Maximum work done in a Carnot Cycle

[BrianSauce]

Given that in a Carnot Cycle the two adiabatic processes are essentially equal and opposite in magnitude the total work done by the cycle is in the two isotherms. The total work of the system is generally given as -NR(Th-Tc)ln(Vb/Va). Does this mean that the work done by a monatomic ideal gas is the same as the work done by a diatomic ideal gas since there is no dependence on gamma?

 

[Chestermiller]

Given that in a Carnot Cycle the two adiabatic processes are essentially equal and opposite in magnitude the total work done by the cycle is in the two isotherms. The total work of the system is generally given as -NR(Th-Tc)ln(Vb/Va). Does this mean that the work done by a monatomic ideal gas is the same as the work done by a diatomic ideal gas since there is no dependence on gamma?

Yes. So...?

 

[BrianSauce]

I'm trying to show the ratio of W'/W where W' is the work done by a diatomic gas in one carnot cycle and W is the work done by a monatomic gas in one carnot cycle. Based on the above, shouldn't this ratio be 1? However I see in my textbook that this is 1/3.

 

[Chestermiller]

They must be assuming.toat something is being held constant. What is the exact wording?

 

[BrianSauce]

The exact wording is that the T_h is 4 times larger than T_c and the ratio of maximum to minimum pressure is 64, i.e. P₁/P₃ = 64.

 

[Chestermiller]

The exact wording is that the T_h is 4 times larger than T_c and the ratio of maximum to minimum pressure is 64, i.e. P₁/P₃ = 64.

Okay. The key to this is P1/P3 is fixed. This is going to change things between the monoatomic and the diatomic cases. Work out all the equations in terms of the heat capacities, and see what you get.

Chet

 

[BrianSauce]

This is the work I have done so far. I have established that W'/W = ln(V'_B/V_a)/ln(V_B/V_A) and I also found that the ratio of V_C/V_A = 16 using the pressure ratio. Points A, B, C, and D refer to the same points as 1, 2, 3, and 4 respectively.