Question about Flow between Parallel Plates

[Red_CCF]

Forgot about the attachment again...

 

[Chestermiller]

From assuming Δx = 0 and d(Δx)/dt = 0 at t = 0, for the constants I got:

c_1 = \frac{(F_{kin} - F_{stat})(1-\sqrt{1-\frac{4mk}{C^2}})}{2k\sqrt{1-\frac{4mk}{C^2}}}
c_2 = \frac{(F_{stat} - F_{kin})(1+\sqrt{1-\frac{4mk}{C^2}})}{2k\sqrt{1-\frac{4mk}{C^2}}}

I'm not sure if I'm doing this correctly, but the square root would be linearized, for f(x) = \sqrt{1\pm x} :

L_1(x) = \sqrt{1-\frac{4mk}{C^2}} - 1/2(x-\frac{4mk}{C^2})(1-\frac{4mk}{C^2})^{-0.5} for \sqrt{1-\frac{4mk}{C^2}}

and

L_2(x) = \sqrt{1+\frac{4mk}{C^2}} + 1/2(x-\frac{4mk}{C^2})(1+\frac{4mk}{C^2})^{-0.5} for \sqrt{1+\frac{4mk}{C^2}}

I've corrected some sign errors in the equations for c1 and c2 (the discriminant should have a minus sign; I also made this same typo in one of my previous responses).

Linearizing the equations with respect to \frac{4mk}{C^2} involves making use of the following linearization approximations:
\sqrt{1-x}≈1-\frac{x}{2}
\frac{1}{1-x}≈1+x
After applying these approximations, I obtain:

Δx_1≈-\frac{(F_{stat}-F_{kin})}{k}\left[\left(1+\frac{mk}{C^2}\right)(1-e^{r_1t})-\left(\frac{mk}{C^2}\right)(1-e^{r_2t})\right]
where, in this linearized approximation,
r_1≈-\frac{k}{C}\left(1+\frac{mk}{C^2}\right)
r_2≈-\frac{C}{m}\left(1-\frac{mk}{C^2}\right)
Note the similarity between this solution, and the solution obtained when m = 0. In both cases, the piston displacement at long times Δx₁ is exactly the same value. Note also that for small values of \frac{4mk}{C^2}, the piston does not oscillate, but its displacement simply increases monotonically and asymptotically to the final value.

Chet

 

[Chestermiller]

For the H2O example what kind of effect would a finite rate have on the properties inside reaction chamber?

I want to think about this question some more.

I'm not sure if they help but I have attached the cover sheet of the appendix which outlines the source and calculation method that is slightly different than the above. It appears that the entropy value is absolute and at least close to some other ones on the internet; however I am not sure why it is varying with temperature while the enthalpy/free energies do not.

The entropy, enthalpy, and free energy of a pure substance at constant pressure all increase with temperature. I looked over your Tables A1 and A2 description, and can see what they are doing.

My understanding of the variables is as follows:

\bar{h_i^o}(T)=enthalpy of pure species i at 1 atm and temperature T
\bar{h_i^o}(T_{ref})=enthalpy of pure species i at 1 atm and temperature T_ref
\bar{h_i^o}(T) and \bar{h_i^o}(T_{ref}) are related by:
\bar{h_i^o}(T)=\bar{h_i^o}(T_{ref})+\int_{T_{ref}}^T{\bar{c}_p(T)dT}
These equations apply both to the elements and to compounds. However, for the elements at T_ref, \bar{h_i^o}(T_{ref}) is taken as zero.
\bar{h_{f,i}^o}(T)=heat of formation of pure species i at 1 atm and temperature T from the elements. If

\bar{s_i^o}(T)=absolute entropy of pure species i at 1 atm and temperature T
\bar{s_i^o}(T_{ref})=absolute entropy of pure species i at 1 atm and temperature T_ref
\bar{s_i^o}(T) and \bar{s_i^o}(T_{ref}) are related by:
\bar{s_i^o}(T)=\bar{s_i^o}(T_{ref})+\int_{T_{ref}}^T{\frac{\bar{c}_p(T)}{T}dT}

\bar{g_i^o}(T)=free energy of pure species i at 1 atm and temperature T
\bar{g_i^o}(T_{ref})=absolute entropy of pure species i at 1 atm and temperature T_ref
\bar{g_i^o}(T)=\bar{h_i^o}(T)-T\bar{s_i^o}(T)
This all looks pretty straightforward. Does this help answer your questions?

How do we maintain a constant acceleration while adding different increments of δP, as it is essentially a net force that will always add to the acceleration?

You're thinking about it backwards. We are applying whatever force necessary to maintain the acceleration of the piston constant (at a very small value). Are you saying that you can't figure out how to compel the piston to exhibit a desired acceleration?
Chet

 

[Red_CCF]

I've corrected some sign errors in the equations for c1 and c2 (the discriminant should have a minus sign; I also made this same typo in one of my previous responses).

Linearizing the equations with respect to \frac{4mk}{C^2} involves making use of the following linearization approximations:
\sqrt{1-x}≈1-\frac{x}{2}
\frac{1}{1-x}≈1+x
After applying these approximations, I obtain:

Δx_1≈-\frac{(F_{stat}-F_{kin})}{k}\left[\left(1+\frac{mk}{C^2}\right)(1-e^{r_1t})-\left(\frac{mk}{C^2}\right)(1-e^{r_2t})\right]
where, in this linearized approximation,
r_1≈-\frac{k}{C}\left(1+\frac{mk}{C^2}\right)
r_2≈-\frac{C}{m}\left(1-\frac{mk}{C^2}\right)
Note the similarity between this solution, and the solution obtained when m = 0. In both cases, the piston displacement at long times Δx₁ is exactly the same value. Note also that for small values of \frac{4mk}{C^2}, the piston does not oscillate, but its displacement simply increases monotonically and asymptotically to the final value.

Chet

Is there a restriction somewhere that mk/C^2 be less than 1?

The entropy, enthalpy, and free energy of a pure substance at constant pressure all increase with temperature. I looked over your Tables A1 and A2 description, and can see what they are doing.

My understanding of the variables is as follows:

\bar{h_i^o}(T)=enthalpy of pure species i at 1 atm and temperature T
\bar{h_i^o}(T_{ref})=enthalpy of pure species i at 1 atm and temperature T_ref
\bar{h_i^o}(T) and \bar{h_i^o}(T_{ref}) are related by:
\bar{h_i^o}(T)=\bar{h_i^o}(T_{ref})+\int_{T_{ref}}^T{\bar{c}_p(T)dT}
These equations apply both to the elements and to compounds. However, for the elements at T_ref, \bar{h_i^o}(T_{ref}) is taken as zero.
\bar{h_{f,i}^o}(T)=heat of formation of pure species i at 1 atm and temperature T from the elements. If

\bar{s_i^o}(T)=absolute entropy of pure species i at 1 atm and temperature T
\bar{s_i^o}(T_{ref})=absolute entropy of pure species i at 1 atm and temperature T_ref
\bar{s_i^o}(T) and \bar{s_i^o}(T_{ref}) are related by:
\bar{s_i^o}(T)=\bar{s_i^o}(T_{ref})+\int_{T_{ref}}^T{\frac{\bar{c}_p(T)}{T}dT}

\bar{g_i^o}(T)=free energy of pure species i at 1 atm and temperature T
\bar{g_i^o}(T_{ref})=absolute entropy of pure species i at 1 atm and temperature T_ref
\bar{g_i^o}(T)=\bar{h_i^o}(T)-T\bar{s_i^o}(T)
This all looks pretty straightforward. Does this help answer your questions?

With regards to the entropy equation, is there supposed to be a pressure component on the right hand side as well?

My main confusion is that they apparently put absolute entropy values next to enthalpy and free energy that is w.r.t the reference state. Is there a general relationship between the absolute entropy in the table and the entropy of formation, as I expect that for elements that 0 = 0 - T*0 (G_f = H_f - TS_f)?

