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
T₁V₁^γ-1=T₂V₂^γ-1
p=1.013x10⁵
mol= 1 mol

Given monoatomic gas, therefore f=3
C_v=f/2R=3/2(8.31)=12.465 J mol^-1 K^-1
C_p=C_v+R= 20.775 J mol^-1 K^-1
γ=C_p/C_v=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 T₁V₁^γ-1=T₂V₂^γ-1
therefore V₂^γ-1=T₁V₁^γ-1/T₂
therefore V₂= (T₁V₁^2/3/T₂)^3/2= 0.07586 m³

therefore P₂=nRT/v₂=4.2613x10⁴Pa

3) Under Isovolumetric process (ΔV=0; ΔW=0)
Q= nC_vΔT
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) ΔE_int=-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

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=nC_pΔT=(20.775)(231.3638-389)=-3274.8921J
ΔE_int=Q-W=-1964.9353J

3) ΔV=0; W=0
Since W=0
then ΔE_int=Q=nC_vΔT=(12.465)(550-231.3638)=3971.8Jc.
Since the whole thing is a cyclical process,
W=Q
which also means
W-Q=0 or the cumulative ΔE_int=0

I can't answer this one because,
after adding all the ΔE_int 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.

 

[Mindscrape]

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.

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

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.

 

[Archy]

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.