Reversible expansion according to a law pv^2 = constant

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The discussion revolves around a thermodynamic cycle involving a fluid undergoing reversible processes defined by specific pressure-volume relationships. Initially, the fluid is compressed from 0.7 bar to 3.5 bar, and the value of n for this process is calculated to be 1.847. Following this, the fluid is heated at constant volume until reaching 4 bar, resulting in a specific volume of 0.5 m3/kg. A reversible expansion according to the law pv^2 = constant then returns the fluid to its original state. The mass of the fluid is determined to be 0.0753 kg, and the net work done during the cycle is calculated to be -640.
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A fluid at 0.7bar occupying 0.009m3 is compressed reversibly to a pressure of 3.5bar according to a law pv^n = constant. The fluid is then heated reversibly at constant volume until the pressure is 4bar; the specific volume is then 0.5m3/kg . A reversible expansion according to a law pv^2 = constant restores the fluid to its initial state. Sketch the cycle on a p-v diagram and calculate:
I. The mass of fluid present
II. The value of n in the first process
III. The net work of the cycle
Answers:
0.0753kg; 1.847; -640
 
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