How come that entropy here is temperature dependent but G_f and H_f are not? Are the formation entropy, enthalpy, free energy values in general dependent where we set the reference state (T,p)?

You're thinking about it backwards. We are applying whatever force necessary to maintain the acceleration of the piston constant (at a very small value). Are you saying that you can't figure out how to compel the piston to exhibit a desired acceleration?
Chet

Are we trying to "match" the P_I increase with δP increase such that there's always a constant (small) net force difference on the piston? Is it theoretically possible to have the piston maintained at a = 0 (outside some starting acceleration to get it moving) and is the premise of a small acceleration that velocity of the piston is small enough to not induce viscous stresses?

Thanks

 

[Chestermiller]

Is there a restriction somewhere that mk/C^2 be less than 1?

No, but we are just trying to address some the doubts you had regarding the transition from small finite piston mass to the limit of zero piston mass, such as:

1. How can you have an accelerating piston if the piston mass is zero?
2. Does the piston oscillate if its mass is very small, but not zero?
3. Is the transition smooth from small finite piston mass to zero piston mass?
This solution answers all these questions.

We can continue the solution to larger mass, if you'd like. If mk/C^2 is equal to 1, the system is critically damped. If it is greater than 1, then the system may oscillate, depending on whether static friction is re-established at the moment that the piston reaches its maximum compression (for the first time).

With regards to the entropy equation, is there supposed to be a pressure component on the right hand side as well?

Those superscript o's refer specifically to the "standard state" where the pressure is 1 atm.

My main confusion is that they apparently put absolute entropy values next to enthalpy and free energy that is w.r.t the reference state. Is there a general relationship between the absolute entropy in the table and the entropy of formation, as I expect that for elements that 0 = 0 - T*0 (G_f = H_f - TS_f)?

How come that entropy here is temperature dependent but G_f and H_f are not? Are the formation entropy, enthalpy, free energy values in general dependent where we set the reference state (T,p)?

The approach they've described in your book is very confusing, as evidenced by the fact that it has even confused a smart guy like you. I even find it a little confusing myself, and would never work with the kind of notation that they have used. Here is that approach that I learned.

1. Let h(T,p), s(T,p), and g(T,p) represent the enthalpy, entropy, and free energy of a pure species at temperature T and p.
2. Let h°(T), s°(T), and g°(T) represent the values of h(T,p), s(T,p), and g(T,p) at p = 1 atm
3. Let h°(298), s°(298), and g°(298) represent the enthalpy, entropy, and free energy of formation of the pure species from its elements at 298 K and 1 atm. So, for an elemental species, h°(298), s°(298), and g°(298) are equal to zero, and for a compound, they are not. In the case of the entropy of a compound, for example, s°(298) = (absolute entropy of compound at 298 and 1 atm) - (absolute entropies of elements comprising the compound at 298 and 1 atm).
We also have:
h°(T)=h°(298)+\int_{298}^T{C_p(T')dT'}
s°(T)=s°(298)+\int_{298}^T{\frac{C_p(T')}{T'}dT'}
g°(T)=h°-Ts°
For a species in an ideal gas mixture (or a pure species), if p is the partial pressure of a species, then,
h(T,p)=h°(T)=h°(298)+\int_{298}^T{C_p(T')dT'}
s(T,p)=s°(T)-R\ln(p)=s°(298)+\int_{298}^T{\frac{C_p(T')}{T'}dT'}-R\ln(p)
g(T,p)=h(T,p)-Ts(T,p)

The standard heat of a reaction ΔH°(T,1atm)=(sum of h°(T) of the pure reactants)-(sum of h°(T) of the pure products)
The standard free energy of reaction ΔG°(T,1atm)=(sum of g°(T) of the pure reactants)-(sum of g°(T) of the pure products)

Are we trying to "match" the P_I increase with δP increase such that there's always a constant (small) net force difference on the piston?

Not exactly. We are trying "match" the δP to the the P_I increase such that there's always a constant(small) net force difference on the piston.

Is it theoretically possible to have the piston maintained at a = 0 (outside some starting acceleration to get it moving) and is the premise of a small acceleration that velocity of the piston is small enough to not induce viscous stresses?

We won't be able get the pressure increase we want by maintaining the net force exactly equal to zero after the piston starts moving, since the piston will slow down and stop. But, yes, the idea is to make the velocity of the piston small enough for viscous stresses to be negligible, and we can do this by maintaining a small constant velocity until we want the piston to stop. We have total control over the kinematics of the piston motion (i.e., its displacement vs time). We can do that by driving the piston with a great big gigantic motor that is insensitive to the back pressure of the gas, and provides whatever displacement history we desire to the piston.

Chet

 

[Red_CCF]

Those superscript o's refer specifically to the "standard state" where the pressure is 1 atm.

The approach they've described in your book is very confusing, as evidenced by the fact that it has even confused a smart guy like you. I even find it a little confusing myself, and would never work with the kind of notation that they have used. Here is that approach that I learned.

1. Let h(T,p), s(T,p), and g(T,p) represent the enthalpy, entropy, and free energy of a pure species at temperature T and p.
2. Let h°(T), s°(T), and g°(T) represent the values of h(T,p), s(T,p), and g(T,p) at p = 1 atm
3. Let h°(298), s°(298), and g°(298) represent the enthalpy, entropy, and free energy of formation of the pure species from its elements at 298 K and 1 atm. So, for an elemental species, h°(298), s°(298), and g°(298) are equal to zero, and for a compound, they are not. In the case of the entropy of a compound, for example, s°(298) = (absolute entropy of compound at 298 and 1 atm) - (absolute entropies of elements comprising the compound at 298 and 1 atm).
We also have:
h°(T)=h°(298)+\int_{298}^T{C_p(T')dT'}
s°(T)=s°(298)+\int_{298}^T{\frac{C_p(T')}{T'}dT'}
g°(T)=h°-Ts°
For a species in an ideal gas mixture (or a pure species), if p is the partial pressure of a species, then,
h(T,p)=h°(T)=h°(298)+\int_{298}^T{C_p(T')dT'}
s(T,p)=s°(T)-R\ln(p)=s°(298)+\int_{298}^T{\frac{C_p(T')}{T'}dT'}-R\ln(p)
g(T,p)=h(T,p)-Ts(T,p)

The standard heat of a reaction ΔH°(T,1atm)=(sum of h°(T) of the pure reactants)-(sum of h°(T) of the pure products)
The standard free energy of reaction ΔG°(T,1atm)=(sum of g°(T) of the pure reactants)-(sum of g°(T) of the pure products)

My current understanding is, there are three sets of properties to consider, absolute h, g, s (relative to absolute zero), relative h^o,g^o,s^o (relative to standard state 298K, 1 atm), and also formation h_f, g_f, s_f, which is found from the formation reaction of a compound. If this is correct, then is it true that:

1. The formation h,g,s are independent of the reference state and I should get the same value whether absolute or relative h, g, s are used (and for water was calculated as -44kJ/kmolK) as it appears to be the difference of reactant and product h, g, or s?

2. Assuming absolute properties are used, Δh_f = ∑n_ih_prod - ∑n_ih_reac (and the same for Δs_f and Δg_f ) and h,g,s here can be either relative to absolute zero or reference state and reactants are elements only

3. I take the standard state h^o = 0 (formation) + h_sensible for elements in their natural state and using this as reference, h^o = h_f + h_sensible for compounds, where sensible heat = 0 at 298K (and same for g^o and s^o). For converting absolute and standard state properties, for enthalpy would it be h_abs = h^o + c_p(298-0).

4. For elements at absolute zero, I assume that absolute h, g, s are zero, but does this hold for compounds?

5. How come in the entropy equation there is a -Rln(P) as opposed to a ratio of P_i/P_o as in my book?

Not exactly. We are trying "match" the δP to the the P_I increase such that there's always a constant(small) net force difference on the piston.

