Physics

Physics

Problem

A fixed quantity of ideal gas is taken around the cycle A→B→C→D→A with the following corner states: A: PA=1.0×105 Pa,  VA=0.02 m3P_A = 1.0\times10^{5}\,\text{Pa},\; V_A = 0.02\,\text{m}^3 B: PB=1.0×105 Pa,  VB=0.06 m3P_B = 1.0\times10^{5}\,\text{Pa},\; V_B = 0.06\,\text{m}^3 C: PC=3.0×105 Pa,  VC=0.06 m3P_C = 3.0\times10^{5}\,\text{Pa},\; V_C = 0.06\,\text{m}^3 D: PD=3.0×105 Pa,  VD=0.02 m3P_D = 3.0\times10^{5}\,\text{Pa},\; V_D = 0.02\,\text{m}^3 The paths AB and CD are isobars, while BC and DA are isochores. a. (i) Draw a qualitative PP–VV diagram of the cycle showing the four states and the direction of traversal. b. (ii) Calculate the work done by the gas in each of the four processes. c. (iii) Hence obtain the net work done over one complete cycle. d. (iv) Determine both the magnitude and the direction (absorbed or rejected) of the net heat exchanged by the gas in this cycle.
Intuitive explanation

This cycle traces a rectangle on a PP–VV diagram, alternating between constant-pressure and constant-volume processes. The key insight is that for a closed cycle, the net work done by the gas equals the area enclosed by the loop on the PP–VV plot—and since internal energy returns to its starting point, this work equals the net heat absorbed. Calculate the rectangular area to find the cycle's efficiency and energy flow.

Intuitive explanation

This cycle traces a rectangle on a PP–VV diagram, alternating between constant-pressure and constant-volume processes. The key insight is that for a closed cycle, the net work done by the gas equals the area enclosed by the loop on the PP–VV plot—and since internal energy returns to its starting point, this work equals the net heat absorbed. Calculate the rectangular area to find the cycle's efficiency and energy flow.

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