Chapter 18: The Laws of Thermodynamics Selected Solutions

Selected Solutions

25. (a) At constant pressure the work done by the gas is W = P(DV)

W = P(V_f - V_i) = P(2V_i - V_i) = PV_i = (160 kPa)(0.83 m³) = 130 kJ

(b)

37. (a) Since the system is thermally insulated the process is adiabatic, therefore

(b) From the ideal gas equation of state we have

49. (a) For a Carnot engine

T_c = T_h(1 - e_max) = (545 K)(1 - 0.300) = 382 K

(b) The efficiency of a heat engine increases as the difference in temperature of the hot and cold reservoirs increases. Therefore, the temperature of the low temperature reservoir must be decreased.

(c) T_c = T_h(1-e_max) = 545 K(1 - 0.400) = 327 K

63. The change in entropy is given by

79. From Problem 37, for the adiabatic process,

Dividing these equations gives

For the isothermal processes,

Selected Solutions by David Reid, Eastern Michigan University. ©2002 by Prentice Hall, Inc.

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