We won't be able get the pressure increase we want by maintaining the net force exactly equal to zero after the piston starts moving, since the piston will slow down and stop. But, yes, the idea is to make the velocity of the piston small enough for viscous stresses to be negligible, and we can do this by maintaining a small constant velocity until we want the piston to stop. We have total control over the kinematics of the piston motion (i.e., its displacement vs time). We can do that by driving the piston with a great big gigantic motor that is insensitive to the back pressure of the gas, and provides whatever displacement history we desire to the piston.

Chet

If the piston is initially accelerated to some small velocity, and we control from this point on the external pressure such that the net force is always zero, what would slow the piston down?

Thanks very much

 

[Chestermiller]

My current understanding is, there are three sets of properties to consider, absolute h, g, s (relative to absolute zero), relative h^o,g^o,s^o (relative to standard state 298K, 1 atm), and also formation h_f, g_f, s_f, which is found from the formation reaction of a compound. If this is correct, then is it true that:

1. The formation h,g,s are independent of the reference state and I should get the same value whether absolute or relative h, g, s are used (and for water was calculated as -44kJ/kmolK) as it appears to be the difference of reactant and product h, g, or s?

2. Assuming absolute properties are used, Δh_f = ∑n_ih_prod - ∑n_ih_reac (and the same for Δs_f and Δg_f ) and h,g,s here can be either relative to absolute zero or reference state and reactants are elements only

3. I take the standard state h^o = 0 (formation) + h_sensible for elements in their natural state and using this as reference, h^o = h_f + h_sensible for compounds, where sensible heat = 0 at 298K (and same for g^o and s^o). For converting absolute and standard state properties, for enthalpy would it be h_abs = h^o + c_p(298-0).

I can't follow all the notation, so I'm going to take another shot at explaining it in the (simpler) way that I understand it. I hope this works for you.

Let's first focus an a compound, rather than an element. Let's talk about two equilibrium states:

State 1: The separate pure elements comprising the compound in stoichiometric proportions (to produce 1 mole of the compound) in their natural states at 298 and 1 atm. Let's call this the standard reference state.
State 2: The compound at temperature T and pressure p.

Let h(T,p) represent the change in enthalpy between State 1 and State 2. From now on, we will call this simply the enthalpy of the compound. The same goes for s(T,p) and g(T,p).

Now for an element:

State 1: The pure element in its natural state at 298 and 1 atm.
State 2: The pure element at temperature T and pressure p.

Note that
h(298,1 atm)= 0
s(298,1 atm) = 0
g(298,1 atm) = 0

In addition, we define the following for states at 1 atm (we call states at 1 atm standard states, and use a superscript o to identify them):

h(298,1 atm) = h°(298) = standard enthalpy of compound or element in the reference state = standard enthalpy of formation of compound or element in the reference state

h(T, 1 atm) = h°(T) = standard enthalpy of compound or element at temperature T.

∑n_ih°(T)_prod - ∑n_ih°(T)_reac= standard heat of formation of compound or element at temperature T

The same definitions apply to s and g.

4. For elements at absolute zero, I assume that absolute h, g, s are zero, but does this hold for compounds?

No. There is an entropy change involved in forming the compound.

5. How come in the entropy equation there is a -Rln(P) as opposed to a ratio of P_i/P_o as in my book?

I was assuming that the pressure was expressed in atm, and that P_o was 1 atm.

If the piston is initially accelerated to some small velocity, and we control from this point on the external pressure such that the net force is always zero, what would slow the piston down?

You apply a small deceleration at the end just as you applied a small acceleration at the beginning.

Chet

Thanks very much[/QUOTE]

 

[Red_CCF]

I can't follow all the notation, so I'm going to take another shot at explaining it in the (simpler) way that I understand it. I hope this works for you.

Let's first focus an a compound, rather than an element. Let's talk about two equilibrium states:

State 1: The separate pure elements comprising the compound in stoichiometric proportions (to produce 1 mole of the compound) in their natural states at 298 and 1 atm. Let's call this the standard reference state.
State 2: The compound at temperature T and pressure p.

Let h(T,p) represent the change in enthalpy between State 1 and State 2. From now on, we will call this simply the enthalpy of the compound. The same goes for s(T,p) and g(T,p).

Now for an element:

State 1: The pure element in its natural state at 298 and 1 atm.
State 2: The pure element at temperature T and pressure p.

Note that
h(298,1 atm)= 0
s(298,1 atm) = 0
g(298,1 atm) = 0

In addition, we define the following for states at 1 atm (we call states at 1 atm standard states, and use a superscript o to identify them):

h(298,1 atm) = h°(298) = standard enthalpy of compound or element in the reference state = standard enthalpy of formation of compound or element in the reference state

h(T, 1 atm) = h°(T) = standard enthalpy of compound or element at temperature T.

∑n_ih°(T)_prod - ∑n_ih°(T)_reac= standard heat of formation of compound or element at temperature T

Using the above notation, from what I understand:
h(T,1atm) = h^o(T) = h_formation + h_sensible = ∑n_ih°(T)_prod - ∑n_ih°(T)_reac + c_p(T-298)

So h(T,p) from the above notation is not absolute enthalpy of a compound or element, its reference is still set at standard state, but is the difference here that h^o essentially a subset of h(T,p) and equivalent to h(T,p = 1atm)?

If I were to define an absolute enthalpy, h_abs(T,p) reference at absolute zero, can I expect ∑n_ih_abs(T)_prod - ∑n_ih_abs(T)_reac to be the equal to the heat of formation in the last line above? If so, does this mean that h_abs(T,1atm)= h°(T) + ∫c_pdT (integral from 0 to 298K)?

No. There is an entropy change involved in forming the compound.

Is the absolute enthalpy (and therefore free energy) and formation enthalpy also equal to zero for elements and compounds?

You apply a small deceleration at the end just as you applied a small acceleration at the beginning.

Chet

Outside of the beginning and the end, why is it necessary to apply a constant small acceleration through the process vs. zero acceleration?

Thanks very much

 

[Chestermiller]

Using the above notation, from what I understand:
h(T,1atm) = h^o(T) = h_formation + h_sensible = ∑n_ih°(T)_prod - ∑n_ih°(T)_reac + c_p(T-298)

No. It's
h(T,1atm) = h^o(T) = h_formation(298) + h_sensible = h°_compound(298) - ∑n_ih°_elements(298) + c_p(T-298)=h°_compound(298)+ c_p(T-298)

So h(T,p) from the above notation is not absolute enthalpy of a compound or element, its reference is still set at standard state, but is the difference here that h^o essentially a subset of h(T,p) and equivalent to h(T,p = 1atm)?

Yes.

If I were to define an absolute enthalpy, h_abs(T,p) reference at absolute zero, can I expect ∑n_ih_abs(T)_prod - ∑n_ih_abs(T)_reac to be the equal to the heat of formation in the last line above? If so, does this mean that h_abs(T,1atm)= h°(T) + ∫c_pdT (integral from 0 to 298K)?

Well, since the last line above was not correct, and it's not a heat of formation, I don't know how to address these questions.

Is the absolute enthalpy (and therefore free energy) and formation enthalpy also equal to zero for elements and compounds?

I don't know exactly how to answer this question because I don't understand it. But, in the framework that I've described, the enthalpy of formation, entropy of formation, and free energy of formation of the elements at 298 and 1 atm are zero. The enthalpy of formation, entropy of formation, and free energy of formation of compounds at 298 and 1 atm are not.

Outside of the beginning and the end, why is it necessary to apply a constant small acceleration through the process vs. zero acceleration?

It's not. This was just an example that I conceived of which I thought would help you understand things better. But, if it works better to understand it in terms of your constant velocity example, that's fine with me.

Chet

 

[Chestermiller]

I think I have another way of explaining it in a way that you can relate to.

