kelvin)300 K
Volume before V110.0 L
Volume after V220.0 L
Ideal gas on a reversible path unless the free-expansion button is pressed · R = 8.314 J/(mol·K), this lab's value — the SI 2019 figure 8.314462618 would print 1728.9 J where this prints 1728.8 J.
T is absolute; the Celsius figure is for reading only, because T multiplies the answer and a Celsius number in its place returns exactly zero work at 0 °C. The area is the work only for a quasi-static path — a free expansion has none, which is why nothing is shaded. Both pressures come from PV = nRT, so multiplying them back is arithmetic, not a check: at 2 dp it disagrees between the two states in 5.99 % of reachable settings. V2/V1 and P1/P2 are one number read two ways, and in 0.46 % of settings their 4-figure strings differ in the last digit.