# Thermodynamics - efficiency from pv diagram

portofino

## Homework Statement

Determine the efficiency for the cycle shown in the figure, using the definition. Assume that the gas in the cycle is ideal monatomic gas.

Compare with the efficiency of a Carnot engine operating between the same temperature extremes.

see attachment

## Homework Equations

the cycle is composed of two isobaric and two isovolumetric paths.

isovolumetric/constant volume:
work, W = 0
heat, Q = deltaU = nC_v(deltaT) where deltaU is change in internal energy, n is number of moles, C_v is constant volume (monotomic so = 3R/2 where R = 8.314), delta T is change in temp

isobaric/constant pressure
work, W = p(V_2 - V_1) where p is pressure in pascals, V_2 is final volume, V_1 is initial volume
heat, Q = deltaU + W = nC_p(deltaT) where C_p is constant pressure = C_v + R

ideal gas law: PV = nRT

efficiency, e = W/Q_h = (Q_h - Q_c)/Q_h = 1 - (Q_c/Q_h) where W is work, Q_h is heat energy (high), and Q_c is cool heat energy (low)

## The Attempt at a Solution

i intend to use efficiency, e = W/Q_h, since W = 0 for each of the two isovolumetric paths, the work only comes from the two isobaric paths, i calculated:

top horizontal path to be W = 2431.8 joules, and the bottom horizontal path to be W = 1215.9 joules, so net work would be W_net = 3647.7 joules

now in order to solve for Q_h or even Q_c for that matter I need to be able to solve for heat Q, in order for that i need to know the number of moles and the temp in kelvin.

since i am only concerned with Q_h, do i only need to know the heat energy Q from the top horizontal (isobaric) path?

so Q = nC_p(deltaT) and using ideal gas law and solving for T and subsituting in for T is the constant pressure heat equation i get:

Q = nC_p(PV/nR) so the n's cancel out, and C_P = C_v + R = 3(8.314)/2 + 8.314 = 20.785, here is where i am confused. since the top horizontal path is isobaric, the volume changes thus what volume do i use for V?

so currently my equation for Q would be Q = nC_p(PV/nR) = 20.785[((607950Pa)(V))/8.314]

any help/clarification appreciated

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Gold Member
so Q = nC_p(deltaT) and using ideal gas law and solving for T and subsituting in for T is the constant pressure heat equation i get:

Q = nC_p(PV/nR) so the n's cancel out, and C_P = C_v + R = 3(8.314)/2 + 8.314 = 20.785, here is where i am confused. since the top horizontal path is isobaric, the volume changes thus what volume do i use for V?

so currently my equation for Q would be Q = nC_p(PV/nR) = 20.785[((607950Pa)(V))/8.314]

any help/clarification appreciated

Remember you're looking for ΔT. You need to find the temperature at each point using V1 and V2 then take the difference in temperature.

scrplyr
i can't see the attachment for some reason, but I am pretty sure its the same problem i have already.
*i think you calculated the work wrong - the work for the second (bottom isobaric path) should be negative, so the total work will end up being 1215 J.

Then you need to find Qh (heat absorbed) in order to get the efficiency; e=W/Qh

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scrplyr
oh and for finding Q (for the top isobaric path) you would use the equation Q=n*Cp*deltaT
You know it's a monatomic ideal gas, so Cp = (5/2)R
Since you don't have n or deltaT, you use the ideal gas law: P*deltaV=n*R*deltaT
Solve for deltaT and plug in the Q equation (the n, and R will end up canceling out)
The rest should follow the same way... I got the right answers, but for the other Q values, I think I assumed some things and I just happened to get lucky with the other numbers - so I cant' really help you out on the rest of it =/

portofino
i tried what you said and got Q = 6079.5 joules for top isobaric, and Q = -1823.85 joules for bottom isobaric

so e = W/Q_h = 1215/(6079.5-1823.85) = 1215/4255.65 = 28.5% which was wrong, what now?

scrplyr
remember that Qh is heat absorbed. So if you got a negative number for Q of the bottom isobaric path, that would be Qc not Qh. I can't really compare my numbers with yours because I kept the units in atm. and liters. But solve Q for the isovolumetric paths (one will be negative and the other will be positive - use the positive number for Qh)

portofino
i still can't seem to get it, how do i find the q for the isvolumetric paths when Q = 3/2(P*deltaV) where pressure changes over each pf the two isovolumetric paths?

Gold Member
For an ideal gas ΔU = nCvΔT

For a monatomic ideal gas Cv = (3/2)R

Therefore, ΔU = (3/2)nRΔT

Because of the ideal gas law, nRΔT = Δ(PV) if n is fixed.

Therefore, ΔU = (3/2)Δ(PV)

portofino
let 6atm, 2L = 1, 6atm,6L = 2, 3atm, 6L = 3, 3atm, 2L = 4:

1-->2 isobaric expansion
Q = nC_pT = n(5R/2)(PV/nR) = 5/2(PV) = 5/2(607950)(0.006-0.002) = 6079.5

2-->3 isovolumetric compression
Q = n(3R/2)(PV/nR) = 3/2(PV) = 3/2(303975-607950)(0.006) = -2735.8

3-->4 isbaric compression
Q = n(5R/2)(PV/nR) = 5/2(PV) = 5/2(303975)(0.002-0.006) = -3039.75

4-->1 isovolumetric expansion
Q = 3/2(PV) = 3/2(6-7950 - 303975)(0.002) = 911.93

work, W = (607950*0.004) - (303975*0.004) = 1215.9

e = W/Q_h = 1215.9/6079.5 = 20% which is incorrect

now what?

and assuming carnot efficiency e_c = 1 - T_c/T_h

at 1, T = PV/nR = 607950(0.002) = 1215.9 holding nR constant
at 2, T = 607950(0.006) = 3647.7 ------> T_h
at 3, T = 303975(0.006) = 1823.85
at 4, T = 303975(0.002) = 607.95 ----> T_c

so carnot efficiency = 1 - (T_c/T_h) = 1 - (607.95/3647.7) = 0.83 = 83.33%

correct??

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farhant1
Your efficiency for the first part should come out to be 17%.
But now I am stuck at "Compare with the efficiency of a Carnot engine operating between the same temperature extremes". How do I do that?

portofino
i think it is a ratio of e_carnot/e where e = 17%

^^^^that ratio is correct, e_carnot/e = 83.33/17 = 4.90

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Gold Member
For qh, you didn't take into account the heat absorbed during the 4-->1 isometric heating.

The Carnot efficiency that you calculated is correct.

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