Let H(T,p), S(T,p), and G(T,p) represent the absolute enthalpy, absolute entropy, and absolute free energy of a pure species at temperature T and pressure p (assuming that such things can be determined). Let H°(T), S°(T), and G°(T) represent the absolute enthalpy, absolute entropy, and absolute free energy of a pure species at temperature T and pressure 1 atm. Then

h°(298)=H°(298)-∑n_iH°_i(298)
s°(298)=S°(298)-∑n_iS°_i(298)
g°(298)=G°(298)-∑n_iG°_i(298)

where the summation on the RHS is over the stoichiometric number of moles of the elements comprising the species. From this, it follows that, for the elements,

h°(298)=0
s°(298)=0
g°(298)=0

We know that

H°(T)=H°(298)+\int_{298}^T{C_p}dT
S°(T)=H°(298)+\int_{298}^T{\frac{C_p}{T}}dT
G°(T)=H°(298)-TS°(T)

From this, it follows that

h°(T)=h°(298)+\int_{298}^T{C_p}dT
s°(T)=s°(298)+\int_{298}^T{\frac{C_p}{T}}dT
g°(T)=h°(298)-Ts°(T)

Formation Properties:

At 298:

h_f°(298)=H°(298)-∑n_iH°_i(298)=h°(298)
s_f°(298)=S°(298)-∑n_iS°_i(298)=s°(298)
g_f°(298)=G°(298)-∑n_iG°_i(298)=g°(298)

At temperature T:

h_f°(T)=H°(T)-∑n_iH°_i(T)=h°(T)-∑n_ih°_i(T)
s_f°(T)=S°(T)-∑n_iS°_i(T)=s°(T)-∑n_is°_i(T)
g_f°(T)=G°(T)-∑n_iG°_i(T)=g°(T)-∑n_ig°_i(T)

Hope this helps.

Chet

 

[Red_CCF]

I think I have another way of explaining it in a way that you can relate to.

Let H(T,p), S(T,p), and G(T,p) represent the absolute enthalpy, absolute entropy, and absolute free energy of a pure species at temperature T and pressure p (assuming that such things can be determined). Let H°(T), S°(T), and G°(T) represent the absolute enthalpy, absolute entropy, and absolute free energy of a pure species at temperature T and pressure 1 atm. Then

h°(298)=H°(298)-∑n_iH°_i(298)
s°(298)=S°(298)-∑n_iS°_i(298)
g°(298)=G°(298)-∑n_iG°_i(298)

where the summation on the RHS is over the stoichiometric number of moles of the elements comprising the species. From this, it follows that, for the elements,

h°(298)=0
s°(298)=0
g°(298)=0

We know that

H°(T)=H°(298)+\int_{298}^T{C_p}dT
S°(T)=H°(298)+\int_{298}^T{\frac{C_p}{T}}dT
G°(T)=H°(298)-TS°(T)

From this, it follows that

h°(T)=h°(298)+\int_{298}^T{C_p}dT
s°(T)=s°(298)+\int_{298}^T{\frac{C_p}{T}}dT
g°(T)=h°(298)-Ts°(T)

Formation Properties:

At 298:

h_f°(298)=H°(298)-∑n_iH°_i(298)=h°(298)
s_f°(298)=S°(298)-∑n_iS°_i(298)=s°(298)
g_f°(298)=G°(298)-∑n_iG°_i(298)=g°(298)

At temperature T:

h_f°(T)=H°(T)-∑n_iH°_i(T)=h°(T)-∑n_ih°_i(T)
s_f°(T)=S°(T)-∑n_iS°_i(T)=s°(T)-∑n_is°_i(T)
g_f°(T)=G°(T)-∑n_iG°_i(T)=g°(T)-∑n_ig°_i(T)

Hope this helps.

Chet

Specific to the above, what is the difference between H and h; is the latter just a per mole basis of the former? Also, what is the actual meaning behind the word "absolute", since the absolute h^o above is equal to 0 at 298K for elements, while I am currently taking absolute to mean the property is taken relative to absolute zero.

I am trying to reorganize my question and hopefully communicate it better but I think a number of my confusions have already been answered. What I am trying understand is the mathematical relationship between: 1) the standard state (298K, 1 atm) properties (enthalpy/entropy/free energy) 2) formation properties, and 3) absolute (relative to absolute zero) properties for an element/compound (assuming this is possible).

If I use the formation reaction H2 + 1/2O2 -> H2O as an example and look at enthalpy only:

At 298K and 1atm, the standard state enthalpies of both elements are 0, but I think their absolute enthalpies are greater than zero. Is there a general relationship between the absolute and standard state enthalpies?

If the reaction takes place at 298K, 1 atm, 241,845kJ/kmol is released and equal to h^o(298)=h_f(298) and for arbitrary temperature h_f(T)= h°_H2O(T) - ∑n_ih°_element(T). Is h_f reference state dependent (intuitively I think it isn’t)? Another way of asking is, would I get the same h_f(298) = -241,845kJ/kmol using absolute enthalpy values (or even that from some arbitrary reference state) instead of standard state enthalpy values?

Thanks very much

 

[Chestermiller]

Specific to the above, what is the difference between H and h; is the latter just a per mole basis of the former?

H is the absolute enthalpy per mole, and h is the enthalpy per mole relative to stoichiometric proportions of the elements at the same temperature and pressure. h is defined by the first equation in my most recent respnse.

Also, what is the actual meaning behind the word "absolute", since the absolute h^o above is equal to 0 at 298K for elements, while I am currently taking absolute to mean the property is taken relative to absolute zero.

There is no such thing as "the absolute h^o". Absolute means the property taken relative to the elements at absolute zero.

I am trying to reorganize my question and hopefully communicate it better but I think a number of my confusions have already been answered. What I am trying understand is the mathematical relationship between: 1) the standard state (298K, 1 atm) properties (enthalpy/entropy/free energy) 2) formation properties, and 3) absolute (relative to absolute zero) properties for an element/compound (assuming this is possible).

That's what I tried to do in my most recent response.

If I use the formation reaction H2 + 1/2O2 -> H2O as an example and look at enthalpy only:

At 298K and 1atm, the standard state enthalpies of both elements are 0, but I think their absolute enthalpies are greater than zero. Is there a general relationship between the absolute and standard state enthalpies?

That was covered in my most recent response. I was hoping that it would make sense to you. Apparently not. I'm running out of ideas on how to convey it.

If the reaction takes place at 298K, 1 atm, 241,845kJ/kmol is released and equal to h^o(298)=h_f(298) and for arbitrary temperature h_f(T)= h°_H2O(T) - ∑n_ih°_element(T). Is h_f reference state dependent (intuitively I think it isn’t)?

h_f°(T) involves two equilibrium states; it is the change in enthalpy between these two equilibrium states: (1) 1 mole of H2O at T and 1 atm and (2) 1/2 mole of O2 and 1 mole of H2 (in separate containers) at T and 1 atm.

Another way of asking is, would I get the same h_f(298) = -241,845kJ/kmol using absolute enthalpy values (or even that from some arbitrary reference state) instead of standard state enthalpy values?

Do the algebra, and see what you get.

Chet

 

[Red_CCF]

I think I have another way of explaining it in a way that you can relate to.

Let H(T,p), S(T,p), and G(T,p) represent the absolute enthalpy, absolute entropy, and absolute free energy of a pure species at temperature T and pressure p (assuming that such things can be determined). Let H°(T), S°(T), and G°(T) represent the absolute enthalpy, absolute entropy, and absolute free energy of a pure species at temperature T and pressure 1 atm.

I just want to make sure that I understood the notations correctly (and using enthalpy as an example):

H(T,p) is absolute entropy per mole, relative to absolute zero. H^o(T) is a subset of H and = H(T,1atm). Both of these quantities have nothing to do with the standard state reference (298K, 1atm) where element enthalpy value is zero. Is it correct to say that H^o(298K) for any pure species (element/compound) is > 0 and ≠ 0 for elements.

h°(298)=H°(298)-∑n_iH°_i(298)
s°(298)=S°(298)-∑n_iS°_i(298)
g°(298)=G°(298)-∑n_iG°_i(298)

where the summation on the RHS is over the stoichiometric number of moles of the elements comprising the species. From this, it follows that, for the elements,

h°(298)=0
s°(298)=0
g°(298)=0

h_f°(298)=H°(298)-∑n_iH°_i(298)=h°(298)
s_f°(298)=S°(298)-∑n_iS°_i(298)=s°(298)
g_f°(298)=G°(298)-∑n_iG°_i(298)=g°(298)

With regards to h^o(T), am I correct to take this to be equivalent to enthalpy defined relative to standard state (298K, 1atm)? I came to this conclusion based on the fact that h^o(298) = 0 for elements and h^o(298) = h_f°(298).

