In thermodynamics and Heat Capicitance why does q = delta H when pressure is constant?
In a problem with calorimeters it states that in a coffee cup calorimeter, q = delta H because of constant pressure. I know how to do these problems i guess but i just want to understand WHY at constant pressure q = delta H.
I understand why q = delta U for a bomb calorimeter because volume is constant so no work is done but the above w/ constant pressure stumps me.
Thanks!
Answer:
Dem2003,
This is a great question, and it's actually one that I spent time pondering in my early days as a chem major. There is an answer, but it is not very intuitive and really takes sitting down and reading through the derivation several times before it starts to sink in. You're right that the reason for q=deltaH at constant V is much more intuitive. I'm going to give this a shot, and see how much it helps.
Enthalpy is defined as H = U + PV. So if you remember your product rule from calculus, a small change in enthalpy is
dH = dU + PdV + VdP.
Imposing the constraint of constant pressure, we can eliminate the last term (since dP = 0), and the equation becomes
dH = dU + PdV
Now, from the 1st law we know that U = q + w, so do this replacement and you get
dH = dq + dw + PdV
Finally, w = -PdV, so we get
dH = dq - PdV + PdV = dq
So essentially, enthalpy is defined so that it measures total changes in the thermal motion of molecules. In an expansion at constant pressure, some of this thermal motion is actually lost in accomplishing the expansion. When P is constant, the PV term exactly compensates for the work that was lost, such that the enthalpy is exactly the same as the heat. (So, it sort of "throws" this energy back in so that it's included in the enthalpy). In a system where expansion is not allowed and the pressure increases, the enthalpy is larger than the heat because there is thermal motion that is included in the enthalpy, but not included in the heat because the temperature increase is "suppressed" by the pressure.
It's a more difficult thing to wrap your head around, so I wouldn't blame you for not really seeing what I'm saying. It took me some time as well. Go through my derivation a few times and it will start to gel a little. Hope this helps!!
If I remember correctly, delta H is by definition, energy at constant pressure. A convenient way to measure and calculate.
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I understand why q = delta U for a bomb calorimeter because volume is constant so no work is done but the above w/ constant pressure stumps me.
Thanks!
Answer:
Dem2003,
This is a great question, and it's actually one that I spent time pondering in my early days as a chem major. There is an answer, but it is not very intuitive and really takes sitting down and reading through the derivation several times before it starts to sink in. You're right that the reason for q=deltaH at constant V is much more intuitive. I'm going to give this a shot, and see how much it helps.
Enthalpy is defined as H = U + PV. So if you remember your product rule from calculus, a small change in enthalpy is
dH = dU + PdV + VdP.
Imposing the constraint of constant pressure, we can eliminate the last term (since dP = 0), and the equation becomes
dH = dU + PdV
Now, from the 1st law we know that U = q + w, so do this replacement and you get
dH = dq + dw + PdV
Finally, w = -PdV, so we get
dH = dq - PdV + PdV = dq
So essentially, enthalpy is defined so that it measures total changes in the thermal motion of molecules. In an expansion at constant pressure, some of this thermal motion is actually lost in accomplishing the expansion. When P is constant, the PV term exactly compensates for the work that was lost, such that the enthalpy is exactly the same as the heat. (So, it sort of "throws" this energy back in so that it's included in the enthalpy). In a system where expansion is not allowed and the pressure increases, the enthalpy is larger than the heat because there is thermal motion that is included in the enthalpy, but not included in the heat because the temperature increase is "suppressed" by the pressure.
It's a more difficult thing to wrap your head around, so I wouldn't blame you for not really seeing what I'm saying. It took me some time as well. Go through my derivation a few times and it will start to gel a little. Hope this helps!!
If I remember correctly, delta H is by definition, energy at constant pressure. A convenient way to measure and calculate.
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