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Emspak
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Homework Statement
Let the radiant energy in a cylinder be carried through a Carnot cycle. The cycle consists of an isothermal expansion at temperature T, an infinitesimal adiabatic expansion where the temperature drops to T-dT, ad returning to the original state via an isothermal compression and adiabatic compression. What is the total work in the cycle? Assume P= u / 3 where u is a function of T only.
Homework Equations
To find the work in a Carnot cycle [itex] W= \int_{V_1}^{V_2} PdV[/itex], and a similar expression for points 2 to 3, 3 to 4 and 4 to 1.
The Attempt at a Solution
The expansion from V1 to V2 would therefore look like this:
[itex]P = u/3 [/itex]
so for the first leg of the cycle [itex]W = \int \frac{u}{3} dV[/itex] and since the first leg is isothermal that would mean P is constant. So [itex]W=\frac{1}{3}u(T)(V_2-V_1)[/itex]
for the second leg of the cycle we can work from the total energy of the system. [itex]U = u(T)V[/itex]. And [itex]dU=u(T)dV + V\frac{du}{dT}dT[/itex]. Since the change of energy is equal to the work done, we can say [itex] W = u(T)dV + V\frac{du}{dT}dT[/itex] and if we add that to the first work expression and plug in the relevant values for V we have [itex] u(T)dV + (V_3-V_2)\frac{du}{dT}dT+ \frac{1}{3}u(T)(V_2-V_1)[/itex].
Since the cycle's third and fourth leg look the same as the first two (except with opposite signs and T-dT in the temperature) we can add up all four and we have:
[itex] -u(T)dV + (V_1-V_4)\frac{du}{dT}dT- \frac{1}{3}u(T)(V_4-V_3) + u(T)dV + (V_3-V_2)\frac{du}{dT}dT+ \frac{1}{3}u(T)(V_2-V_1).[/itex].
The dT terms cancel out
[itex] -u(T)dV + (V_1-V_4)du- \frac{1}{3}u(T)(V_4-V_3) + u(T)dV + (V_3-V_2)du+ \frac{1}{3}u(T)(V_2-V_1).[/itex].
and the u(T)dV terms seem to as well. But I wanted to check if I was going about this the right way. The answer we are supposed to get has a du in the nominator...