Efficiency of Carnot engine

[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.

 

[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.

 

[Von Neumann]

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.

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.

 

[phyzguy]

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

 

[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.

 

[phyzguy]

You got it. Glad to help.