銀月遊俠 2013-10-27 11:53 AM

Also, some partial derivative terms are confusing such as this one (dH/dP)T, which is from the total differential of H, howcome a constant pressure heat change H can change with pressure P?? Isn't that H itself already implies constant pressure condition, then there should be no change in pressure. Another one is (dP/dS)v howcome the pressure change and the volume is kept constant?

If the Helmholtz free energy is defined as dA=d(U-TS), then I can write it as dA=δq+δw-TdS and leads to dA=δq-TdS by imposing the constant volume and temperature condition. Then does that mean the Helmholtz energy is the heat that is not contributed to entropy so that this heat can be used to do work under constant volume, but if the condition is under constant volume how a system does PV work?

Or does the “maximum work definition” of free energy terms only talking about condition other than constant volume?

For the third law of thermodynamics, there exists residual entropy that is even at absolute zero (which cannot be reached) some substances still have entropy. I know the 3rd Law only applies to pure substances so this situation will not bother the Law. Then, how can we set a reference line for the entropy of such impure substances?

[[i] 本帖最後由 銀月遊俠 於 2013-10-27 09:58 PM 編輯 [/i]]

jmlo 2013-10-29 01:48 PM

Here you have to recall the definition of H:

H = U + PV

The total differential of enthalpy is

dH = dU + d(PV)

Therefore, if pressure and volume are constant, then there will be no mechanical work done, implying that dH = dU. In this case, dH = Cv dT

However, there is a subtle difference. By definition, under constant pressure, a positive change of H (i.e., dH > 0) equals the heat absorbed by the system; similarly, a negative change of dH (i.e., dH < 0) is equal to the heat released. But under constant volume, the change of H is not equal to the heat absorbed/released.

Such case usually occurs for solids and liquids where volume and pressure are essentially constant during reactions. This is not the case for ideal gas systems since

Cp = Cv + nR

and

dH = dU + RT dn

dH = dU only if dn = 0; i.e., the number of moles of species is conserved in the reaction.

[quote] Also, some partial derivative terms are confusing such as this one (dH/dP)T, which is from the total differential of H, howcome a constant pressure heat change H can change with pressure P?? Isn't that H itself already implies constant pressure condition, then there should be no change in pressure. Another one is (dP/dS)v howcome the pressure change and the volume is kept constant? [/quote]

As stated above, enthalpy can vary with pressure; i.e., it is a function of pressure. It is only that when pressure is constant, the change of enthalpy is equal to the heat change of the system. Hence, don't calculate dH using dH = dq when P is not constant.

Do you mean (dP/dT)v instead? For ideal gas, it is equal to nR/V. When n is constant, increasing the temperature will lead to a higher average kinetic energy of the atoms/molecules, and thus a higher pressure of the gas.

[quote] If the Helmholtz free energy is defined as dA=d(U-TS), then I can write it as dA=δq+δw-TdS and leads to dA=δq-TdS by imposing the constant volume and temperature condition. Then does that mean the Helmholtz energy is the heat that is not contributed to entropy so that this heat can be used to do work under constant volume, but if the condition is under constant volume how a system does PV work?

Or does the “maximum work definition” of free energy terms only talking about condition other than constant volume? [/quote]

Your arithmetic here is not right. Starting from A = U - TS and knowing that U = q + w, we have (under reversible conditions)

dA = dU - TdS - SdT = dq + dw - TdS - SdT = SdT + dw - TdS - SdT = -SdT + dw

Here dw represents all possible types of work. More explicitly,

dA = -SdT - PdV + dw'

where dw' is non-pressure-volume work done. Using this equation, we can understand the Helmholtz energy change as a measure of the maximum amount of work the system can do ON ITS SURROUNDING. (Some textbooks define Helmholtz energy in an opposite way; the minimum amount of work that would have to be done ON THE SYSTEM to bring about this change in state.)

By definition, we can see that in general case in which V and T are constant, dA = dw'. That is, the change of Helmholtz free energy is based on the non-pressure-volume work done.

The "maximum work definition" can also be applied to Gibbs free energy in which

dG = -SdT + VdP + dw'

Again, at constant T and P, dG = dw'

[quote] For the third law of thermodynamics, there exists residual entropy that is even at absolute zero (which cannot be reached) some substances still have entropy. I know the 3rd Law only applies to pure substances so this situation will not bother the Law. Then, how can we set a reference line for the entropy of such impure substances? [/quote]

We can't. The residual entropy of a substance with a small amount of impurity is not the sum of the residual entropy of these individual species because it is dependent upon how the impurity is distributed within the crystal lattice of the substance. One may apply statistical thermodynamics to compute the residual entropy under certain approximations, but I highly doubt it.

