Half mole of an ideal gas ( = 5/3 ) is taken through a cycle abcda in a P - V diagram. The states are located at a(5000 cm ^3, 100 kPa ) , b(10000 cm ^3, 100 kPa ) , c(10000 cm ^3, 200 kPa ) , and d(5000 cm ^3, 200 kPa ) . Take R = 25 3 J K ⁻¹ mol ⁻¹ . Determine the amount of heat liberated in the processes cd and da .

Step-by-step solution

Given = 5 3 , the molar heat capacities are C_v = R - 1 = 3 2 R and C_p = C_v + R = 5 2 R . For the isobaric process cd , the pressure is constant at P = 200 kPa = 2 10^5 Pa . The heat exchanged is Q_ cd = n C_p T = C_p R P V = 5 2 P (V_d - V_c) . Substituting the given values: Q_ cd = 5 2 (2 10^5) (5000 - 10000) 10⁻⁶ = -2500 J . The negative sign indicates that 2500 J of heat is liberated. For the isochoric process da , the volume is constant at V = 5000 cm ^3 = 5 10⁻³ m ^3 . The heat exchanged is Q_ da = n C_v T