At point D in the figure below, the pressure and temperature of 2.00 mol of an i

Question

At point D in the figure below, the pressure and temperature of 2.00 mol of an ideal monatomic gas are 2.00 atm and 360 K, respectively. The volume of the gas at point B on the PV diagram is three times that at point D and its pressure is twice that at point Paths AB and CD represent isothermal processes. The gas is carried through a complete cycle along the path DABCD. Determine the total amount of work done by the gas and the heat absorbed by the gas along each portion of the cycle. W_by gas,tot= Kj Q_D rightarrow A = Kj Q_A rightarrow B = Kj Q_B rightarrow C = Kj Q_B rightarrow C = Kj Q_c rightarrow D = Kj

Explanation / Answer

the volume at point D:

   VD = n R TD / PD = 2 X 8.314 X 360 / 2 X 101.3 = 29.5 L

volume at point B: VB = VC = 3VD = 88.5 L

pressure at point C: PC =n R TC / VC = 2 X 8.206 X 10-2 X 360 / 88.5

                                    PC = 0.667 atm

Similarly, B = 2 PC, then PB = 1.33 atm and TD = TC = 360K

temperature: TA = TB = PB VB / n R = 1.33 X 88.5 / 2 X 8.206 X 10-2 = 720K

pressure at A: PA = 2 PD = 4 atm

heat energy: QD-->A = 3 / 2 n R dT = 3/2 X 2 X 8.314 X (720-360) = 8.98 kJ

heat energy: QA-->B = nRT ln( VB / VA ) = 2 X 8.314 X 720 X ln ( 88.5 / 29.5 ) = 13.1 kJ

heat energy: QB-->C = 1 / 2 nR(TC-TB ) = 1 /2 X 2 X 8.314 X (360-720) = -8.98kJ

heat energy: QC-->D =n RTln(VD/VC) = 2 X 8.314 X 360 X ln(29.5/88.5) = -6.58 kJ

hence, the total work done: Wtotal= 6.62 kJ

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