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

Question

A thermodynamic process is shown in the figure. The pressures and volumes corresponding to some points in the figure are Pₐ = 3 × 10⁴ Pa, Vₐ = 2 × 10⁻³ m³, Pᵦ = 8 × 10⁴ Pa, Vᵦ = 5 × 10⁻³ m³. In 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

560 J. The change in internal energy (ΔU) for the entire process AC can be calculated using the first law of thermodynamics: ΔU = Q - W. The heat added (Q) in the entire process AC is the sum of heat added in AB and BC, i.e., Q = 600 J + 200 J = 800 J. The work done (W) is the area under the PV curve, which is the work done in AB (W_AB = PΔV = 3 × 10⁴ × (5 × 10⁻³ - 2 × 10⁻³) = 90 J) plus the work done in BC (W_BC = 0, as volume is constant). Thus, W = 90 J. Therefore, ΔU = 800 J - 90 J = 560 J.

Solution

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