- #1

Kelsi_Jade

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Problem:

One mole of ideal gas (cV=1.5 R) perform a Carnot cycle between the temperature 400 K and 300 K. On the upper isothermal transformation, the initial volume is 1 liter and the final volume is 5 liter. Find the work performed during a cycle and the amount of heat exchanged with the two sources.

I know that since the process is isothermal the temperature remains constant: ΔT=0 but Q≠0.

Vi=1L

Vf=5L

Th=400k

Tc=300k

Qh=Th(Sb-Sa) - where Sb is the maximum system entropy and Sa is the minimum system entropy

Qc=Tc(Sb-Sa)

W=[itex]\int[/itex]PdV=(Th-Tc)(Sb-Sa)

My confusion is where I need to find the entropy to solve for the heat exchanged bt the sources is Q is needed to find entropy??

If someone could let me know if I'm going in the right direction equation-wise, or help explain where to go next it would be appreciated.