The discussion focuses on analyzing a cyclic process of an ideal gas involving calculations of entropy changes and work done. It is established that the change in entropy of the system, ##\Delta S_{system}##, is zero due to the cyclic nature of the process. The total work done by the system is calculated using the equation ##W_{net} = W_{1-2} + W_{2-3} + W_{3-1}##, where ##W_{2-3} = 0## due to the isovolumetric process. The participants clarify that the heat absorbed during the cycle, ##Q_h = 890J##, is crucial for determining the entropy changes in the surroundings, particularly during the reversible processes.
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That sounds right to me.Thermofox said: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##
This means That the entropy of the universe increased by 0.41, right?Philip Koeck said:That sounds right to me.
Yes that sounds reasonable.Thermofox said: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)##?
I have a follow-up question (also to @Chestermiller):Thermofox said:Ok, now I feel like I have a better understanding of all of this. Thanks a lot!
Philip Koeck said:I have a follow-up question (also to @Chestermiller):
W1-2 has to be the same as -Qh so that ΔU becomes zero for a complete cycle.
Philip Koeck said:I have a follow-up question (also to @Chestermiller):
W1-2 has to be the same as -Qh so that ΔU becomes zero for a complete cycle.
If, however, I insist on calculating W1-2 as Pext ΔV I'm a bit stuck.
With Pext 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 Pext 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).Philip Koeck said:If, however, I insist on calculating W1-2 as Pext ΔV I'm a bit stuck.
With Pext 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 Pext can't be higher than P2.