Do the algebra, and see what you get.

Chet

I’m having trouble finding absolute enthalpy values, but my book does have absolute entropy values so I will use those and the formation of water H2+1/2O2 -> H2O as before.

Per the equation given:
S^o_f = S^o_H2O(T) – 1/2S_O2^o(T) – S_H2^o(T) = 188.715 - 0.5*205.043- 130.595= -44.4kJ/kmolK

If the above is correct and my assumption is correct, then if element entropy is taken to equal zero at standard state (298K, 1atm), then the entropy of H2O would be -44.4kJ/kmolK at standard state. However according to Wikipedia it appears it is convention to specify standard molar entropy values relative to absolute zero and thus I haven’t been able to find a table where element entropies are taken as zero at standard state to verify this.

Thanks very much

 

[Chestermiller]

I just want to make sure that I understood the notations correctly (and using enthalpy as an example):

H(T,p) is absolute entropy per mole, relative to absolute zero. H^o(T) is a subset of H and = H(T,1atm).

Yes.

Both of these quantities have nothing to do with the standard state reference (298K, 1atm) where element enthalpy value is zero. Is it correct to say that H^o(298K) for any pure species (element/compound) is > 0 and ≠ 0 for elements.

Yes.

With regards to h^o(T), am I correct to take this to be equivalent to enthalpy defined relative to standard state (298K, 1atm)?

No. It's enthalpy at temperature T defined relative to the enthalpy of the elements comprising the substance at the standard state (298K, 1 atm)

I came to this conclusion based on the fact that h^o(298) = 0 for elements and h^o(298) = h_f°(298).

You would come to the same conclusion using the definition I gave above.

I’m having trouble finding absolute enthalpy values, but my book does have absolute entropy values so I will use those and the formation of water H2+1/2O2 -> H2O as before.

Per the equation given:
S^o_f = S^o_H2O(T) – 1/2S_O2^o(T) – S_H2^o(T) = 188.715 - 0.5*205.043- 130.595= -44.4kJ/kmolK

If the above is correct and my assumption is correct, then if element entropy is taken to equal zero at standard state (298K, 1atm), then the entropy of H2O would be -44.4kJ/kmolK at standard state. However according to Wikipedia it appears it is convention to specify standard molar entropy values relative to absolute zero and thus I haven’t been able to find a table where element entropies are taken as zero at standard state to verify this.

What I had in mind here was not focusing on a specific formation reaction. I wanted you to try to prove it algebraically.

Here's what I'd like you to try to show algebraically:

If h_f°(T)=H°(T)-∑n_iH_i°(T)

then h_f°(T)=h°(T)-∑n_ih_i°(T)

Chet

 

[Red_CCF]

No. It's enthalpy at temperature T defined relative to the enthalpy of the elements comprising the substance at the standard state (298K, 1 atm)

You would come to the same conclusion using the definition I gave above.

H is the absolute enthalpy per mole, and h is the enthalpy per mole relative to stoichiometric proportions of the elements at the same temperature and pressure. h is defined by the first equation in my most recent respnse.

With regards to h^o(T), per the first quote, isn't that the same as the enthalpy defined relative to standard state? For the second quote, my interpretation is that h^o(T) is defined relative to the the stoichiometric proportions of its elements at its actual T (so I need to subtract the sensible heat of the elements from 298K to T of the compound) as opposed to just at 298K (per standard state definition), is this correct? This implies h°(T)=H°(T)-∑n_iH°_i(T) which for elements is 0 for all T which didn't make much sense to me.

H°(T)=H°(298)+\int_{298}^T{C_p}dT

From this, it follows that

h°(T)=h°(298)+\int_{298}^T{C_p}dT

Is h°(T)=H°(T)-∑n_iH°_i(T)+∑n_ih°_i(T) (based on the equality in h_f°(T))? I'm having trouble showing how to arrive at the definition of h°(T) quoted above from H°(T).

What I had in mind here was not focusing on a specific formation reaction. I wanted you to try to prove it algebraically.

Here's what I'd like you to try to show algebraically:

If h_f°(T)=H°(T)-∑n_iH_i°(T)

then h_f°(T)=h°(T)-∑n_ih_i°(T)

Chet

For H°(T)-∑n_iH_i°(T)=h°(T)-∑n_ih_i°(T), per definition of H°(T) and h°(T), the sensible components cancel as

H°(T)-h°(T)=∑n_iH_i°(T)-∑n_ih_i°(T)

and the equation reduces to H°(298)-∑n_iH_i°(298)=h°(298)-∑n_ih_i°(298) = h°(298). What I'm left with is that h_f°(T) is constant for all T?

Thank you very much

 

[Chestermiller]

With regards to h^o(T), per the first quote, isn't that the same as the enthalpy defined relative to standard state?

Yes, provided what you call the standard state refers its elements at the standard state of 298 and 1 atm.

For the second quote, my interpretation is that h^o(T) is defined relative to the the stoichiometric proportions of its elements at its actual T (so I need to subtract the sensible heat of the elements from 298K to T of the compound) as opposed to just at 298K (per standard state definition), is this correct?

I was unable to find the equation I was referring to in the quote. But, in any event, your statement is incorrect. h^o(T) is defined at temperature T relative to the the stoichiometric proportions of its elements at 298 and 1 atm (the standard reference state). On the other hand, h_f° is defined at temperature T relative to the stoichiometric proportions of its elements at its actual T.

This implies h°(T)=H°(T)-∑n_iH°_i(T) which for elements is 0 for all T which didn't make much sense to me.

No. This is the definition of h_f°(T).

Is h°(T)=H°(T)-∑n_iH°_i(T)+∑n_ih°_i(T) (based on the equality in h_f°(T))?

This is incorrect.

For H°(T)-∑n_iH_i°(T)=h°(T)-∑n_ih_i°(T), per definition of H°(T) and h°(T), the sensible components cancel as

H°(T)-h°(T)=∑n_iH_i°(T)-∑n_ih_i°(T)

and the equation reduces to H°(298)-∑n_iH_i°(298)=h°(298)-∑n_ih_i°(298) = h°(298). What I'm left with is that h_f°(T) is constant for all T?

This is not done correctly. And, if done correctly, the sensible heats do not cancel.

Let's start with the equation I gave for h_f°(T):

h_f°(T)=H°(T)-∑n_iH°_i(T)

Now, H°(T)=H°(298)+\int_{298}^T{C_pdT}
and, for its elements, H_i°(T)=H_i°(298)+\int_{298}^T{C_{pi}dT}
If we substitute this into the equation for h_f°(T), we obtain:

h_f°(T)=(H°(298)-∑n_iH_i°(298))+\int_{298}^T{C_pdT}-∑n_i\int_{298}^T{C_{pi}dT}
But, (H°(298)-∑n_iH_i°(298))=h°(298)
So,
h_f°(T)=h°(298)+\int_{298}^T{C_pdT}-∑n_i\int_{298}^T{C_{pi}dT}
But, h°(298)+\int_{298}^T{C_pdT}=h°(T)
and 0+\int_{298}^T{C_{pi}dT}=h_i°(T) for the elements.
So,
h_f°(T)=h°(T)-∑n_ih_i°(T)

Chet

 

[Red_CCF]

Yes, provided what you call the standard state refers its elements at the standard state of 298 and 1 atm.

