Thermodynamics: Cyclic Processes

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Thermodynamics: Cyclic Processes (solved)

Sorry for the false alarm guys. It looked like I did use the wrong equation while finding the internal energy for case 1.
referred to this. Thanks.
https://www.physicsforums.com/showthread.php?t=412577

Given that
1 mol of ideal monoatomic gas at p= 1 atm undergoes a three step process:
1) Adiabetical expansion from T1=550K to T2=389K
2) Compression at constant pressure until T=T3
3) Return to original pressure and temperature by a constant-volume process

Find
a. T3
b. Change in internal energy, work done and heat added to the gas for each process
c. Change in internal energy, work done by the gas and heat added for a complete cycle

Homework Equations


Q=W
PV=nRT
T1V1γ-1=T2V2γ-1
p=1.013x105
mol= 1 mol

Given monoatomic gas, therefore f=3
Cv=f/2R=3/2(8.31)=12.465 J mol-1 K-1
Cp=Cv+R= 20.775 J mol-1 K-1
γ=Cp/Cv=5/3

The Attempt at a Solution



for a.
By process,
1) Under adiabatic expansion (ΔQ=0)
Using PV=nRT
then V1=nRT1/P=0.04512 m3

Using T1V1γ-1=T2V2γ-1
therefore V2γ-1=T1V1γ-1/T2
therefore V2= (T1V12/3/T2)3/2= 0.07586 m3

therefore P2=nRT/v2=4.2613x104Pa

3) Under Isovolumetric process (ΔV=0; ΔW=0)
Q= nCvΔT
P2=P3=4.2613x104Pa since 2) is an isobaric process
P4=1.013x105
T4=550K

Using PV=nRT; ΔV=0
then V3=V4
then nRT3/P3=nRT4/P4
therefore, T3=T4P3/P4=231.3638K

That aside, my real problem is in question b, actually

for b.
By process,
1) ΔEint=-W under adiabatic process,

I think that under this condition, W should probably be
[PLAIN]http://www.texify.com/img/%5Cnormalsize%5C%21%5Cdisplaystyle%20W%3D%20%5Cint_%7BVi%7D%5E%7BVf%7D%20%5Cfrac%20%7BnR%28T_2-T_1%29%7DV.gif [Broken]

But I'm not sure of it's validity.
Anyway eventually it'll lead to W=nR(389-550)(ln 0.07586-ln 0.04512)=-695.1298J
Which I doubt is correct.

2) Under isobaric process (ΔP=0); T3=231.3638K
W=nRΔT=(8.31)(231.3638-389)J=-1309.9568J
Q=nCpΔT=(20.775)(231.3638-389)=-3274.8921J
ΔEint=Q-W=-1964.9353J

3) ΔV=0; W=0
Since W=0
then ΔEint=Q=nCvΔT=(12.465)(550-231.3638)=3971.8J


c.
Since the whole thing is a cyclical process,
W=Q
which also means
W-Q=0 or the cumulative ΔEint=0

I can't answer this one because,
after adding all the ΔEint it appears that it's not 0.
So. Problem there. I still suspect is 1) that's causing the trouble.

Help there?
Also, if you may, please do look through the workings to see if there are any faults in there. Thank you.
 
Last edited by a moderator:

Answers and Replies

  • #2
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Yeah, your work seems fine other than part 1). Though, I cannot attest to your numbers, maybe setting your straight for the adiabatic process will fix it.

[tex]PV^\gamma = C[/tex]
[tex]W=\int P dV[/tex]
[tex]W = \int C V^{- \gamma} dV[/tex]
[tex]W = \frac{1}{- \gamma + 1} (CV_2^{-\gamma +1}- CV_1^{-\gamma +1})[/tex]
we know that
[tex]P_1 V_1^\gamma = C = P_2 V_2^\gamma[/tex]
so plug it in for C in the right spots
[tex]W_{adi}=\frac{1}{1-\gamma}(P_2 V_2 - P_1 V_1)[/tex]

P.S. I am always getting my sign confused and have to rely on physical intuition to get it right, so don't trust my sign.
 
  • #3
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Ohh. So that's how that formula came about.

Hmm. I actually used W= -nCvΔT? It worked sufficiently, in a surprising way.
But I was wondering how, since V1 and V2 are obviously not the same and Cv is the heat capacity at constant volume?

Thanks for the reply though. Really helpful.
 

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