The total work in a Carnot cycle - checking some work

[Emspak]

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 W= \int_{V_1}^{V_2} PdV, 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:

P = u/3
so for the first leg of the cycle W = \int \frac{u}{3} dV and since the first leg is isothermal that would mean P is constant. So W=\frac{1}{3}u(T)(V_2-V_1)

for the second leg of the cycle we can work from the total energy of the system. U = u(T)V. And dU=u(T)dV + V\frac{du}{dT}dT. Since the change of energy is equal to the work done, we can say W = u(T)dV + V\frac{du}{dT}dT and if we add that to the first work expression and plug in the relevant values for V we have u(T)dV + (V_3-V_2)\frac{du}{dT}dT+ \frac{1}{3}u(T)(V_2-V_1).


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:

-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)..

The dT terms cancel out

-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)..

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...

 

[Andrew Mason]

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 W= \int_{V_1}^{V_2} PdV, 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:

P = u/3
so for the first leg of the cycle W = \int \frac{u}{3} dV and since the first leg is isothermal that would mean P is constant. So W=\frac{1}{3}u(T)(V_2-V_1)

for the second leg of the cycle we can work from the total energy of the system. U = u(T)V. And dU=u(T)dV + V\frac{du}{dT}dT. Since the change of energy is equal to the work done, we can say W = u(T)dV + V\frac{du}{dT}dT and if we add that to the first work expression and plug in the relevant values for V we have u(T)dV + (V_3-V_2)\frac{du}{dT}dT+ \frac{1}{3}u(T)(V_2-V_1).


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:

-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)..

The dT terms cancel out

-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)..

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...

Can you use the efficiency of the Carnot engine operating between T and T-dT to determine the work produced as a function of dT and U?

AM