Thermodynamics-when should I use u & h ?

[yecko]

Homework Statement

Homework Equations

E =Q + W
U(internal energy)=m(u2-u1)
H(enthalpy)=m(h2-h1)

The Attempt at a Solution

1st question: I understand in this case as no heat change, the work done should be equal to internal energy change... but why it used m(h2-h1) in this case for enthalpy change in this case?

2nd question: generally speaking, when should i use enthalpy and internal energy for calculating energy change?

Thank you very much!

 

[mjc123]

Enthalpy is used because it is simpler in this case. As you are told, "ΔU + W_b = ΔH during a constant pressure quasi-equilibrium process" and therefore W_e + W_pw = ΔH. Substituting known values gives you an equation with one unknown, V, which is what you want to determine. There is no need to separately calculate ΔU and/or W_b. That would just complicate things without helping to answer the given question.
Generally speaking, you need to take an approach like that above, write down an equation with all the relevant quantities, and determine what ΔU and ΔH are each equal to, and decide which is more appropriate to calculate for your problem.

 

[yecko]

As you are told, "ΔU + Wb = ΔH during a constant pressure quasi-equilibrium process" and therefore We + Wpw = ΔH.

There is no need to separately calculate ΔU and/or Wb.

indeed solution of all these kind of questions would work done to the system equal to change of internal energy first, by m(u2-u1) first... and then it suddenly equals to m(h1-h2) which doubt me... why?

or do you mean just because there are volume change, on top of state and temperature, we should only use enthalpy instead of internal energy?
or should any kind of work done should just use enthalpy instead of internal energy data?

thanks

 

[yecko]

on the other hand, question like this used only internal energy instead of using enthalpy for the total energy... why?

 

[mjc123]

m(u2-u1) does not "suddenly equal" m(h2-h1). Can you not read what's in front of your face? A term is transferred from one side of the equation to the other. As is given in your example:
W_e - W_b = ΔU = m(u2-u1)
Add W_b to each side:
W_e = W_b + m(u2-u1) = m(h2-h1)
ΔH = ΔU + Δ(PV), and as explained above, in a constant-pressure process Δ(PV) = PΔV = W_b. In your second example the pressure is not constant, so W_b = ∫PdV ≠ Δ(PV), and W_b ≠ ΔH - ΔU. In this case, since the process is linear, if you had the enthalpy data you could work from that just as easily.

 

[yecko]

So to conclude, h is only being used when pressure is constant, Q=H,
u is being used when volume is constant (or even it is not constant, whenever when pressure is not constant?), Q=E+W=U+W... right?

 

[mjc123]

Not necessarily. U and H are always both there; they are state functions of the system. (The relation between them is particularly simple at constant pressure.) Which you use in your calculations depends on which is more convenient - and sometimes, on which you have data for. (A lot of tabulated thermodynamic data is given in terms of enthalpy; e.g. enthalpies of formation, phase transitions etc. You could fairly easily calculate ΔU values from them, but ΔH values are what you are given.)
As I said above, do what your examples do - write down all the relevant energy terms in an equation of the form E_in - E_out = ΔE_system. What is known and what is unknown? What assumptions are being made (e.g. constant pressure, insulated system etc.)? How do you calculate terms like W_b in the given scenario?