The different between the work (PV ) in ΔU and the work (PV ) ΔH

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In summary, the equations ΔU = q+w and ΔH = ΔU+Δ(PV) are incorrect. The correct equations are ΔU=q-TΔS and ΔH=ΔU+Δ(PV). Work (w) and heat (q) are dependent on the path between equilibrium states, while U and H are independent of the path. In a constant external pressure process, ΔH=q and when the final pressure equals the initial pressure, ΔH=q. For ideal gases, U and H are only functions of temperature, but for real gases, the effect of pressure on U and H can be calculated using Maxwell Relationships.
  • #1
izen
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By definition
ΔU = q+PV
ΔH = ΔU+PV
Well, what are the different between the work (PV ) in ΔU and the work (PV ) ΔH then ?

thank you
 
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  • #2
Both equations are incorrect.

ΔU=q+w
ΔH=ΔU+Δ(PV)

So, now, what is your question?
 
  • #3
w= -P ΔV
in case of constant Pressure, ΔH= (q -PΔV) +PΔV >> ΔH= q

So w in ΔU will be the same when constant pressure with Δ(PV) in ΔH ?

if the pressure is not constant What is P in ΔU ? is it P external? and what is P in ΔH ? P in ΔH is Pressure of system? in case of ideal gas, Δ(PV) = nRT so where is P ?

and if not ideal gas how can we calculate Δ(PV) in ΔH?

Thank you
 
  • #4
Dear izen,

These are all excellent questions, but I don't have time to respond right now. We are expecting some out-of-town relatives, and I'm pretty tied up for the rest of the day. I will get back with you as soon as possible to answer all your questions.

Chet
 
  • #5
Hi Chet,

I will be waiting for you here. Enjoy your evening. :)

Izen
 
  • #6
izen said:
w= -P ΔV
in case of constant Pressure, ΔH= (q -PΔV) +PΔV >> ΔH= q

So w in ΔU will be the same when constant pressure with Δ(PV) in ΔH ?

if the pressure is not constant What is P in ΔU ? is it P external? and what is P in ΔH ? P in ΔH is Pressure of system? in case of ideal gas, Δ(PV) = nRT so where is P ?

and if not ideal gas how can we calculate Δ(PV) in ΔH?

Thank you

You need to think of U and H as properties of a system characteristic of each state of equilibrium. ΔU and ΔH are the changes in U and H from one equilibrium state (say, state 1) to another equilibrium state (say, state 2). Even though there may be many paths that will take the system between the two equilibrium states, the changes in U and H are independent of the path.

On the other hand, the parameters q and w are quantities that are totally dependent on the path between the two equilibrium states. Only their difference is independent of the path. q and w are sometimes called energy in transit.

For expansion or compression, the work w is always determined by [tex]w = -\int{P_extdV}[/tex], not by w= -P ΔV. If the process is reversible, the pressure P within the system is virtually homogeneous, and, in addition, P = Pext (actually P differs from Pext by an infinitecimal amount). If the process is irreversible, the pressure within the system is not homogeneous, and so there is no one single value of pressure that characterizes the pressure within the system.

For the change in enthalpy, we have
[tex]\Delta H = q -\int{P_{ext}dV}+\Delta (PV)[/tex]
where the pressure P and the volume V in the Δ(PV) correspond to the system values in the initial and final equilibrium states. If a process is carried out at constant external pressure, the change in enthalpy is given by:
[tex]\Delta H = q -P_{ext}(V_2-V_1)+(P_2V_2-P_1V_1)[/tex]

In such a process, for the final state to be an equilibrium state, one must have that
P2=Pext. Under such circumstances:

[tex]\Delta H = q -(P_2-P_1)V_1[/tex]

If the final pressure of the system is equal to the initial pressure (constant pressure process),

[tex]\Delta H = q[/tex]

Examples of when this applies are (a) if you have an ideal gas, and you add heat to the chamber while the piston is moved in such a way as to hold the pressure constant and (b) a combination of saturated vapor and saturated liquid in the chamber, and you add heat while the piston is moved in such a way as to keep the pressure constant.

In the case of an ideal gas, there is no effect of pressure on either U or H. Both parameters are functions only of temperature. For a real gas, if you wish to calculate the effect of pressure on U and H, the formulation is more complicated. For infinitecimal changes between equilibrium states, the changes in U and H are described by the differential equations:

dU = TdS - PdV
dH = TdS+VdP

These equations apply only to differential changes between equilibrium states. The effect of pressure P on U and H are derived starting with these equations, and, by applying the so-called Maxwell Relationships.
 
  • #7
Hi Chet thanks for the answer :)
 

FAQ: The different between the work (PV ) in ΔU and the work (PV ) ΔH

What is the difference between the work (PV) in ΔU and the work (PV) in ΔH?

The work (PV) in ΔU refers to the change in internal energy of a system caused by the work done on or by the system, while the work (PV) in ΔH refers to the change in enthalpy of a system due to work done on or by the system. Both ΔU and ΔH involve a change in the energy of a system, but ΔH also takes into account the change in pressure and volume of the system.

How are the calculations for work (PV) in ΔU and ΔH different?

The calculations for work (PV) in ΔU and ΔH are different because they involve different variables. The work in ΔU is calculated using the change in internal energy, while the work in ΔH is calculated using the change in enthalpy. Additionally, the work in ΔH also takes into account the change in pressure and volume of the system.

What is the significance of work (PV) in ΔU and ΔH in thermodynamics?

The work (PV) in ΔU and ΔH are important concepts in thermodynamics because they help us understand the energy changes in a system. These calculations help us determine the work done on or by a system, and can also give insight into the efficiency of a process. They are also used in various thermodynamic equations and principles.

How do changes in pressure and volume affect the work (PV) in ΔU and ΔH?

The work (PV) in ΔU and ΔH are affected by changes in pressure and volume because they are both related to energy changes in a system. When pressure or volume changes, the internal energy and enthalpy of the system also change, leading to a change in the work done (PV) in ΔU and ΔH. This is why these variables are taken into account in the calculations.

Can the work (PV) in ΔU and ΔH ever be the same?

No, the work (PV) in ΔU and ΔH cannot be the same because they involve different variables and have different calculations. The work (PV) in ΔH takes into account the change in pressure and volume of the system, while the work (PV) in ΔU does not. As a result, the values for work (PV) in ΔU and ΔH will always be different, even if they are calculated for the same system.

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