Let's start with the equation I gave for h_f°(T):

h_f°(T)=H°(T)-∑n_iH°_i(T)

Now, H°(T)=H°(298)+\int_{298}^T{C_pdT}
and, for its elements, H_i°(T)=H_i°(298)+\int_{298}^T{C_{pi}dT}
If we substitute this into the equation for h_f°(T), we obtain:

h_f°(T)=(H°(298)-∑n_iH_i°(298))+\int_{298}^T{C_pdT}-∑n_i\int_{298}^T{C_{pi}dT}
But, (H°(298)-∑n_iH_i°(298))=h°(298)
So,
h_f°(T)=h°(298)+\int_{298}^T{C_pdT}-∑n_i\int_{298}^T{C_{pi}dT}
But, h°(298)+\int_{298}^T{C_pdT}=h°(T)
and 0+\int_{298}^T{C_{pi}dT}=h_i°(T) for the elements.
So,
h_f°(T)=h°(T)-∑n_ih_i°(T)

Per this definition then is h°(T) = H°(T) - ∑n_iH_i(298)? If so, then is the following correct to show the relationship you provided for h°(T):

h°(T) = H°(T) - ∑n_iH_i(298) = H°(298) - ∑n_iH_i(298) + \int_{298}^T{C_{pi}dT}
h°(T) = h°(298) + \int_{298}^T{C_{pi}dT}

Can I interpret this as equivalent to the standard enthalpy conventionally defined and used in textbooks unless there are some minor differences that I am not seeing?

Essentially the process we went through was, define an absolute enthalpy H°(T) relative to absolute zero, then the enthalpy of formation becomes h_f°(T) = H°(T) - ∑n_iH_i(T). This is followed by defining the relative enthalpy h°(T) = H°(T) - ∑n_iH_i(298) that became the standard state enthalpy (if my first line is correct) and expressing h_f°(T) in terms of h°(T) and h°(T) in terms of h°(298) follow from the above definitions, is this correct?

Thanks very much

 

[Chestermiller]

Per this definition then is h°(T) = H°(T) - ∑n_iH_i(298)?

Yes. Yes. and Yes.

If so, then is the following correct to show the relationship you provided for h°(T):

h°(T) = H°(T) - ∑n_iH_i(298) = H°(298) - ∑n_iH_i(298) + \int_{298}^T{C_{pi}dT}
h°(T) = h°(298) + \int_{298}^T{C_{pi}dT}

Yes, if you remove the "i" subscript from C_p, since, in what we have been doing, we have been using the i subscript to refer to the elements.

Can I interpret this as equivalent to the standard enthalpy conventionally defined and used in textbooks unless there are some minor differences that I am not seeing?

I think so.

Essentially the process we went through was, define an absolute enthalpy H°(T) relative to absolute zero, then the enthalpy of formation becomes h_f°(T) = H°(T) - ∑n_iH_i(T). This is followed by defining the relative enthalpy h°(T) = H°(T) - ∑n_iH_i(298) that became the standard state enthalpy (if my first line is correct) and expressing h_f°(T) in terms of h°(T) and h°(T) in terms of h°(298) follow from the above definitions, is this correct?

Yes. I think you finally have it. But please understand that, in typical developments, they don't find it necessary to introduce the absolute enthalpy (whatever the absolute enthalpy means) into the analysis. I only introduced this formalism here because I thought it might make it easier for you to overcome your barrier to understanding. Of course, in the case of entropy, the absolute entropy is often used, and the development exactly parallels the one you described above (except, of course, for using C_p/T in place of C_p).

Chet

 

[Red_CCF]

Yes. I think you finally have it. But please understand that, in typical developments, they don't find it necessary to introduce the absolute enthalpy (whatever the absolute enthalpy means) into the analysis. I only introduced this formalism here because I thought it might make it easier for you to overcome your barrier to understanding. Of course, in the case of entropy, the absolute entropy is often used, and the development exactly parallels the one you described above (except, of course, for using C_p/T in place of C_p).

Chet

Is there a reason why absolute entropy is used but absolute enthalpy/free energy is not, especially since I imagine that for free energy the entropy used is relative to standard state? Is it typical that enthalpy tables (like steam tables) use standard state enthalpy, as I compared the steam table in my thermodynamics book with H₂O in my combustion book and they are very different?

With regards to elements, are we defining it as the lowest energy state/most stable form we encounter so for something like H₂, the formation reaction would be H₂ -> H₂ (not something like 2H -> H₂)? For the formation of H i assume it is 1/2H₂ -> H?

Thanks very much

 

[Chestermiller]

Is there a reason why absolute entropy is used but absolute enthalpy/free energy is not, especially since I imagine that for free energy the entropy used is relative to standard state?

I don't know. I have never seen a single book that presents absolute enthalpy or absolute free energy, and I'm not even sure that such concepts make sense or can be determined. The practicality of the matter is that, to do industrial calculations (e.g., chemical equilibrium), we don't need to know these functions down to absolute zero, but we do need to know them at typical process temperatures, which are above 0C. So why bother to measure and determine these functions all the way down to absolute zero, when our methodology works perfectly well using the function values relative to the elements (in their natural states) at the reference conditions.

In many textbooks, they don't even work with entropy in determining equilibrium constants. They just give the enthalpy of formation and the free energy of formation at the reference state, together with the heat capacity as a function of temperature. This information is sufficient to determine the free energy of formation as a function of temperature at 1 atm., which is sufficient to determine the equilibrium constant as a function of temperature.

Is it typical that enthalpy tables (like steam tables) use standard state enthalpy, as I compared the steam table in my thermodynamics book with H₂O in my combustion book and they are very different?

Yes. For steam calculations, it is convenient to choose as the reference state liquid water at 0C and 1 atm. For chemical equilibrium calculations, the reference state is conventionally chosen as 25C and 1 atm. No big deal. When doing calculations, you just need to understand what reference state they are using.

With regards to elements, are we defining it as the lowest energy state/most stable form we encounter so for something like H₂, the formation reaction would be H₂ -> H₂ (not something like 2H -> H₂)?

Yes. It is the natural state at 25C and 1 atm.

For the formation of H i assume it is 1/2H₂ -> H?

Yes.

 

[Red_CCF]

I don't know. I have never seen a single book that presents absolute enthalpy or absolute free energy, and I'm not even sure that such concepts make sense or can be determined. The practicality of the matter is that, to do industrial calculations (e.g., chemical equilibrium), we don't need to know these functions down to absolute zero, but we do need to know them at typical process temperatures, which are above 0C. So why bother to measure and determine these functions all the way down to absolute zero, when our methodology works perfectly well using the function values relative to the elements (in their natural states) at the reference conditions.

In many textbooks, they don't even work with entropy in determining equilibrium constants. They just give the enthalpy of formation and the free energy of formation at the reference state, together with the heat capacity as a function of temperature. This information is sufficient to determine the free energy of formation as a function of temperature at 1 atm., which is sufficient to determine the equilibrium constant as a function of temperature.

Are there anything entropy is useful for in practice and is there a reason they are not used very much?

So both absolute and standard state references similar in the sense that species' enthalpy are defined relative to their constituent elements and only dissimilar in that the actual state of reference is different?

Just for theoretical purposes, not that it is useful, but if I wanted to find the relationship in enthalpy change from standard state to absolute zero, does one exist (something like ∫c_pdT but accounting for pressure change) or is this something that has to be measured empirically?

No, but we are just trying to address some the doubts you had regarding the transition from small finite piston mass to the limit of zero piston mass, such as:

1. How can you have an accelerating piston if the piston mass is zero?
2. Does the piston oscillate if its mass is very small, but not zero?
3. Is the transition smooth from small finite piston mass to zero piston mass?
This solution answers all these questions.

We can continue the solution to larger mass, if you'd like. If mk/C^2 is equal to 1, the system is critically damped. If it is greater than 1, then the system may oscillate, depending on whether static friction is re-established at the moment that the piston reaches its maximum compression (for the first time).

With regards to the piston example we did earlier, one key assumption was that ΔP_ext = 0 at all times. If this were not true (say for a quasistatic or a less "violent" non-quasistatic process), how would one approach this problem; is it possible without doing this numerically?

Thank you very much

 

[Chestermiller]

Are there anything entropy is useful for in practice and is there a reason they are not used very much?

