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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
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
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
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)=3971.8Jc.
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.
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
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)=3971.8Jc.
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.
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