Why is delta-H equal to q at constant pressure?

[DiamondV]

Homework Statement

In my lecture notes(beginner thermodnyamics), we just got introduced to the first law(DeltaU=q+w) and two scenarios. One at constant volume which yields the equation Delta U = qv. I understand that
The second scenario is at Constant pressure and it says that At constant pressure q subcript p = DeltaH. DeltaH being change in enthalpy. I don't understnad that bit at all. How is being at constant pressure tell you that the heat energy is the change in enthalpy?

Homework Equations

The Attempt at a Solution

 

[Chestermiller]

If the pressure is constant, what is w equal to?

 

[DiamondV]

If the pressure is constant, what is w equal to?

w = -p times deltaV. note that p is the external pressure
So if pressure is constant, the formula is the same.

 

[Chestermiller]

w = -p times deltaV. note that p is the external pressure
So if pressure is constant, the formula is the same.

OK. What do you get if you substitute that into your first law equation?

 

[DiamondV]

OK. What do you get if you substitute that into your first law equation?

w = -p times deltaV. First law is DeltaU = q+w. So subbing it in would give me:
DeltaU = q-pDeltaV

 

[Chestermiller]

w = -p times deltaV. First law is DeltaU = q+w. So subbing it in would give me:
DeltaU = q-pDeltaV

OK. Now how is ΔH defined in terms of ΔU and Δ(pV)?
What does this defining equation reduce to if p is constant?

Chet

 

[DiamondV]

OK. Now how is ΔH defined in terms of ΔU and Δ(pV)?
What does this defining equation reduce to if p is constant?

Chet

Well DeltaH is change in enthalphy which is change in heat energy, so shouldn't DeltaH always equal to q no matter if its at constant pressure or not since the only other energy transfer is work and that doesn't involve heat. So at all times DeltaH =q? I've just confused myself somehow.

 

[Chestermiller]

Well DeltaH is change in enthalphy which is change in heat energy, so shouldn't DeltaH always equal to q no matter if its at constant pressure or not since the only other energy transfer is work and that doesn't involve heat. So at all times DeltaH =q? I've just confused myself somehow.

No, because the work is not always -pΔV, it is more generally the integral of -pdV. And ΔH is not always ΔU+pΔV, it is more generally ΔU+Δ(pV). Try it using that information and see what you get.

 

[Mark Harder]

Remember the multiplication rule from differential calculus: d(f(x)*g(x))/dx =g(x)* df(x)/dx + f(x)* dg(x)/dx. Or, in the differential form (shortcut):
d(f(x)*g(x)) = g(x)* df(x) + f(x)* dg(x). and so, d(p*v) = ?
Now integrate that, given p=constant.

 

[Chestermiller]

Remember the multiplication rule from differential calculus: d(f(x)*g(x))/dx =g(x)* df(x)/dx + f(x)* dg(x)/dx. Or, in the differential form (shortcut):
d(f(x)*g(x)) = g(x)* df(x) + f(x)* dg(x). and so, d(p*v) = ?
Now integrate that, given p=constant.

Do you find an error in what DiamondV and I have done so far?

Chet

 

[DiamondV]

Do you find an error in what DiamondV and I have done so far?

Chet

Thanks. I think I've gotten a grasp of it anyways. had my final thermodynamics exam, never seeing it again thank god.

 

[Mark Harder]

Do you find an error in what DiamondV and I have done so far?

Chet

No, not at all. Reading over the posts in this thread, I couldn't find an explanation for why d(PV) = PdV + VdP, so I provided one, just in case the OP didn't know that. It's a necessary step in the derivation for delta-H at constant pressure.
Mark