# Homework Help: Efficiency of Carnot engine

1. Feb 28, 2013

### Von Neumann

The problem:
An engine puts an ideal monatomic gas through a clockwise rectangular cycle on a PV diagram with horizontal and vertical sides. The lower left point has a pressure of 1 atm and a volume of 1m^3 and the upper right point has pressure and volume three times greater. Calculate the efficiency of a Carnot engine operating between the highest and lowest temperatures.

Solution (so far):
I know that for a Carnot engine e=1-T_c/T_h, but without being given the temperature differences I'm not exactly sure how you'd begin. I calculated the efficiency of the engine itself to be 22.2%, the work done in the cycle to be 4.04x10^7 J, and the heat absorbed in the cycle to be 434 kcal; if any of those quantities can be related to T_c/T_h. The answer comes to be 88.9%. Thanks in advance for any help.

Last edited: Feb 28, 2013
2. Feb 28, 2013

### phyzguy

Try using the ideal gas law. Since you are given the ratios of pressures and volumes, you should be able to calculate the ratio of temperatures.

3. Feb 28, 2013

### Von Neumann

Since there are 2 unknowns, namely the temperature difference and the number of moles, I don't see an obvious way that I could use the ideal gas equation to solve this problem.

4. Feb 28, 2013

### phyzguy

I promise you it will work. Can you write the ideal gas equation at Tc and Th?

5. Feb 28, 2013

### Von Neumann

Oh nevermind,

(P_3*V_3)/(P_1*V_1)=T_3/T_1
(3*P*3*V)/(PV)=T_3/T_1
9=T_3/T_1=T_h/T_c

Therefore,

T_c/T_h=1/9

So,

e=1-1/9=8/9=88.9%

Thanks! Sorry, for posting such an obvious question; just a little low on sleep.

6. Feb 28, 2013

### phyzguy

You got it. Glad to help.

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