A real gas expands and is brought back
30 October 2026
Statement The real gas under study expands freely, in thermal contact with a heat reservoir at T = 300\ \mathrm{K}. In this process it absorbs Q = 50\ \mathrm{J}. After this first process the gas is compressed isothermally and reversibly until it recovers its initial state. In this second process the gas gives off 300\ \mathrm{J}.
Obtain: 1. the entropy change of the gas in the first process; 2. the work done by the gas in the second process.
Original (Basque) Aztertu beharreko gas erreala era askean zabaldu da, T = 300\ \mathrm{K}-eko bero-iturriarekin ukipen termikoan. Prozesu horretan xurgatu duen beroa da Q = 50\ \mathrm{J}. Lehenengo prozesu horren ondoren, gasa konprimitu dugu, era isotermo itzulgarrian; hasierako egoera berreskuratu arte. Bigarren prozesu horretan, gasak 300 J kanporatu du. Lortu honako hauek: gasaren entropia-aldaketa, lehenengo prozesuan; gasak egindako lana, bigarren prozesuan.
What to watch The two states i and f are fixed, and so is everything the gas owns: \Delta U = +50\ \mathrm{J} and \Delta S = +1\ \mathrm{J\,K^{-1}}. What the expansion does to its surroundings is not fixed. With no load on the piston it is the statement: nothing is lifted, the reservoir supplies the 50 J, and the universe gains 0.83\ \mathrm{J\,K^{-1}}. With a load of mass M, the gas lifts it, the reservoir pays for that work as extra heat, and the universe gains less. The release is computed: the piston hits a stop, the cord goes slack and snaps taut, the load bobs until the cord’s losses still it, and all that kinetic energy ends as heat at 300 K. The reversible return adds sand grain by grain, so the load always balances the gas.
The statement gives no equation of state, so the page uses an abstract van der Waals gas, not any particular substance: n = 0.125\ \mathrm{mol}, a = 0.2\ \mathrm{Pa\,m^6\,mol^{-2}}, b = 5\times10^{-5}\ \mathrm{m^3\,mol^{-1}}, with V_\mathrm{i} = 35.73\ \mathrm{cm^3} and V_\mathrm{f} = 83.40\ \mathrm{cm^3} chosen so that its 300 K isotherm gives \Delta U = 50\ \mathrm{J}, \Delta S = 1\ \mathrm{J\,K^{-1}} and 250 J of work exactly. Another gas would draw other curves; the balance would be the same, because it uses only the statement’s numbers and Mg\,\Delta h. The piston’s area is 1 cm², so one load step stores at most p_\mathrm{f}(V_\mathrm{f} - V_\mathrm{i}) = 171.3\ \mathrm{J}, short of the 250 J of the reversible expansion; the rest is work lost, T\,\Delta S_\text{univ}.
The course page prints \Delta S_1 = 10/3\ \mathrm{J\,K^{-1}} for this exercise; the statement’s own numbers give 1\ \mathrm{J\,K^{-1}}, since the reversible return gives off 300 J at 300 K. The page says so.