CHEMISTRY - THERMOCHEMISTRY

Question

A gas expands against a variable opposing pressure given by P = 10/V atm, where V is the volume of the gas at each stage of expansion. Further, in expanding from 10 to 100 liters, the gas undergoes a change in internal energy of ∆E = 100 cal. How much heat, in calories, is absorbed by the gas during the process? [NOTE: 1 atm = 24.22 cal]

Answer 1

Short answer: about 658 calories are absorbed. The whole problem turns on one thing: the opposing pressure is not constant, so you cannot use w = P*(V2 - V1). You have to integrate. Step 1 - work done by the gas. w = integral of P dV from V1 to V2, with P = 10/V w = integral from 10 to 100 of (10/V) dV w = 10 * ln(V), evaluated from 10 to 100 w = 10 * ln(100/10) = 10 * ln(10) = 10 * 2.3026 w = 23.03 liter-atm Step 2 - convert to calories. The note 1 atm = 24.22 cal is shorthand for 1 liter-atm = 24.22 cal. A pressure by itself cannot equal an energy; it is the pressure-volume product that carries energy units. w = 23.03 * 24.22 = 557.7 cal Step 3 - apply the first law. Delta E = q - w, where w is work done BY the gas. q = Delta E + w = 100 + 557.7 q = 657.7 cal, or about 658 cal absorbed. Two things worth noticing. If your textbook writes the first law as Delta E = q + w, with w meaning work done ON the gas, then w = -557.7 cal and q = 100 - (-557.7) = 657.7 cal. Same number. The sign convention changes the bookkeeping, never the physics. The form P = 10/V is not arbitrary. For an ideal gas P = nRT/V, so P = 10/V is quietly telling you that nRT = 10 liter-atm and the expansion is isothermal. That is why a natural log appears at all. The traps on this one: Using w = P*(change in V) with the starting pressure. P1 = 10/10 = 1 atm, giving w = 1 * 90 = 90 liter-atm = 2180 cal. Too big by roughly a factor of 4. Using the final pressure instead. P2 = 10/100 = 0.1 atm, giving w = 0.1 * 90 = 9 liter-atm = 218 cal. Now too small. Both fail for the same reason: P falls continuously as the gas expands, and the integral is what averages it properly. The effective average pressure here is 23.03/90 = 0.256 atm, which is not the average of 1 and 0.1. Finally, sanity-check the size of the answer. The gas does 557.7 cal of work on the surroundings and still finishes 100 cal higher in internal energy, so it must have taken in the sum of the two. Heat absorbed being much larger than Delta E is exactly what you should expect here, not a red flag.