Entropy and internal energy are fundamental to the whole development. The other thermodynamic functions, H, G, and A are derived from them. However, G and A are useful with regard to solving problems involving thermodynamic equilibrium (phase equilibrium and chemical equilibrium).

So both absolute and standard state references similar in the sense that species' enthalpy are defined relative to their constituent elements and only dissimilar in that the actual state of reference is different?

Yes.

Just for theoretical purposes, not that it is useful, but if I wanted to find the relationship in enthalpy change from standard state to absolute zero, does one exist (something like ∫c_pdT but accounting for pressure change) or is this something that has to be measured empirically?

Yes. Are you aware of the relationship between dH, and dT and dP for a single phase pure substance? Also, there may be some phase changes along the way.

With regards to the piston example we did earlier, one key assumption was that ΔP_ext = 0 at all times. If this were not true (say for a quasistatic or a less "violent" non-quasistatic process), how would one approach this problem; is it possible without doing this numerically?

If you specified what ΔP_ext was as a function of time, then you could solve for the corresponding piston displacement analytically. A numerical solution would not be required because, in our simplified model, the system is linear.

Chet

 

[Red_CCF]

Entropy and internal energy are fundamental to the whole development. The other thermodynamic functions, H, G, and A are derived from them. However, G and A are useful with regard to solving problems involving thermodynamic equilibrium (phase equilibrium and chemical equilibrium).

How are entropy and internal energy measured empirically as tabulated in textbooks?

Yes. Are you aware of the relationship between dH, and dT and dP for a single phase pure substance? Also, there may be some phase changes along the way.

Is it the following relationship:

dh = (\frac{\partial{h}}{\partial{T}})_PdT+(\frac{\partial{h}}{\partial{P}})_TdP

assuming integrating this is possible, the absolute enthalpy H(T, P) of a compound or element would be the integral of the first term on the RHS from 0K to T and the second term from 0atm to P and for H(T, 1 atm) = H^o(T) the pressure integral is always from 0 to 1 atm? I imagine that phase changes would be accounted for by separating integral into three parts and adding h_sublimation and h_vaporization (assuming we are ending up with a gas).

If you specified what ΔP_ext was as a function of time, then you could solve for the corresponding piston displacement analytically. A numerical solution would not be required because, in our simplified model, the system is linear.

Chet

If the process was dictated to be quasistatic, is it possible to predict a ΔP_ext(t) which will allow such process to occur, or must ΔP_ext be determined empirically? Also, does the addition of this term only change the particular/steady state solution of the ODE?

Thank you very much

 

[Chestermiller]

How are entropy and internal energy measured empirically as tabulated in textbooks?
You already did many problems about this when you studied the first and second laws. You just have to conceive of experiments that focus on these parameters. In the case of entropy, your experiment has to be done under nearly reversible conditions.

Is it the following relationship:

dh = (\frac{\partial{h}}{\partial{T}})_PdT+(\frac{\partial{h}}{\partial{P}})_TdP

Look this up in Smith and Van Ness. The expression for the partial of h with respect to P is obtained using a Maxwell relationship, and this derivative can be calculated from the P-V-T behavior of the material.

assuming integrating this is possible, the absolute enthalpy H(T, P) of a compound or element would be the integral of the first term on the RHS from 0K to T and the second term from 0atm to P and for H(T, 1 atm) = H^o(T) the pressure integral is always from 0 to 1 atm?

As I said, the second term is calculated from the P-V-T behavior of the material.

I imagine that phase changes would be accounted for by separating integral into three parts and adding h_sublimation and h_vaporization (assuming we are ending up with a gas).

The integrations would be done backwards, starting from the ideal gas state at normal temperatures and low pressures. For the ideal gas region, h is not a function of pressure.

If the process was dictated to be quasistatic, is it possible to predict a ΔP_ext(t) which will allow such process to occur, or must ΔP_ext be determined empirically?

We already did quasistatic cases in which the acceleration and the velocity approach zero. You yourself discussed a case where you apply a short initial small acceleration, followed by a constant small velocity, followed by a short final deceleration. I discussed a case of a very small acceleration applied over the entire deformation, followed by a short deceleration to stop the deformation. For both these cases, ΔP_ext(t) is slightly different, but the difference is insignificant in the quasistatic limit. Both these cases give the exact same result for ΔP_ext as a function of Δx in the limit in which the acceleration and the velocity approach zero. This is the quasistatic case.

Also, does the addition of this term only change the particular/steady state solution of the ODE?

No. It changes the entire history of the deformation. However, in the limit of quasistatic deformation, every point along the transition is essentially steady state.

Chet

 

[Red_CCF]

Look this up in Smith and Van Ness. The expression for the partial of h with respect to P is obtained using a Maxwell relationship, and this derivative can be calculated from the P-V-T behavior of the material.

As I said, the second term is calculated from the P-V-T behavior of the material.

The integrations would be done backwards, starting from the ideal gas state at normal temperatures and low pressures. For the ideal gas region, h is not a function of pressure.

Were you referring to the following relationship for ∂h/∂P:

\frac{∂h}{∂P} = [V-T(\frac{∂V}{∂T})_P]dP

Does the above only applied to fluids (the book appear to imply this) or any phased material?

Realistically, what is the "path" that is acceptable to perform the integral? I assume that we cannot just integrate from 1 to 0 atm at constant 298K and then from 298K to 0K at constant 0 atm or vice versa? Typically at 1 atm, how low must the temperature go before we begin to deviate from ideal gas?

We already did quasistatic cases in which the acceleration and the velocity approach zero. You yourself discussed a case where you apply a short initial small acceleration, followed by a constant small velocity, followed by a short final deceleration. I discussed a case of a very small acceleration applied over the entire deformation, followed by a short deceleration to stop the deformation. For both these cases, ΔP_ext(t) is slightly different, but the difference is insignificant in the quasistatic limit. Both these cases give the exact same result for ΔP_ext as a function of Δx in the limit in which the acceleration and the velocity approach zero. This is the quasistatic case.

For quasistatic compression, many textbooks describes it generally as increasing external pressure by dP -> piston compresses -> wait until equilibrium is established -> repeat. By this logic I feel that the piston must always be accelerating since there is always a net external dP acting on the piston, is this correct? I had a feeling after further thought that a constant velocity quasistatic compression is not achievable since there is always a need for continuous net force?

No. It changes the entire history of the deformation. However, in the limit of quasistatic deformation, every point along the transition is essentially steady state.

Chet

With regards to the original equation copied below:

m\frac{d^2(Δx_1)}{dt^2}+C\frac{d(Δx_1)}{dt}+kΔx_1= -AΔP_{ext}(t)-(F_{stat}-F_{kin})

I'm having trouble seeing the significance of the change of including ΔP_ext(t), since from what I see the homogeneous equation would stay the same as if ΔP_ext(t) = 0, and only steady state solution changes which will only affect the "shift" of the transient stage/initial compression.

Thank you very much

 

[Chestermiller]

Were you referring to the following relationship for ∂h/∂P:

\frac{∂h}{∂P} = [V-T(\frac{∂V}{∂T})_P]dP

Does the above only applied to fluids (the book appear to imply this) or any phased material?

This applies to solids, liquids, and gases.

Realistically, what is the "path" that is acceptable to perform the integral? I assume that we cannot just integrate from 1 to 0 atm at constant 298K and then from 298K to 0K at constant 0 atm or vice versa? Typically at 1 atm, how low must the temperature go before we begin to deviate from ideal gas?

You can integrate over whatever path you wish. At 1 atm and below, the behavior of most gases is ideal, and for liquids and solids, the pressure term is usually small.

I don't understand your obsession with trying to integrate down to absolute zero. You seem to be spending a lot of your valuable thinking time on something that, to me, doesn't seem very important practically.

For quasistatic compression, many textbooks describes it generally as increasing external pressure by dP -> piston compresses -> wait until equilibrium is established -> repeat.

