The state of an ideal gas can be represented by a point on a PV (pressure-volume

Question

The state of an ideal gas can be represented by a point on a PV (pressure-volume) diagram. If you know the quantity of gas, n, a unique point in pressure (P) and volume (V) can be used to determine a temperature (T). Each point on a PV diagram also has a single internal energy (U) assigned to it. If a process starts at a point and returns to that same point on a PV diagram, it returns to the same P, V, T, and U.

The PV diagram below shows four different states, A, B, C, and D. The lines connecting the states represent processes or transitions. For example, the line connecting states A and B represents an expansion of the gas (transition to larger volume) while the pressure is kept constant. In the case of this diagram, the pressure at states A and B is 4.68105 Pa. The pressure at states C and D is 1.92105 Pa. Likewise the volume at states A and D is 1.9310-3 m3 and the volume at states B and C is 7.5310-3 m3.

What is the work done by the gas for the transition BC?

What is the work done by the gas for the transition AB?

What is the work done by the gas for the transition CD?

What is the change in internal energy, ?U, if you follow the system all the way from A to B to C to D and back to A?

Explanation / Answer

Solution:

a) The process B to C is Isochoric. So work done is 0 since the change in volume is 0.

b) The process from A to B is isobaric since the pressure remains constant.

Work done in an isobaric process:

WA-->B = P delta V

             = ( 4.68 e5) [(7.53 e-3) - (1.93 e-3) ]

              = 2.62 x103 J

c) Work done for the transition CD:

W C--D = P delta V

            = (1.92 e5)[(1.93 e-3 ) - (7.5 e-3)]

            = -1.07 e3 J

here the transition is from a higher volume state C to a lower volume state D. (compression)

To find internal energy find the temperatures at A,B,C,D using PV = RT

T = PV/R

TA = Pa Va/R = (4.68 e5)(1.93 e-3) /(8.314) = 108.64 K

TB = (4.68 e5)(7.53 e-3)/(8.314) = 423.9 K

TC = (1.92 e5)(7.53 e-3) / 8.314 = 173. 9 K

TD = (1.92 e5) ( 1.93 e-3) / 8.314 = 44.57 K

delta U A-->B = 3/2 R (TB - TA) = 3/2 (8.314) (423.9 - 108.64) = 3931.607 J

UB-->C = 3/2 (8.314)(TC - TB) = 3/2 (8.314)(173.9 - 423.9 ) = -3117.75 J

UC-->D = 3/2(8.314)(TD - TC) = 3/2(8.314)(44.57 - 173.9) = -1612.8669 J

UD-->A = 3/2(8.314)(TA-TD) = 3/2(8.314) (108.64 - 44.57) = 799.017 J

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