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

In summary, In the conversation, the topic of the first law of thermodynamics and its application in two different scenarios, constant volume and constant pressure, was discussed. It was established that at constant pressure, q subcript p = DeltaH, which is the change in enthalpy. The concept of enthalpy was further explained in terms of the first law equation and its relation to work and heat energy. The conversation also touched on the use of differential calculus in understanding the relationship between pressure, volume, and enthalpy. Ultimately, the conversation ended with a clarification on the concept of d(PV) and its role in the derivation of delta-H at constant pressure.
  • #1
DiamondV
103
0

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

 
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  • #2
If the pressure is constant, what is w equal to?
 
  • #3
Chestermiller said:
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.
 
  • #4
DiamondV said:
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?
 
  • #5
Chestermiller said:
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
 
  • #6
DiamondV said:
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
 
  • #7
Chestermiller said:
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.
 
  • #8
DiamondV said:
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.
 
  • #9
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.
 
  • #10
Mark Harder said:
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
 
  • #11
Chestermiller said:
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.
 
  • #12
Chestermiller said:
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
 

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

What is the first law of thermodynamics?

The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed, only transformed from one form to another.

How does the first law of thermodynamics relate to energy balance?

The first law of thermodynamics explains that the total energy within a closed system remains constant. This means that any energy that enters or leaves the system must be accounted for, resulting in an energy balance.

What are the different forms of energy involved in the first law of thermodynamics?

The first law of thermodynamics involves multiple forms of energy, including mechanical energy, thermal energy, chemical energy, nuclear energy, and electromagnetic energy.

Can the first law of thermodynamics be violated?

No, the first law of thermodynamics is a fundamental law of physics and cannot be violated. However, it is possible for energy to be lost due to inefficiencies in energy transformation processes.

How is the first law of thermodynamics applied in real-life situations?

The first law of thermodynamics is applied in a wide range of real-life situations, such as the operation of engines, power plants, and refrigerators. It also explains the conversion of energy in biological systems, such as cellular respiration in living organisms.

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