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Thermal (PV diagram)

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Consider the cycle shown below for n moles of an ideal gas. The gas has a heat capacity of Cv=(3/2)nR and is initially at a temperature T0, a pressure P0 and a volume V0. The process a->b consists of a quasistatic, isobaric expansion to twice the initial volume. The process b->c consists of a quasistatic, isochoric decrease in pressure and the process c->a is a quasistatic , adiabatic compression.

pv.JPG

(a) In terms of the initial pressure at point a, P0, what is the pressure at the point c?

Worked this out to be Pc = 2P0

(b) In terms of the initial temperature at point a, T0, what is the temperature at points b and c.

Worked these out to be Tb = 2T0 and Tc = 4-1/fT0

(c) Construct a table which displays for each of the three processes in this cycle:
The work done, the heat transfer, the change in internal energy, the change in enthalpy and the change in entropy. Express your answers in terms of R, the universal gas constant, n, the number of moles of the gas present and T0, the initial temperature at point a.

For a->b:

Work done: W=-PΔV=-P0(2V0-V0)=-P0V0=nRT0

Heat transfer: 0?

Change in internal energy: (f/2)nRΔT=(f/2)nR(Tb-Ta)=(f/2)nR(2T0-T0)

Change in enthalpy: ΔH=ΔU+PΔV=(f/2)nR(4-1/fT0-2T0)+nRT0

Change in entropy: ΔS=nRln(Vf/Vi)=nRln(2)

For b->c:

Work done: 0

Heat transfer: Does this = ΔU?

Change in internal energy: ΔU=(f/2)nRΔT=(f/2)nR(Tc-Tb)=(f/2)nR(4-1/fT0-2T0)

Change in enthalpy: ΔH=ΔU+ΔPV=(f/2)nR(4-1/fT0-2T0)+0

Change in entropy: ΔS=nRln(Vf/Vi) Does this work here?

I'm just not sure if I'm on the right track... Especially for part (c). Any help with this would be appreciated.
 

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  • #2
Redbelly98
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Congrats on reaching 100 posts :smile:

(a) looks good, though they probably want you to use the actual value of γ to calculate a completely numerical value x Po.

(b) I agree with Pb, but get something else for Pc. Also, what is f? I'm not familiar with that quantity.

(c)

Wab: it's nRTo for work done by the gas, or -nRTo done on the gas. Not sure which convention you're using.

Qab is not zero. Path a-b is not adiabatic.

ΔUab: Since U = (3/2) nRT, and you know the temperatures, this is pretty straightforward. Again, not sure what f is (is it 3?). Also, there is an obvious simplification for the expression (2To-To)

I'm going to stop here for now. If you work out the above stuff, then we can continue with the rest.
 
  • #3
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f is the number of degrees of freedom, so f = 3.

and for Pc I had the power up the wrong way.. I end up with Pc = T0/root(8) does this look better?

So as you said Qab is not 0. But ΔUab = (3/2)nRΔT and

ΔT=T0 so ΔUab = (3/2)nRT0. Then can you use this to

solve for Qab? U=Q+W so (3/2)nRT0=-nRT0+Q =>

(Our convention is that W=-PΔV)

Q=(5/2)nRT0


So how am I going now?
 
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  • #4
Redbelly98
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f is the number of degrees of freedom, so f = 3.
Okay, got it. :smile:

and for Pc I had the power up the wrong way.. I end up with Pc = T0/root(8) does this look better?
(Guess you meant Tc here.)
Not quite, it looks like something happened with the exponent on 2. I.e., it's not 2-3/2T0 as you have, but the correct answer does not look very different from 2-3/2T0.

So as you said Qab is not 0. But ΔUab = (3/2)nRΔT and

ΔT=T0 so ΔUab = (3/2)nRT0. Then can you use this to

solve for Qab? U=Q+W so (3/2)nRT0=-nRT0+Q =>

(Our convention is that W=-PΔV)

Q=(5/2)nRT0
Yes.
 
  • #5
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don't tell me Tc=2T0

My logic for this is..

