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Topic: In Thermodynamics, dw = Pdv?  (Read 27706 times)

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Offline Dev

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In Thermodynamics, dw = Pdv?
« on: January 22, 2013, 04:25:36 AM »
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?

Offline formaldehyde23

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Re: In Thermodynamics, dw = Pdv?
« Reply #1 on: January 22, 2013, 11:28:21 AM »
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.

Offline ramboacid

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Re: In Thermodynamics, dw = Pdv?
« Reply #2 on: January 22, 2013, 09:08:29 PM »
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.
"Opportunity is missed by most people because it is dressed in overalls and looks like work." - Thomas Edison

Offline Dev

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Re: In Thermodynamics, dw = Pdv?
« Reply #3 on: January 23, 2013, 07:54:21 AM »
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?


Offline Dev

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Re: In Thermodynamics, dw = Pdv?
« Reply #4 on: January 23, 2013, 10:41:46 AM »
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.


Offline sankalpmittal

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Re: In Thermodynamics, dw = Pdv?
« Reply #5 on: January 24, 2013, 02:34:52 PM »
Hi!

I have a basic doubt regarding thermodynamics,

dU = dQ + dW

where, we right dW = PdV

Yes , but in this , pressure is constant.

Quote
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 !!
Quote
What does d(PV) mean?

Do you know about "partial differentiation" ?

Offline curiouscat

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Re: In Thermodynamics, dw = Pdv?
« Reply #6 on: January 24, 2013, 02:52:57 PM »

Quote
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.

Offline sankalpmittal

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Re: In Thermodynamics, dw = Pdv?
« Reply #7 on: January 25, 2013, 05:51:42 AM »

Quote
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.

Offline curiouscat

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Re: In Thermodynamics, dw = Pdv?
« Reply #8 on: January 25, 2013, 05:57:52 AM »

Quote
Writing this : dW = d(PV)  is wrong.

Yes! I think now, everything's in order.

Quote
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.

Offline curiouscat

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Re: In Thermodynamics, dw = Pdv?
« Reply #9 on: January 25, 2013, 06:08:34 AM »
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?

Offline curiouscat

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Re: In Thermodynamics, dw = Pdv?
« Reply #10 on: January 26, 2013, 07:01:23 AM »
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.


Offline Dev

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Re: In Thermodynamics, dw = Pdv?
« Reply #11 on: January 28, 2013, 08:57:30 AM »
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.

Offline curiouscat

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Re: In Thermodynamics, dw = Pdv?
« Reply #12 on: January 28, 2013, 09:11:37 AM »
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.

Offline Dev

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Re: In Thermodynamics, dw = Pdv?
« Reply #13 on: January 28, 2013, 09:37:39 AM »
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)

Offline curiouscat

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Re: In Thermodynamics, dw = Pdv?
« Reply #14 on: January 28, 2013, 09:48:10 AM »
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?

Quote
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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