1st law: Thermo equation 6.1 from Smith, Van Ness, Abbott

In summary: The reason you need to reference a reversible change between the two states to derive the relationship is that it is the only kind of path we can use to determine the change in entropy. However, irrespective of how convoluted the actual path between the two neighboring states was, including irreversible paths and wide excursions, once the two neighboring states have been established, the relationship between d(nU), d(nS), and d(nV) between the two states has to satisfy this equation.In summary, the equation d(nU) = T d(nS) - P d(nV) describes the inter-relationship between changes in state functions of a material between two closely neighboring thermodynamic equilibrium states. The reason for referencing a reversible change is because
  • #1
kayan
37
0
On page 200 in the 7th edition of Intro to ChemE Thermodynamics by Smith, Van Ness, and Abbott, there is an equation that has always bugged me since reading it (or rather the interpretation of it, not the derivation). It is equation 6.1 and states:
d(nU) = T d(nS) - P d(nV)​
with n = moles in the system, U = molar internal energy, T = temp, S = molar entropy, P = pressure, and V = molar volume. This equation was simply derived from the 1st law assuming a reversible process for the work and heat terms. Then, it goes on to state that this eqn was "derived for the special case of a reversible process. However, it contains only properties of the system...Therefore, it is not restricted in application to only reversible processes..." There is some other filler info that you should read too, but I've quoted the main message.

My question is, how can this reasoning make more sense to me? It seems like a big leap of faith to assume that something derived from a specific process (a reversible one) can then be applied to any process (even irreversible ones) just by stating the claim that it only involves state variables. I totally understand what a state variable is, but this doesn't help my acceptance of this idea. Please help!

UPDATE: As an example, on page 74 eqn 3.20a:
dU = Cv dT​
was totally derived for only an ideal gas, however, I see only thermodynamic state properties appearing in this equation, so why is it wrong to assume here that this eqn only can be applied to an ideal gas, whereas 6.1 can be unrestricted from the assumptions under which it was derived. I hope you see my confusion in the analogy.
 
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  • #2
kayan said:
it contains only properties of the system

kayan said:
leap of faith

kayan said:
the claim that it only involves state variables.
"Claim?" Are there other kinds of variables present?
kayan said:
I totally understand what a state variable is
Are you comfortable using state variables in both reversible and irreversible processes?

The question is not totally clear in terms of the source of your confusion; state variables, or, reversible processes used in derivations?
 
  • #3
kayan said:
On page 200 in the 7th edition of Intro to ChemE Thermodynamics by Smith, Van Ness, and Abbott, there is an equation that has always bugged me since reading it (or rather the interpretation of it, not the derivation). It is equation 6.1 and states:
d(nU) = T d(nS) - P d(nV)​
with n = moles in the system, U = molar internal energy, T = temp, S = molar entropy, P = pressure, and V = molar volume. This equation was simply derived from the 1st law assuming a reversible process for the work and heat terms. Then, it goes on to state that this eqn was "derived for the special case of a reversible process. However, it contains only properties of the system...Therefore, it is not restricted in application to only reversible processes..." There is some other filler info that you should read too, but I've quoted the main message.

My question is, how can this reasoning make more sense to me? It seems like a big leap of faith to assume that something derived from a specific process (a reversible one) can then be applied to any process (even irreversible ones) just by stating the claim that it only involves state variables. I totally understand what a state variable is, but this doesn't help my acceptance of this idea. Please help!

