A thermodynamic process is shown in the following figure. The pressure and volumes corresponding to some points in the figure are: P A = 2 × 10 5 Pa, P B = 5 × 10 5 Pa, V A = 1 × 10 -3 m 3 , V D = 4 ×

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A thermodynamic process is shown in the following figure. The pressure and volumes corresponding to some points in the figure are: P_A = 2 × 10⁵ Pa, P_B = 5 × 10⁵ Pa, V_A = 1 × 10^-3 m³, V_D = 4 × 10^-3 m³. In the process AD, 1000 J of heat is added to the system and in DC, 750 J of heat is added. The heat absorbed by the system in the direct process AC (along the straight line) would be

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AD is at constant pressure P A : W AD = 2 × 10 5 × 3 × 10 -3 = 600 J, so Δ U AD = 1000 - 600 = 400 J. DC is at constant volume: W = 0, Δ U DC = 750 J. Δ U AC = 1150 J (state function). Along the straight line AC, work = trapezoid area = ((2+5) × 10 5 )/2 × 3 × 10 -3 = 1050 J. Q AC = Δ U + W = 1150 + 1050 = 2200 J. Matches (a).

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