銀月遊俠 2013-10-29 08:04 PM

1.Actually, I am not sure if I get it right. Do you mean that I misunderstand that if a process have ΔH then the system is under constant pressure?

2.For that question, I just curious about why the pressure of the system changes while the volume is fixed.

New questions

In the mathematical statement of 2nd Law of thermodynamics, it says ΔSsys≥q/T howcome in the 3rd law(or I should say the Nearst heat theorm: the entropy change accompanying any phy or chem process approaches zero as the temp approaches zero) it says ΔS->0 as T->0. It seems I mixed up something here but I don't know what it is.

How to prove this one

[img]http://i1189.photobucket.com/albums/z428/AnthonyCChem/Capture_zps22da8f62.jpg[/img]alpha is expansion coefficient Kt is isothermal compressibility

I don't understand ΔG. If a system is at equilibrium, ΔG=Wmax or ΔG=0 ??? Or a process is reversible then ΔG=0. When a system is at equilibrium ΔG=Wmax? But isn't that a reversible process implies the system is at equlibrium throughtout the process??

[img]http://i1189.photobucket.com/albums/z428/AnthonyCChem/Capture_zps551328bf.jpg[/img]

I don't understan how to calculate the ΔfusH(T). Isn't that it should be ΔfusH(T)=ΔH1+ΔfusH(tf)+ΔHs??

[[i] 本帖最後由 銀月遊俠 於 2013-10-29 11:11 PM 編輯 [/i]]

2.For that question, I just curious about why the pressure of the system changes while the volume is fixed.

New questions

In the mathematical statement of 2nd Law of thermodynamics, it says ΔSsys≥q/T howcome in the 3rd law(or I should say the Nearst heat theorm: the entropy change accompanying any phy or chem process approaches zero as the temp approaches zero) it says ΔS->0 as T->0. It seems I mixed up something here but I don't know what it is.

How to prove this one

[img]http://i1189.photobucket.com/albums/z428/AnthonyCChem/Capture_zps22da8f62.jpg[/img]alpha is expansion coefficient Kt is isothermal compressibility

I don't understand ΔG. If a system is at equilibrium, ΔG=Wmax or ΔG=0 ??? Or a process is reversible then ΔG=0. When a system is at equilibrium ΔG=Wmax? But isn't that a reversible process implies the system is at equlibrium throughtout the process??

[img]http://i1189.photobucket.com/albums/z428/AnthonyCChem/Capture_zps551328bf.jpg[/img]

I don't understan how to calculate the ΔfusH(T). Isn't that it should be ΔfusH(T)=ΔH1+ΔfusH(tf)+ΔHs??

[[i] 本帖最後由 銀月遊俠 於 2013-10-29 11:11 PM 編輯 [/i]]

jmlo 2013-10-30 08:42 AM

Yes. By definition, since H = U + PV and enthalpy is a state function, the change of H is just

dH = (U2 + P2V2) - (U1 + P1V1)

There is no restriction whether pressure has to be constant in order for H to be defined.

[quote] 2.For that question, I just curious about why the pressure of the system changes while the volume is fixed. [/quote]

For an ideal gas, PV = nRT. Hence,

P = (nR/V)*T

When V is constant, any change in T will lead to a change in P, and vice versa. For example, if a gas is injected into a hard container whose boundary is not flexible. In this case, V is constant. When the gas is heated up, the average kinetic energy of gas particles is increased. So is the momentum. Since pressure is defined as

P = F/A

and F = ma = d(mv)/dt

If the momentum of gas particles is increased, then F will be increased, and in turn P will be increased.

* I made a mistake in post #2. dq should be equal to TdS instead of SdT since

dS = dq/T

by the second law of thermodynamics.

[quote] In the mathematical statement of 2nd Law of thermodynamics, it says ΔSsys≥q/T howcome in the 3rd law(or I should say the Nearst heat theorm: the entropy change accompanying any phy or chem process approaches zero as the temp approaches zero) it says ΔS->0 as T->0. It seems I mixed up something here but I don't know what it is. [/quote]

There is no contradiction between the 2nd and 3rd laws. While the 3rd law tells you that entropy of a pure substance under a complete internal equilibrium is zero in the limit T -> 0, the 2nd law tells you how the change of entropy of a system is associated with the heat change of the system at a certain temperature.

Under the Nernst heat theorem, the system is under an internal equilibrium (i.e., reversible). Since all possible allotropic forms of a substance have the zero entropy, the change of entropy from one form to another is zero. The situation is similar to an adiabatic reversible process in which dS = 0.

[quote] How to prove this one

alpha is expansion coefficient Kt is isothermal compressibility [/quote]

The proof is a bit lengthy.

