Reproduce a drawing of a Carnot cycle.
On your Carnot cycle drawing, shade the region that represents the heat consumed by the system at the higher temperature.
On the same drawing, now shade in a distinct manner the region that represents the heat that is unable to be converted to work.
Suppose we have a heat engine in which σ H - σ L = N 0 , where N 0 is Avogadro's number. Suppose further that T l = 300K and T h = 600K . Give the work done per cycle.
The work done is the product of the temperature difference and the entropy difference, so W = 300k B N 0 = 2493 Joules.
For an isothermal and isobaric process, what is the change in the Gibbs free energy?
We simply use the identity associated with the Gibbs free energy, and letting dτ and dp be zero, we find: dG = μ dN .
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