Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Dev on January 22, 2013, 04:25:36 AM
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Hi!
I have a basic doubt regarding thermodynamics,
dU = dQ + dW
where, we right dW = PdV
but someone recently said that dW = d(PV) = VdP + PdV.
Is this correct?
Which one is correct?
What does d(PV) mean?
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The formula states dU = dH + dW. Since dW = PdV, we can substitute this into the first equation and get dU = dH + PdV.
Here, U is the internal energy and H is the enthalpy, P is pressure and V is the volume.
dW = d(PV) = VdP + PdV is incorrect. The mistake here is that he might be thinking that d(PV) is the derivative. d(PV) = VdP + PdV is correct if you are taking the derivative of PV, where P and V are variables.
d(PV) is not the derivative in this case; rather, it is the delta or change in the quantity of PV. (By the way, you wouldn't really write d(PV) in chemistry. You'd rather write PdV since pressure is kept constant.
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Also consider that if you're dealing with a change in enthalpy, dP is zero because of constant pressure and thus doesn't appear in the expression. If you were dealing with change in internal energy dV would be zero because of constant volume, and thus PdV wouldn't appear in the expression.
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Actually I was talking about the formula associated with the first law,
dU = dQ + dW
So far, I haven't dealt with enthalpy. But I think dH is not equal to dQ.
In this formula, we substitute dW = PdV
I think that I PdV + I VdP = I d(PV) = P1V1 - P2V2
where I stands for integration. P1 = initial pressure P2 = final pressure , same for V
Also, I didn't understand why you claimed that Pressure is kept constant. Which pressure are you talking about?
What about the case where an ideal gas is reversibly compressed/expands?
What actually is d(PV) (or delta PV) ? What does this term mean?
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In case you need to know where I'm coming from,
The only exposure to thermodynamics that i have is my high school chemistry.
My class is going through most of the special thermodynamic processes
(isothermal, isochoric, isobaric, adiabatic - irreversible and isothermal and adiabatic - reversible)
My class hasn't started with enthalpy yet. Only heat exchanged, Internal Energy and work done has been discussed.
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Hi!
I have a basic doubt regarding thermodynamics,
dU = dQ + dW
where, we right dW = PdV
Yes , but in this , pressure is constant.
but someone recently said that dW = d(PV) = VdP + PdV.
Is this correct?
Which one is correct?
Yes again. But in this , pressure is not constant and volume is obviously not !!
What does d(PV) mean?
Do you know about "partial differentiation" ?
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but someone recently said that dW = d(PV) = VdP + PdV.
Is this correct?
Which one is correct?
Yes again. But in this , pressure is not constant and volume is obviously not !!
So if volume were indeed constant would work be VdP?
I think your explanation's wrong.
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but someone recently said that dW = d(PV) = VdP + PdV.
Is this correct?
Which one is correct?
Yes again. But in this , pressure is not constant and volume is obviously not !!
So if volume were indeed constant would work be VdP?
I think your explanation's wrong.
Yes... W=P(V2-V1)
Oh , I blundered.
Work is always given by pressure times change in volume.
W=PΔV
OR
dW=PdV , if pressure is constant.
Writing this : dW = d(PV) = VdP + PdV is wrong. It should be ,
ΔW = PΔ2V+ΔP.ΔV
Now if work is infinitesimally small,
dW=Pd2V + dP.dV
Now here both pressure and volume change.
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Writing this : dW = d(PV) is wrong.
Yes! I think now, everything's in order.
ΔW = PΔ2V+ΔP.ΔV
Now if work is infinitesimally small,
dW=Pd2V + dP.dV
Now here both pressure and volume change
Interesting. Never tried to write it that way.
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Now if work is infinitesimally small,
dW=Pd2V + dP.dV
Now here both pressure and volume change.
Actually, I cannot seem to see how you'd get this equation either. So, I'll guess you are wrong again with this equation. But I might be mistaken. Can you show a derivation?
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Now if work is infinitesimally small,
dW=Pd2V + dP.dV
Now here both pressure and volume change.
Actually, I cannot seem to see how you'd get this equation either. So, I'll guess you are wrong again with this equation. But I might be mistaken. Can you show a derivation?
Also this part of your expression is mathematically ill defined I think.
dW=Pd2V
Irrespective of what W, P, V are I don't think combining a d2 with a d makes sense.
Maybe someone else will comment.
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Hi, my problem is my class has derived values of work for almost all standard processes, which include isothermal, adiabatic, isochoric, isobaric in irreversible and adiabatic,isothermal for reversible.
and nowhere in these derivations, did we ever use the term VdP.
Here are a few results we found:
Irreversible:
Isochoric : W = 0
Isobaric W = P(V2 - V1)
Isothermal W = nRT ln(V2/V1)
I believe these 3 are enough to make my doubt clear.
can you say that in all these cases, Pressure is constant?
I don't think so.
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Hi, my problem is my class has derived values of work for almost all standard processes, which include isothermal, adiabatic, isochoric, isobaric in irreversible and adiabatic,isothermal for reversible.
and nowhere in these derivations, did we ever use the term VdP.
Here are a few results we found:
Irreversible:
Isochoric : W = 0
Isobaric W = P(V2 - V1)
Isothermal W = nRT ln(V2/V1)
I believe these 3 are enough to make my doubt clear.
can you say that in all these cases, Pressure is constant?
I don't think so.
Not at all. In the isochoric case pressure need not be constant but it can be.
In the isothermal case there is no way for the pressure to be constant.
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Exactly.
That's my concern.
In all these derivations, we never considered the term of VdP. Which basically means we assumed that pressure is constant. Which is absolutely wrong.
Still, these results which we got are standard results.
Unless, anyone would like to argue those results, we have to accept that W = I PdV and not I d(PV)
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Exactly.
That's my concern.
In all these derivations, we never considered the term of VdP. Which basically means we assumed that pressure is constant.
Huh? Why?
Still, these results which we got are standard results.
Unless, anyone would like to argue those results, we have to accept that W = I PdV and not I d(PV)
Of course it is. dW = PdV
Show me a reference that says dW = d(PV) ?
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Hi.
Well, I started off believing that W = I PdV.
It was someone who suggested that W is actually I d(PV)
The thing that would make me happiest is finding out W = I PdV,
but no one here who has posted before this said that W = PdV. They always added that this is true only when pressure is constant.
What I need to hear is that W = I PdV for everything.
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The thing that would make me happiest is finding out W = I PdV,
but no one here who has posted before this said that W = PdV.
Re-read my Reply#8 in this thread above.
They always added that this is true only when pressure is constant.
What I need to hear is that W = I PdV for everything.
I'll say it. But I may be wrong.
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While I wish you wouldn't have said that last part, I'm going to take it.
W is indeed equal to I PdV for everything.
Thanks.
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You are confusing 2 or three processes taking place on or by the system,
the correct equation is for non isenthalpic process
combining 1st and second law of thermodynamics
dH=dU+pdV+Vdp
this can be easily transformed from equation of enthalpy
H=U+pV
taking differential operator on both sides
dH=dU+d(PV)
dH=dU+pdV+Vdp
work can also be written in terms of enthalpy (In case of Steam Cycles)
dW=H-H=dU+pdV+Vdp
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