Carnot Cycle: Isothermal Expansion Phase - Work Done?

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SUMMARY

The work done during the Isothermal Expansion Phase of a Carnot cycle is equal to the energy extracted from the hot reservoir, as the temperature remains constant. This is established by the equation ΔQ = ΔU + ΔW, where ΔU equals zero, leading to ΔQ = ∫PdV. The work done can be calculated as the area under the PV graph, resulting in the formula nRTln(Vi/Vf), where n is the number of moles, R is the ideal gas constant, and T is the constant temperature.

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  • Thermodynamics: Isothermal processes
  • PV diagrams and area calculations
  • Ideal gas law (PV = nRT)
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Hi. I have a qualitative question about the work done from state 1 to state 2 of a Carnot cycle. This is the Isothermal Expansion Phase. Is work done by the engine on the environment during THIS PHASE ONLY equal to the energy extracted from the hot reservoir to keep the temperature constant. I feel that it should be less. Is my reasoning incorrect?
 
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G01 said:
Hi. I have a qualitative question about the work done from state 1 to state 2 of a Carnot cycle. This is the Isothermal Expansion Phase. Is work done by the engine on the environment during THIS PHASE ONLY equal to the energy extracted from the hot reservoir to keep the temperature constant. I feel that it should be less. Is my reasoning incorrect?
Yes.

Since the temperature does not change, the heat flow into the gas is equal to the work done: (\Delta Q = \Delta U + \Delta W = \Delta U + \int PdV; Since \Delta U = 0, \Delta Q = \int PdV)

The work done is the area under the PV graph for the isothermal expansion phase. Since the isotherm is T = PV/nR = constant, it is of the form P = K/V where K is constant = nRT. The area under that graph (\int PdV = K\int dV/V) is nRTln(Vi/Vf).

AM
 
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Thanks a lot Andrew. Man I love having this website:smile:
 

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