This seems to be referring to scenarios in which the external pressure is applied in small discrete increments. If you work this out using our equations, you will find that, as you consider cases in which the increments become smaller and smaller, the amount of irreversibility from the damping combined with the inertia of the piston becomes less and less and approaches the exact quasistatic case of no net force.

By this logic I feel that the piston must always be accelerating since there is always a net external dP acting on the piston, is this correct?

Try doing the analysis using small incremental increases in the external pressure followed by re-equilibration in-between, and see for yourself what you get. (To make things simple, do it for the case of no friction). You have a math model to work with. See what it tells you.

I had a feeling after further thought that a constant velocity quasistatic compression is not achievable since there is always a need for continuous net force?

Just plug in a constant velocity into the differential equation, and see what it tells you. You will find that you can maintain a constant velocity without having a net force. Understand that Pext is not the net force.

With regards to the original equation copied below:

m\frac{d^2(Δx_1)}{dt^2}+C\frac{d(Δx_1)}{dt}+kΔx_1= -AΔP_{ext}(t)-(F_{stat}-F_{kin})

I'm having trouble seeing the significance of the change of including ΔP_ext(t), since from what I see the homogeneous equation would stay the same as if ΔP_ext(t) = 0, and only steady state solution changes which will only affect the "shift" of the transient stage/initial compression.

Sorry, I don't understand this question.

Chet

 

[Red_CCF]

You can integrate over whatever path you wish. At 1 atm and below, the behavior of most gases is ideal, and for liquids and solids, the pressure term is usually small.

I don't understand your obsession with trying to integrate down to absolute zero. You seem to be spending a lot of your valuable thinking time on something that, to me, doesn't seem very important practically.

I just wanted to get an theoretical understanding of how properties are related between standard reference state and another arbitrary reference state (absolute zero just came to mind first as an example). When you say that I can integrate along any path, does that mean the path I proposed is valid, since it goes through some states where I don't think is physically possible (i.e. no pressure at finite temperature)?

This seems to be referring to scenarios in which the external pressure is applied in small discrete increments. If you work this out using our equations, you will find that, as you consider cases in which the increments become smaller and smaller, the amount of irreversibility from the damping combined with the inertia of the piston becomes less and less and approaches the exact quasistatic case of no net force.

Try doing the analysis using small incremental increases in the external pressure followed by re-equilibration in-between, and see for yourself what you get. (To make things simple, do it for the case of no friction). You have a math model to work with. See what it tells you.

How come the quasistatic case is said to have no net force?

For the case of no friction, at the first time increment when ΔP_ext(0) = 0 and ΔP_ext(dt) = δP I have the following from reducing the equation:

m\frac{d^2(Δx_1)}{dt^2}+C\frac{d(Δx_1)}{dt}+kΔx_1= -AΔP_ext(t) = -AδP

I am a little lost after this stage, as I am unsure how to add the re-equilibration stage into the equation and that as δP -> dP that ma + cv -> 0.

Just plug in a constant velocity into the differential equation, and see what it tells you. You will find that you can maintain a constant velocity without having a net force. Understand that Pext is not the net force.

Chet

The net force I was referring to was the difference between P_ext and the internal gas pressure. From previous equations at the first time increment, I concluded that for constant velocity, AdP = -kΔx₁ - Cv_piston so v = -1/C(AdP + kΔx₁). I believe that Δx₁ < 0 as t > 0, and for v to stay constant this implies that the difference P_ext and Δx is constant (plotted over time the two have the same trend). This seems to make sense intuitively since kΔx is analogous to gas pressure which for quasistatic process should balance the applied pressure.

Thank you very much

 

[Chestermiller]

I just wanted to get an theoretical understanding of how properties are related between standard reference state and another arbitrary reference state (absolute zero just came to mind first as an example). When you say that I can integrate along any path, does that mean the path I proposed is valid, since it goes through some states where I don't think is physically possible (i.e. no pressure at finite temperature)?



How come the quasistatic case is said to have no net force?

For the case of no friction, at the first time increment when ΔP_ext(0) = 0 and ΔP_ext(dt) = δP I have the following from reducing the equation:

m\frac{d^2(Δx_1)}{dt^2}+C\frac{d(Δx_1)}{dt}+kΔx_1= -AΔP_ext(t) = -AδP

I am a little lost after this stage, as I am unsure how to add the re-equilibration stage into the equation and that as δP -> dP that ma + cv -> 0.



The net force I was referring to was the difference between P_ext and the internal gas pressure. From previous equations at the first time increment, I concluded that for constant velocity, AdP = -kΔx₁ - Cv_piston so v = -1/C(AdP + kΔx₁). I believe that Δx₁ < 0 as t > 0, and for v to stay constant this implies that the difference P_ext and Δx is constant (plotted over time the two have the same trend). This seems to make sense intuitively since kΔx is analogous to gas pressure which for quasistatic process should balance the applied pressure.

Thank you very much

I can help you figure out the answers to all these questions, but it will be in a couple of days. Grandchildren visiting!

Chet

 

[Chestermiller]

I just wanted to get an theoretical understanding of how properties are related between standard reference state and another arbitrary reference state (absolute zero just came to mind first as an example). When you say that I can integrate along any path, does that mean the path I proposed is valid, since it goes through some states where I don't think is physically possible (i.e. no pressure at finite temperature)?

How come the quasistatic case is said to have no net force?

For the case of no friction, at the first time increment when ΔP_ext(0) = 0 and ΔP_ext(dt) = δP I have the following from reducing the equation:

m\frac{d^2(Δx_1)}{dt^2}+C\frac{d(Δx_1)}{dt}+kΔx_1= -AΔP_ext(t) = -AδP

I am a little lost after this stage, as I am unsure how to add the re-equilibration stage into the equation and that as δP -> dP that ma + cv -> 0.

The net force I was referring to was the difference between P_ext and the internal gas pressure. From previous equations at the first time increment, I concluded that for constant velocity, AdP = -kΔx₁ - Cv_piston so v = -1/C(AdP + kΔx₁). I believe that Δx₁ < 0 as t > 0, and for v to stay constant this implies that the difference P_ext and Δx is constant (plotted over time the two have the same trend). This seems to make sense intuitively since kΔx is analogous to gas pressure which for quasistatic process should balance the applied pressure.

Thank you very much

We are going to answer your questions by doing a little modeling. There are three basic principles that are important in doing modeling:

1. Start simple
2. Start simple
3. Start simple

Why is it so important to start simple? Because, if you can't solve the simpler versions of your problem, you certainly won't be able to solve the more complicated versions. And, after you solve a simpler version of a problem, you will already have some results under your belt to compare against.

I'm going to formulate two problems for you to work on, representing simpler versions of what you are asking.

Problem A: In your thermo tables, they give two different values for the heat of formation of water at 25 C and 1 atm. One of these is for liquid water, and the other is for the hypothetical state of water vapor. How do they get the value for the hypothetical state of water vapor from the value for liquid water? That is, how do they get the change in enthalpy from liquid water at 1 and 25C to the hypothetical state of water vapor at 1 atm and 25C?

Problem B: (part 1)
I'm going to reformulate the equation relating Pext(t) to x(t) (no friction and no piston mass) in a slightly different form:
AP_{ext}(t)=AP_{ext}(0)-C\frac{dx}{dt}-kx
where now, x is the displacement relative to the length of the spring at equilibrium at P_ext(0), and AP_ext(0)=P_Ei is the imposed external force at time zero. At time t = 0, the imposed external load is suddenly changed to P_Ef, and held at that value for all subsequent time. Please solve for x(t) as a function of P_Ei, P_Ef, t, k, and C.

After you solve for this, we will look at the amount of work done by the external force, and the contributions of the spring and the damper to that work. We will then subdivide the load change P_Ef-P_Ei into smaller incremental steps, with equilibration between the sequential steps to see how the total amount of work done changes as the number of steps increases. We will compare this with the quasistatic result.

Chet

 

[Chestermiller]

It looks like Red_CCF is no longer participating in this thread. Is there anyone else out there who has been following this thread, and who would like me to complete the solution to Problem B? If not, I'll end it here.

Chet