Entropy and the second law of thermodynamics

[denniszhao]

Homework Statement

A Stirling engine operates between a hot reservoir at temperature TH=400K and a cold reservoir at temperature TC=300K. The working substance is n=0.15mol of ideal gas with gamma factor γ=1.4. Assume all heat going into the engine comes from the hot reservoir and all heat dissipated by the engine goes to the cold reservoir and compute the change of entropy for the universe each cycle.
I wanna first figure out the heat but I don't know their volume at that moment and how is gamma factor used in this problem.

Relevant Equations

Change of entropy for the universe=change of entropy for cold reservoir+that for hot reservoir=-QH/TH+QC/TC

 

[DEvens]

Maybe you can get some hints in the following link?

http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/shegas.html

 

[denniszhao]

Maybe you can get some hints in the following link?

http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/shegas.html

Thanks! It is to solve for the heat of the isochoric processes and now i still need the heat of those two isothermal processes which need the volume change.

 

[denniszhao]

What does "isochoric" mean? Where do you see such a process?

the stirling cycle contains two isothermal processes (constant temperature) and two isochoric processes (constant volume). you can check out the diagram attached above.

 

[Chestermiller]

Let the smaller volume be V1 and the larger volume be V2. In terms of V1 and V2, what are QH and QC? What are $\Delta S_H$ and $\Delta S_C$?

 

[Chestermiller]

Here's a hint: The net change in entropy of the system plus surroundings is zero for the two isothermal steps.

Also, pay particular attention to this: "Assume all heat going into the engine comes from the hot reservoir and all heat dissipated by the engine goes to the cold reservoir"

 

[rude man]

I wonder if this is doable.
I mean, the ratio of Vb/Va could be any number without changing the text of the problem, yet the change in entropy per cyccle is a function of this ratio.

the stirling cycle contains two isothermal processes (constant temperature) and two isochoric processes (constant volume). you can check out the diagram attached above.

I thought I deleted this post almost immediately I posted it; sorry, didn't look at the diagream carfully.

 

[Chestermiller]

I wonder if this is doable.
I mean, the ratio of Vb/Va could be any number without changing the text of the problem, yet the change in entropy per cyccle is a function of this ratio.

I thought I deleted this post almost immediately I posted it; sorry, didn't look at the diagream carfully.

Here is my final hint: The initial and final volumes are irrelevant.

 

[rude man]

Here is my final hint: The initial and final volumes are irrelevant.

That's because ln(VH/VL) - ln(VL/VH) = 0 right?

 

[Chestermiller]

That's because ln(VH/VL) - ln(VL/VH) = 0 right?

Yes, aside from the minus sign.

 

[rude man]

Yes, aside from the minus sign.

10-4.

 

[denniszhao]

Here is my final hint: The initial and final volumes are irrelevant.

but i can't cancel that part tho

 

[rude man]

@denniszhao,
EDIT: sorry, got my signs wrong. Here is correct:

$\Delta Q_C$ > 0. The cold reservoir gains entropy.
$\Delta Q_H$ < 0. The hot reservoir loses entropy.

 

[Chestermiller]

but i can't cancel that part tho

You made an algebra mistake. The log terms cancel.