Maximum work done in a Carnot Cycle

In summary, the work done by a diatomic gas in one Carnot cycle is 16 times the work done by a monatomic gas, but this is only true if the temperature is held constant.
  • #1
BrianSauce
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1
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?
 
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  • #2
BrianSauce said:
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...?
 
  • #3
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.
 
  • #4
They must be assuming.toat something is being held constant. What is the exact wording?
 
  • #5
The exact wording is that the Th is 4 times larger than Tc and the ratio of maximum to minimum pressure is 64, i.e. P1/P3 = 64.
 
  • #6
BrianSauce said:
The exact wording is that the Th is 4 times larger than Tc 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
 
  • #7
This is the work I have done so far. I have established that W'/W = ln(V'B/Va)/ln(VB/VA) and I also found that the ratio of VC/VA = 16 using the pressure ratio. Points A, B, C, and D refer to the same points as 1, 2, 3, and 4 respectively.
 

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What is a Carnot Cycle?

A Carnot Cycle is a theoretical thermodynamic cycle that involves a reversible process of a working fluid between two heat reservoirs at different temperatures. It is named after French physicist Nicolas Léonard Sadi Carnot.

What is the maximum work done in a Carnot Cycle?

The maximum work done in a Carnot Cycle is given by the difference in temperature between the two heat reservoirs, divided by the temperature of the hotter reservoir, multiplied by the heat absorbed from the hotter reservoir. This is known as the Carnot efficiency and is a fundamental limit for the efficiency of any heat engine.

What factors affect the maximum work done in a Carnot Cycle?

The maximum work done in a Carnot Cycle is affected by the temperature difference between the two heat reservoirs, the type of working fluid used, and the efficiency of the heat engine. It also depends on the reversibility of the process, as a reversible process would result in a higher maximum work done.

How is the maximum work done in a Carnot Cycle calculated?

The maximum work done in a Carnot Cycle is calculated using the Carnot efficiency formula: W = (Th - Tl)/Th * Qh, where W is the work done, Th is the temperature of the hotter reservoir, Tl is the temperature of the colder reservoir, and Qh is the heat absorbed from the hotter reservoir.

What is the practical application of understanding the maximum work done in a Carnot Cycle?

Understanding the maximum work done in a Carnot Cycle is important in thermodynamics as it helps in the design and optimization of heat engines. It also provides a theoretical limit for the efficiency of any heat engine, which is useful in practical applications such as power plants and refrigeration systems.

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