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Thermodynamics Homework Question of a piston cylinder?
Question:
A gas contained in a piston-cylinder assembly undergoes two processes, A and B, between the same end states, 1 and 2, where p1 = 1 bar, V1 = 1 m3, U1 = 400 kJ and p2 = 10 bar, V2 = 0.1 m3, U2 = 450 kJ:
Process A: Constant-volume process from state 1 to a pressure of 10 bar, followed by a constant-pressure process to state 2.
Process B: Process from 1 to 2 during which the pressure-volume relation is pV = constant.
Kinetic and potential effects can be ignored. For each of the processes A and B, (a) sketch the process on p-V coordinates, (b) evaluate the work, in kJ, and (c) evaluate the heat transfer, in kJ.
I have an Idea of the equations that I have to use but the set up of the question is confusing me.How would I work out process A and B?Full details would be appreciated.
2 Answers
- az_lenderLv 75 years agoFavorite Answer
The gas violates the ideal gas law fairly dramatically, since p1 V1 = p2 V2. If the gas were ideal, then T1 and T2 would have to be equal, so U1 and U2 would be equal. I'll do the question anyway; turns out the ideal gas law is not needed to complete this particular problem.
(b) W = (10 bar)(-0.9 m^3) = -900 kJ. (Process A)
For Process B, you have:
W = integral from V1 to V2 of p dV, where p = p1 V1/V,
so W = p1 V1 ln |1/10| = 100kJ * ln(1/10).= -230 kJ.
(c) Heat transfer: in Process A, the gas must lose 850 kJ of heat, since it gains only 50 kJ of internal energy when 900 kJ of work are done on the gas.
In Process B, the gas must lose 180 kJ of heat; same reasoning as Process A.