Two gases that do not see each other: the route of the calculation
30 October 2026
Statement Two compartments of 10 L and 15 L are separated by a fixed, adiabatic and impermeable wall. In the first we put O₂ at 288 K and 2 atm, that is 0.846\ \mathrm{mol}; in the second, N₂ at 300 K and 1 atm, 0.609\ \mathrm{mol}. Both diatomic ideal gases, c_v = 5R/2. The separating wall is suddenly removed; obtain the entropy change of the whole system. If the gases were the same, how would the result change? A switch on the page gives the nitrogen the oxygen’s amount (1.39 atm), for the case where the second step ends with equal pressures.
Original (Basque) 10 eta 15 l-ko gordailuak finkoa, adiabatikoa eta iragaztezina den hormak banandu ditu. Lehenengo gunean O_{2} gasa sartu dugu: 288 K-ean eta 2 atm-n. Bigarrenean, N_{2} gasa; kasu honetan, 300 K-ean eta 1 atm-n. Biak gas ideal diatomikotzat hartuko ditugu (c_{v} = 5R/2). Hasierako baldintzetatik abiatuz, bereizte-horma bat-batean kendu dugu. Lortu sistema osoari dagokion entropia-aldaketa. Gordailuetako gasak berdinak izanik, nola aldatuko lirateke aurreko emaitzak?
What to watch The route to the entropy is three changes of state, each with a known \Delta S. A double free expansion: each gas alone, into vacuum, to 25 L; ideal, so its temperature does not move, and the points’ speeds do not change. For ideal gases this step is the mixing: \sum nR\ln(V/V_\mathrm{i}) = 9.04\ \mathrm{J\,K^{-1}}. Thermal contact through a shared wall: a point that hits it leaves at 293.0\ \mathrm{K}; the oxygen warms, the nitrogen cools, 0.006\ \mathrm{J\,K^{-1}}. Superposition: the two vessels slide into one; the points pass through each other and nothing either gas owns changes, \Delta S = 0; each gas keeps its pressure, 0.81 and 0.59 atm, the partial pressures of the final 1.40 atm. With the equal-amounts switch both gases are at 0.82 atm after step 2 and the superposition is seen at its plainest. With one gas on both sides the picture is the same but the count is not: two portions of one gas in one vessel are one gas, and step 3 carries the Gibbs correction \sum n_iR\ln x_i = -8.23\ \mathrm{J\,K^{-1}}, so the total falls to 0.81\ \mathrm{J\,K^{-1}}, the freed-wall value. Two views: colours, or one colour with a label on each point.
The oxygen is 600 material points, the nitrogen as many as its amount gives, with speeds drawn from the Maxwell distribution at its temperature (molar masses 32 and 28), straight flights, elastic bounces off the walls and no collisions with each other. The kinetic temperature on the page is read from the mean of v^2 over the points. In step 2 the shared wall re-emits every point that hits it at T_\mathrm{f}; the sample’s mean energy follows the expectation of that process, so the read temperatures converge without sampling noise. With equal amounts the route gives 10.05\ \mathrm{J\,K^{-1}} and 0.29\ \mathrm{J\,K^{-1}}; with the statement’s, 9.04\ \mathrm{J\,K^{-1}} and 0.81\ \mathrm{J\,K^{-1}}, the results of the statement’s page.