A thermodynamic process is shown in the figure. The pressures and volumes corresponding to some points in the figure are P_A=3×10⁴ Pa, V_A=2×10⁻³ m³, P_B=8×10⁴ Pa, V_D=5×10⁻³ m³. In the process AB, 600 J of heat is added to the system and in process BC, 200 J of heat is added to the system. The cha…

Question

A thermodynamic process is shown in the figure. The pressures and volumes corresponding to some points in the figure are P_A=3×10⁴ Pa, V_A=2×10⁻³ m³, P_B=8×10⁴ Pa, V_D=5×10⁻³ m³. In the process AB, 600 J of heat is added to the system and in process BC, 200 J of heat is added to the system. The change in internal energy of the system in process AC would be

Accepted Answer

(1) 560 J. Given: P_A=3×10⁴ Pa, V_A=2×10⁻³ m³, P_B=8×10⁴ Pa, V_D=5×10⁻³ m³; Q_AB=600 J and Q_BC=200 J. Idea: Δ U depends only on the endpoints, so the change along AC equals the change along ABC — and that is easier to compute. AB is vertical on the diagram, so the volume is unchanged and W_AB=0. BC is horizontal at P_B, from V_A to V_D: W_BC=8×10⁴×(5-2)×10⁻³=240 J. Total heat along ABC: Q=600+200=800 J. Δ U=Q-W=800-240=560 J.

Step-by-step solution

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