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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 T

_{1}=550K to T

_{2}=389K

2) Compression at constant pressure until T=T

_{3}

3) Return to original pressure and temperature by a constant-volume process

Find

a. T

_{3}

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

T

_{1}V

_{1}

^{γ-1}=T

_{2}V

_{2}

^{γ-1}

p=1.013x10

^{5}

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 V

_{1}=nRT

_{1}/P=0.04512 m

^{3}

Using T

_{1}V

_{1}

^{γ-1}=T

_{2}V

_{2}

^{γ-1}

therefore V

_{2}

^{γ-1}=T

_{1}V

_{1}

^{γ-1}/T

_{2}

therefore V

_{2}= (T

_{1}V

_{1}

^{2/3}/T

_{2})

^{3/2}= 0.07586 m

^{3}

therefore P

_{2}=nRT/v

_{2}=4.2613x10

^{4}Pa

3) Under Isovolumetric process (ΔV=0; ΔW=0)

Q= nC

_{v}ΔT

P

_{2}=P

_{3}=4.2613x10

^{4}Pa since 2) is an isobaric process

P

_{4}=1.013x10

^{5}

T

_{4}=550K

Using PV=nRT; ΔV=0

then V

_{3}=V

_{4}

then nRT

_{3}/P

_{3}=nRT

_{4}/P

_{4}

therefore, T

_{3}=T

_{4}P

_{3}/P

_{4}=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 [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); T

_{3}=231.3638K

W=nRΔT=(8.31)(231.3638-389)J=-1309.9568J

Q=nC

_{p}ΔT=(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

_{v}ΔT=(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 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.

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