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 C. 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. kJ kJ kJ by gas,tot kJ

Explanation / Answer

Given,

Pressure, P = 2 atm

Temperature, T = 360 K

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 = 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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