Canot Cycle, Work done

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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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Andrew Mason
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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?
Yes.

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

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 ([itex]\int PdV = K\int dV/V[/itex]) is nRTln(Vi/Vf).

AM
 
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G01
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Thanks alot Andrew. Man I love having this website:smile:
 

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