How Is Net Work and Heat Calculated in a Thermodynamic Cycle?

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The discussion focuses on calculating net work and heat in a thermodynamic cycle involving an ideal gas, with two isobaric and two isothermal legs. For the high-pressure isobaric leg, the net work is determined as 2PV. To find the total work done in the complete cycle, participants discuss summing the work for each leg while considering the signs of work done. The heat supplied to the gas is also derived from the properties of the isothermal processes, using the relationship W = nRT ln(V_B/V_A). The importance of interpreting the P-V diagram accurately to determine volume values for calculations is emphasized throughout the discussion.
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Homework Statement



A sample of an ideal gas is taken through the thermodynamic cycle
shown on the P-V diagram at attached. Two of the “legs” of the cycle are
isobaric and the other two are isothermal.


a) Find the net work done by the gas along the high pressure isobaric
leg in terms of p0 and Vo.


b) Find the net work done by the gas in one complete cycle in terms of
p0 and Vo.

c) Find the net heat supplied to the gas in one cycle in terms of p0 and
Vo.



Homework Equations





The Attempt at a Solution



Part A)

W = integral fr v to 3v of Pdv = P(3v - v) = 2PV

Part B) Not sure if this is crrect do i just find the work for each leg of the of the cycle and then sum them up

Ill start with the high pressure isbaric leg and work around the cycle clockwise

Leg 1 = 2PV (Part a)

Leg 2 = not sure how to integrate this if i don't knw what the v final is... i guess I am doing this wrong
 

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joemama69 said:
Part B) Not sure if this is crrect do i just find the work for each leg of the of the cycle and then sum them up

Kind of, but you have to consider that sometimes the work is positive and sometimes it's negative. Remember that work is equal to the area enclosed by the PV curve and you can adjust your positives and negatives accordingly.

Leg 2 = not sure how to integrate this if i don't knw what the v final is... i guess I am doing this wrong

You do know what the final v is because PV=nRT, and if temperature is constant, then P=C/V.
 
so to find the area, i woud add the first two integrals then subracts the second two.

for the second integral my book says fr a isothermal process W = nRT(ln V_B/V_A), so does that simplify to ln V_B/V_A because it is constant and nRT is essentially 1

for the third leg, the pressure does nt change but the volue does so it is nt constand, how do i find the v final fr this integral
 
joemama69 said:
so to find the area, i woud add the first two integrals then subracts the second two.
Yes.

for the second integral my book says fr a isothermal process W = nRT(ln V_B/V_A), so does that simplify to ln V_B/V_A because it is constant and nRT is essentially 1

How do you know nRT is 1? nRT is equal to PV, and you know PV.

for the third leg, the pressure does nt change but the volue does so it is nt constand, how do i find the v final fr this integral

You're supposed to read it from the graph. It looks like 2Vo to me.
 
of first off V_b is the secnd point in the cycle and V_c is the third point

W = nRT ln V_c/V_b = PV ln Vc/Vb

P(3V) = .5P(Vc) therefore Vc = 6V and Vb = 3V

W = PV ln Vc/Vb = PV ln 6V/3V = PV ln2
 
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