Entropy change of surroundings in a cyclic ideal gas process

[Philip Koeck]

Ok.
Q_{2-3}=Q_h
=> $Q_h= nc_v \Delta T= 0.5 \frac 3 2 8.31 (T_3 - T_2)$, hence $$ T_2 = \frac {nc_vT_3 - Q} {nc_v}= 150.36 \frac J K$$

If I use this temperature I obtain that $\Delta S_{4-2}= -11\frac J K$
=>$\Delta S_{1-2}= 0.41\frac J K$

That sounds right to me.

 

[Thermofox]

That sounds right to me.

This means That the entropy of the universe increased by 0.41, right?
I found $\Delta S_{2-3,sys}= 4.16 \frac J K$ and $\Delta S_{3-1,sys}= -4.56\frac J K$
If I sum them I obtain that $\Delta S_{sys}=0.01 \frac J K$, but I think I can consider this as an error of approximation.
Moreover does this mean that $\Delta S_{surr.}= - \Delta S_{sys,2-3} + (-\Delta S_{sys, 3-1})= -4.16 + 4.56= 0.40 (\frac J K)$?

 

[Philip Koeck]

This means That the entropy of the universe increased by 0.41, right?
I found $\Delta S_{2-3,sys}= 4.16 \frac J K$ and $\Delta S_{3-1,sys}= -4.56\frac J K$
If I sum them I obtain that $\Delta S_{sys}=0.01 \frac J K$, but I think I can consider this as an error of approximation.
Moreover does this mean that $\Delta S_{surr.}= - \Delta S_{sys,2-3} + (-\Delta S_{sys, 3-1})= -4.16 + 4.56= 0.40 (\frac J K)$?

Yes that sounds reasonable.
Without rounding errors the entropy for the system should be unchanged after one cycle, whereas the entropy of the surroundings should have increased (by about 0.41 J/K)

 

[Thermofox]

Ok, now I feel like I have a better understanding of all of this. Thanks a lot!

 

[Philip Koeck]

Ok, now I feel like I have a better understanding of all of this. Thanks a lot!

I have a follow-up question (also to @Chestermiller):

W_1-2 has to be the same as -Q_h so that ΔU becomes zero for a complete cycle.

If, however, I insist on calculating W_1-2 as P_ext ΔV I'm a bit stuck.
With P_ext equal to P₂, the internal pressure in state 2 (which seemed the obvious choice to me), I get only about half of -890 J.
Surely P_ext can't be higher than P₂.

 

[Chestermiller]

I have a follow-up question (also to @Chestermiller):

W_1-2 has to be the same as -Q_h so that ΔU becomes zero for a complete cycle.

I have a follow-up question (also to @Chestermiller):

W_1-2 has to be the same as -Q_h so that ΔU becomes zero for a complete cycle.

If, however, I insist on calculating W_1-2 as P_ext ΔV I'm a bit stuck.
With P_ext equal to P₂, the internal pressure in state 2 (which seemed the obvious choice to me), I get only about half of -890 J.
Surely P_ext can't be higher than P₂.

If, however, I insist on calculating W_1-2 as P_ext ΔV I'm a bit stuck.
With P_ext equal to P₂, the internal pressure in state 2 (which seemed the obvious choice to me), I get only about half of -890 J.
Surely P_ext can't be higher than P₂.

Unless you know $P_{ext}$ explicitly for an irreversible expansion or compression, you can not determine the work (without solving the internal Navier Stokes equation, the continuity equation, and the differential thermal energy balances equation simultaneously).