PiViɣ=PfVfɣ

then replacing P with T using the ideal gas law => P=nRT/V

TiViɣ-1=TfVfɣ-1

but Vi = Vf so they just cancel? which leaves Ti=Tf and Ti=Tb and Tf=Tc so Tc=Tb

But if the Volume is constant and the pressure is changing then the temperature must change. What am I doing wrong? Maybe its just late and I can't think properly.
 
  • #6
Redbelly98
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Be careful. Which path are you talking about, a-c or b-c?

If you mean path a-c, then
Vi Vf

Or if you meant b-c, then
PiViγ PfVfγ

Perhaps use subscripts a, b, c instead of i, f so that it is clearer.
 
  • #7
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How does Tc=2-2/3T0 look?

using VaTaf/2=VcTcf/2
 
  • #8
Redbelly98
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One last question!

If this cycle is operated as a heat engine, what would be its efficiency?

From memory this is 1 - Tc/Th

There Tc is the temp of the cold reservoir and Th the temp of the hot reservoir.

I have no idea on where to start for this.
 
  • #10
Andrew Mason
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One last question!

If this cycle is operated as a heat engine, what would be its efficiency?

From memory this is 1 - Tc/Th

There Tc is the temp of the cold reservoir and Th the temp of the hot reservoir.

I have no idea on where to start for this.
This is not a Carnot engine. So:

[tex]\eta = W/Q_h =1-Q_c/Q_h \ne 1-T_c/T_h[/tex]

You have to determine the heat flow into the gas in a->b and b->c to find Qh. You then have to calculate W (the area under a->b minus the area under c->a).

AM
 
  • #11
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So that means I need to know the answer to the following (what this originally started as)

Construct a table which displays for each of the three processes in this cycle:
The work done, the heat transfer, the change in internal energy, the change in enthalpy and the change in entropy. Express your answers in terms of R, the universal gas constant, n, the number of moles of the gas present and T0, the initial temperature at point a.

(Cv=(3/2)nR)

Now I've cleared up some problems let see if i can get this right!

So a->b

Work done Wab: Just area under the graph P0V0=-nRT0

Change in internal energy Uab: (3/2)nRΔT = (3/2)nRΔT = (3/2)nRT0

Heat transfer Qab: Using U=Q+W (know U and W) => Qab=(5/2)nRT0

Change in enthalpy ΔHab: ΔH=ΔU+PΔV=(5/2)nRT0

change in entropy ΔSab: ΔS=nRln(Vb/Va)=nRln(2) (I'm not sure this works when the internal energy changes)
Or can I use the fact that its at constant pressure so ΔS = Integral(Ta->Tb) Cp/T dT ?


So b->c

Work done Wbc: Area under graph = 0

Change in internal energy Ubc: For this can I use an integral from Tb to Tc I get an answer of (3/2)nRT0(2-2/3-2) Is this right?

Heat transfer Qbc: As W = 0, ΔU=ΔQ so Q just is the same as above? (3/2)nRT0(2-2/3-2)

Change in enthalpy ΔHbc: ΔH=ΔU+ΔPV Can I use this? Or is it just the change in Q? ΔH=(3/2)nRT0(2-2/3-2)

change in entropy ΔSbc: ΔS=Cv/T dT Can I use this here as well?


So c->a

Heat transfer Qca: This is 0 as it's adiabatic.

Change in internal energy Uca: ΔU=(3/2)nR * Integral(Tc->Ta) 1 dT = (3/2)nRT0(1-2-2/3) I have a feeling this is very wrong.

Work done Wca: As Q = 0 then W=U so ΔW=(3/2)nRT0(1-2-2/3)

Change in enthalpy ΔHca: No idea.

change in entropy ΔSca: Is this 0?

Now if any of this is even remotely correct does Qh = Qab+Qbc = (5/2)nRT0 + (3/2)nRT0(2-2/3-2)

And the work done W = nRT0 - (3/2)nRT0(1-2-2/3)

I'm very sketchy on all of this, especially c->a. I can understand if you are over helping me with this but either way thanks for all the help!
 