Starting from the definition of H:

H = U + PV

dH = dU + PdV + VdP

Since dU = TdS - PdV, we have

dH = (TdS - PdV) + PdV + VdP = TdS + VdP

Now, we take the partial derivative of dH with respect to T at constant V:

(dH/dT)_v = T(dS/dT)_v + V(dP/dT)_v

From thermodynamic relations we know that

(dS/dT)_v = Cv/T

Also, by cyclic relation:

(dP/dT)_v (dT/dV)_P (dV/dP)_T = -1 so that (dP/dT)_v = -(dV/dT)_P/(dV/dP)_T

We can plug in these expressions into (dH/dT)_V to obtain

(dH/dT)_v = T(Cv/T) + V[-(dV/dT)_P/(dV/dP)_T]

= Cv + V{ [(1/V)(dV/dT)_P] / [-(1/V)(dV/dP)_T] }

By definition, we know that

alpha = (1/V)(dV/dT)_P (thermal expansion coefficient)

and

K_T = -(1/V)(dV/dP)_T (isothermal compressibility)

Hence, we can rewrite (dH/dT)_V:

(dH/dT)_V = Cv + V[alpha/K_T] ………. (1)

Since Cp - Cv = VT(alpha^2/K_T), we can write (1) in terms of Cp:

(dH/dT)_V = Cp - VT (alpha^2/K_T) + V(alpha/K_T)

= Cp - V(alpha/K_T) (alpha*T - 1)

Now, we have to recall the definition of Joules-Thomson coefficient

u = (V/Cp)(alpha*T - 1)

It gives Cp*u = V(alpha*T - 1). we substitute this into (dH/dT)_v to yield

(dH/dT)_v = Cp - (alpha/K_T)*Cp*u = Cp*(1 - alpha*u/K_T) ….. (2)

[quote] I don't understand ΔG. If a system is at equilibrium, ΔG=Wmax or ΔG=0 ??? Or a process is reversible then ΔG=0. When a system is at equilibrium ΔG=Wmax? But isn't that a reversible process implies the system is at equlibrium throughtout the process?? [/quote]

When a system is at equilibrium, ΔG=0. However, when the system is far from this point, ΔG=-Wmax, and it tells you how much work in maximum the system can recover when the system moves from this point to the equilibrium.

[quote] I don't understan how to calculate the ΔfusH(T). Isn't that it should be ΔfusH(T)=ΔH1+ΔfusH(tf)+ΔHs?? [/quote]

You can use this definition, but when you write down ΔH1 and ΔHs, the integrals will be reversed. You should end up having the same final expression. It makes sense since Tf > T.

銀月遊俠 2013-10-30 10:11 PM

I know what is the content of the two laws, but from the mathematical relation

dS= q/T, if T is very low close to 0 the dS will be very large. So why there is still Nernst thorem saying as T tend to 0 ΔS accompanying any change tends to 0.

[quote]When a system is at equilibrium, ΔG=0. However, when the system is far from this point, ΔG=-Wmax, and it tells you how much work in maximum the system can recover when the system moves from this point to the equilibrium.[/quote]

For this one, I think I have got the meaning from the book which says that at equilibrium the ΔG=Wnon-exp but if Wnon-exp is not possible at equilibrium ΔG=0. For A, at equilibrium, ΔA it is the max work output W which includes Wexp and Wnon-exp, if W is not possible ΔA=0 at equilibrium. Is that right?

Thank you all the effort questions are still going on and on.

[[i] 本帖最後由 銀月遊俠 於 2013-10-30 10:29 PM 編輯 [/i]]

jmlo 2013-10-30 11:26 PM

dS= q/T, if T is very low close to 0 the dS will be very large. So why there is still Nernst thorem saying as T tend to 0 ΔS accompanying any change tends to 0. [/quote]

Here you have to recall how the 2nd law is defined. The heat change in the expression

dS = q/T

is indeed not a constant but a function of T. When we calculate S we compute the following integral from T0 to T1:

S = int_T0^T1 Cp dT/T

Apparently, there is a singularity problem at T=0. Lucky enough, it is found experimentally that the heat capacity at constant pressure Cp is a function of T in the form

Cp(T) = C0*T + C1*T^2 + C2*T^3 + ....

which is approaching zero when T -> 0. Therefore, the integral is still valid even when T -> 0.

Hence, the Nernst heat theorem does not contradict with the second law indeed.

[quote] For this one, I think I have got the meaning from the book which says that at equilibrium the ΔG=Wnon-exp but if Wnon-exp is not possible at equilibrium ΔG=0. For A, at equilibrium, ΔA it is the max work output W which includes Wexp and Wnon-exp, if W is not possible ΔA=0 at equilibrium. Is that right? [/quote]

Yes.

[quote] Thank you all the effort questions are still going on and on. [/quote]

You are welcome. :loveliness:

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