CALC A cylinder with a piston contains 0.250 mol of oxygen at 2.40 X 105 Pa and 355 K. The oxygen may be treated as an ideal gas. The gas first expands isobarically to twice its original volume. It is then compressed isothermally back to its original volume, and finally it is cooled isochorically to its original pressure. (a) Show the series of processes on a pV-diagram. Compute (b) the temperature during the isothermal compression; (c) the maximum pressure; (d) the total work done by the piston on the gas during the series of processes.

Solution 64P Step 1: Here, n = 0.250 mol, Initial pressure is,1P = 2.4 × 10 Pa, Initial temperature is, T = 355 K , 1 The gas then expands isobarically (const pressure) to twice the original volume. This means, V 2 2V wi1h P = P .1 2 We know from ideal gas equation that, PV = nRT ----------(1) V = nR T P Here n, R and P are constant, which means V is directly proportional to the temperature. When the volume gets doubled, temperature also will be doubled. So, T 2 355 × 2 = 710 K . Step 2: The gas then compressed to it’s original volume isothermally (temperature constant). This means, T3= T =2710 K . So, the final temperature after the compression is 710 k. And V =3V . 1 Here first we need to find the initial volume, which can be found by the formula given below. P 1 =1nRT 1 2.4 × 10 × V = 0.250 × 8.314 × 355 1 5 3 V = 10.250 × 8.314 × 355) / (2.4 × 10 ) = 0.00307 = 3.07 × 10 m . Step 3: Then the gas cooled to it’s actual pressure isochorically (const. volume). The maximum pressure is achieved when the gas compressed isothermally to the original volume V . this means the volume gets halved. 1 P maxV /1 = nRT P max = 2nRT = 2 × P = 2 × 2.4 × 10 = 4.8 × 10 Pa. V1 1 The maximum pressure is 4.8 × 10 Pa.5