enthalpy

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chiddler

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ΔH = ΔU + PΔV

when P is constant, then

ΔH = q

Ok I understand that because you can use the internal energy equation and substitute stuff.

So what about when pressure is not constant. Can you not measure enthalpy?

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ΔH = ΔU + PΔV

when P is constant, then

ΔH = q

Ok I understand that because you can use the internal energy equation and substitute stuff.

So what about when pressure is not constant. Can you not measure enthalpy?


H=U+PV and U=Q-W and W=PV

So H=U-PV+PV, as you say.

If P is not constant, the change in volume is not easily predictable. I think this relates to PV work in the sense that when P is not constant, you need to use calculus to calculate the work done. So you can measure enthalpy, just not in a way that we would be able to/required to on the mcat.
 
now that i think about it,

internal energy, ΔU = Q - PΔV. Rearranging this to: Q = ΔU + PΔV

enthalpy, ΔH = ΔU + PΔV

Q = ΔU + PΔV
ΔH = ΔU + PΔV

uh. what's the difference? it looks like ΔH = ΔQ under all conditions.
 
now that i think about it,

internal energy, ΔU = Q - PΔV. Rearranging this to: Q = ΔU + PΔV

enthalpy, ΔH = ΔU + PΔV

Q = ΔU + PΔV
ΔH = ΔU + PΔV

uh. what's the difference? it looks like ΔH = ΔQ under all conditions.


Yea, U=Q-PV

and H=U+PV so H=Q-PV+PV. PV cancels out even if P is not constant it seems. I guess you can use H=Q even if P isn't constant? But you just can't calculate PV work easily if P isn't constant?
 
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Yea, U=Q-PV

and H=U+PV so H=Q-PV+PV. PV cancels out even if P is not constant it seems. I guess you can use H=Q even if P isn't constant? But you just can't calculate PV work easily if P isn't constant?

no there's a reason that they don't cancel out. but my calculus sucks to understand it conceptually. i asked this question a few days ago on reddit:

Me: If you substitute ΔU = Q - PΔV into ΔH = ΔU + PΔV:
ΔH = ΔU + PΔV
ΔH = (Q - PΔV) + PΔV = Q
ΔH = Q

Them: It was bad math, work in differential form:
dU = dQ + dW
Assume we only have PV work
dU = dQ - PdV
H = U + PV
dH = dU + PdV + VdP
dH = dQ + VdP
Thus, dH = dQ only with constant pressure
 
no there's a reason that they don't cancel out. but my calculus sucks to understand it conceptually. i asked this question a few days ago on reddit:

Me: If you substitute ΔU = Q - PΔV into ΔH = ΔU + PΔV:
ΔH = ΔU + PΔV
ΔH = (Q - PΔV) + PΔV = Q
ΔH = Q

Them: It was bad math, work in differential form:
dU = dQ + dW
Assume we only have PV work
dU = dQ - PdV
H = U + PV
dH = dU + PdV + VdP
dH = dQ + VdP
Thus, dH = dQ only with constant pressure


Told you it had something to do with calculus :D

I can't explain it though, I took calculus 6 years ago.
 

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