A sample of an ideal gas goes through the process shown below. From A to B, the

Question

A sample of an ideal gas goes through the process shown below. From A to B, the process is adiabatic; from B to C, it is isobaric with 345 kJ of energy entering the system by heat. From C to D, the process is isothermal; and from D to A, it is isobaric with 371 kJ of energy leaving the system by heat. Determine the difference in internal energy, Eint,B ? Eint,A. I know the answer is 19 kJ. I do not know how you get to that answer. Please show me all formulas and steps taken. Please! Thank you!

Explanation / Answer

Eint,B - EintA = (Eint,B - EintC) + (Eint,C - Eint,D) + (Eint,D - EintA) As the process from C to D is siothermal, Eint,C - EintD = 0 => Eint,B - EintA = (Eint,B - EintC) + (Eint,D - EintA) ... ( 1 ) For the process BC, Q = EintC - Eint,B + W => 345 = EintC - Eint,B + PdV => 345 = EintC - Eint,B + 3 * (0.40 - 0.09) * (101.325) ... [1atm = 101.325 kN/m^2] => Eint,B - EintC = - 250.76775 kJ ... ( 2 ) For the process DA, Q = EintA - Eint,D + W => - 371 = EintA - Eint,D + PdV => - 371 = EintA - Eint,D + 1 * (1.20 - 0.20) * (101.325) ... [1atm = 101.325 kN/m^2] => Eint,D - EintA = 269.675 kJ ... ( 3 ) Plugging from ( 2 ) and ( 3 ) in ( 1 ), Eint,B - EintA = - 250.76775 kJ + 269.675 kJ = 18.90725 kJ.

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