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Use the conditions and processes of .54 to compute (a) the
Chapter 19, Problem 65P(choose chapter or problem)
Use the conditions and processes of 19.54 to compute (a) the work done by the gas, the heat added to it, and its internal energy change during the initial expansion; (b) the work done, the heat added, and the internal energy change during the final cooling; (c) the internal energy change during the isothermal compression. 19.54 .. ?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.
Questions & Answers
QUESTION:
Use the conditions and processes of 19.54 to compute (a) the work done by the gas, the heat added to it, and its internal energy change during the initial expansion; (b) the work done, the heat added, and the internal energy change during the final cooling; (c) the internal energy change during the isothermal compression. 19.54 .. ?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.
ANSWER:Solution 65P 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 .