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  • #12
Andrew Mason
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You have the right idea. For c->a the work done is equal in magnitude to the change in internal energy, since there is no heat flow.

[tex]W_{ca} = -\Delta U = nC_v(T_a-T_c) [/tex]

[tex]T_c = T_a(2)^{1-\gamma}[/tex], so:

[tex]W_{ca} = -nC_v(T_a(1-(2)^{1-\gamma})[/tex]

Add that to your answer for the work done from a->b to get the total work (be careful about the signs. Wca is negative and Wab is positive).

AM
 
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  • #13
Redbelly98
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I agree with the calculations for ΔU, W, and Q for each process. I'm not sure about ΔH and ΔS, I would have to review that or maybe Andrew can comment on it. (ΔS is definitely 0 for c-a, the adiabatic process.)

As for efficiency, I agree with
the work done W = nRT0 - (3/2)nRT0(1-2-2/3)
(though strictly speaking this is -W.)

As for Qh, only those processes with Q>0 will contribute to the calculation of Qh. So you need to check which of Qab and Qbc are positive.

EDIT: didn't see Andrew's recent post before posting this ... had this window open for more than an hour before I got around to composing my response :redface:
 
  • #14
Redbelly98
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Add that to your answer for the work done from a->b to get the total work (be careful about the signs. Wca is negative and Wab is positive).

AM
Hi Andrew,

FYI, in forty's class they define

W = -P dV

i.e., it's the work done on the gas.

[tex]W_{ca} = -nR(T_a(1-(2)^{1-\gamma})[/tex]
Is there a 3/2 factor missing? (Or maybe I am missing something?)
 
  • #15
Andrew Mason
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Is there a 3/2 factor missing? (Or maybe I am missing something?)
Redbelly,

Yes. Thanks for noticing that. That should be Cv not R. The internal energy change from c->a is:

[tex]\Delta U = nC_v(T_a(1-(2)^{1-\gamma}) = \frac{3nR}{2}(T_0(1-(2)^{-2/3})[/tex]

AM
 
  • #16
Andrew Mason
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I agree with the calculations for ΔU, W, and Q for each process. I'm not sure about ΔH and ΔS, I would have to review that or maybe Andrew can comment on it. (ΔS is definitely 0 for c-a, the adiabatic process.)
The entropy calculation is:

[tex]\Delta S_{total} = \Delta S_{ab} + \Delta S_{bc} + \Delta S_{ca}[/tex]

For a->b, P is constant so dQ = nC_pdT

[tex]\Delta S_{ab} = \int_{T_a}^{T_b} dQ/T = nC_p \int_{T_a}^{T_b} dT/T = nC_p\ln\left(\frac{T_b}{T_a}\right)[/tex]

For b->c, V is constant, so dQ = dU = nCvdT

[tex]\Delta S_{bc} = nC_v\int_{T_b}^{T_c}dT/T = nC_v\ln\left(\frac{T_c}{T_b}\right)[/tex]

For c->a, Q = 0 so there is no change in entropy (dQ/T = 0).

AM
 
  • #17
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So for my efficiency I need W/Qh

So work = -nRT0 - (3/2)nRT0(1-2^-2/3)

And Qh is just Qab as Qbc is negative so I get

(-nRT0 - (3/2)nRT0(1-2^-2/3) ) / (5/2)nRT0

How does this look?
 
  • #18
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Anyway guys thanks for all the help, really is appreciated!
 
  • #19
Redbelly98
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You had the correct work done by the gas earlier,

And the work done W = nRT0 - (3/2)nRT0(1-2-2/3)
Looks like you got a "-" sign error in your post #17. If you correct that, you'll be close ... but will need to simplify the expression further.
 
  • #20
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( nRT0 - (3/2)nRT0(1-2-2/3) ) / (5/2)nRT0

Is right it just needs to be simplified?

If i simplify i get an answer of (2/5)-(3/5)(1-2-2/3)=.1779

Please tell me im right >.<
 
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  • #21
Redbelly98
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Yes.
 
  • #22
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I can now actually say thanks for the help, all done! Really thanks again it is appreciated!
 

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