You have asked a really great question. This kind of thing causes huge confusion among students, mainly because it is presented so poorly in textbooks. The equation d(nU) = T d(nS) - P d(nV) merely describes the inter-relationship between the changes in the state functions nU, nS, and nV between two closely neighboring thermodynamic equilibrium states of a material. In one equilibrium state the parameter values are nU, nS, nV, T and P, and in the neighboring equilibrium state the parameter values are nU+d(nU), nS+d(nS), nV+d(nV), T+dT, and P+dP. The reason you need to reference a reversible change between the two states to derive the relationship is that it is the only kind of path we can use to determine the change in entropy. However, irrespective of how convoluted the actual path between the two neighboring states was, including irreversible paths and wide excursions, once the two neighboring states have been established, the relationship between d(nU), d(nS), and d(nV) between the two states has to satisfy this equation.
UPDATE: As an example, on page 74 eqn 3.20a:
dU = Cv dT​
was totally derived for only an ideal gas, however, I see only thermodynamic state properties appearing in this equation, so why is it wrong to assume here that this eqn only can be applied to an ideal gas, whereas 6.1 can be unrestricted from the assumptions under which it was derived. I hope you see my confusion in the analogy.
The equation for a general material (e.g. a non-ideal gas) between two thermodyamic equilibrium states is more complicated than this and includes a term involving dV (or dP) in addition to the dT term. However, for the special case of an ideal gas, the coefficient of dV (or dP) is equal to zero. See Section 6.1 in Smith et al.

Chet
 
  • #4
Bystander said:
The question is not totally clear in terms of the source of your confusion; state variables, or, reversible processes used in derivations?

I tried to be as clear as possible (stating that I understood state variables, hence that was not my source of confusion), and apparently it was communicated well enough for the poster below to perceive the correct question and answer it.

Chestermiller said:
You have asked a really great question. This kind of thing causes huge confusion among students, mainly because it is presented so poorly in textbooks. The equation d(nU) = T d(nS) - P d(nV) merely describes the inter-relationship between the changes in the state functions nU, nS, and nV between two closely neighboring thermodynamic equilibrium states of a material. In one equilibrium state the parameter values are nU, nS, nV, T and P, and in the neighboring equilibrium state the parameter values are nU+d(nU), nS+d(nS), nV+d(nV), T+dT, and P+dP. The reason you need to reference a reversible change between the two states to derive the relationship is that it is the only kind of path we can use to determine the change in entropy. However, irrespective of how convoluted the actual path between the two neighboring states was, including irreversible paths and wide excursions, once the two neighboring states have been established, the relationship between d(nU), d(nS), and d(nV) between the two states has to satisfy this equation.

The equation for a general material (e.g. a non-ideal gas) between two thermodyamic equilibrium states is more complicated than this and includes a term involving dV (or dP) in addition to the dT term. However, for the special case of an ideal gas, the coefficient of dV (or dP) is equal to zero. See Section 6.1 in Smith et al.

Chet

This is what I needed, thanks.
 
  • #5
When you are using equation, you idealized all otherwise, it will not equate the relationship. The actual thing varies with this ideal equation, but as a reference or basis of computation, we use the ideal equation.

R is an ideal Universal Gas Constant Cp-Cv = R, remember Ideal Gas equation as PV =RT, this where PV= const. other wise PVn =const. --> this is no longer ideal
 
  • #6
Legolaz said:
When you are using equation, you idealized all otherwise, it will not equate the relationship. The actual thing varies with this ideal equation, but as a reference or basis of computation, we use the ideal equation.
I have no idea what this means. Can you please elaborate (maybe with some examples)?
R is an ideal Universal Gas Constant Cp-Cv = R, remember Ideal Gas equation as PV =RT, this where PV= const. other wise PVn =const. --> this is no longer ideal
PVn =const. is a valid equation for an ideal gas undergoing a reversible polytropic process.

Chet
 
  • #7
Chestermiller said:
I have no idea what this means. Can you please elaborate (maybe with some examples)?

PVn =const. is a valid equation for an ideal gas undergoing a reversible polytropic process.

Chet

OP's confused using ideal equations to the actual condition, wherein there's no term as "reversible condition" in the real world, he asserted which is true.
My say about this is: these are smoothened plots basing on ideal models. The actual process or states is actually off +/- the computed values from these models.
What is important with this is we can quantify standard or the ideal values and compare actual measurements. Thus, from a mathematical point of view, prediction can be made in advance in the design and prior implementation of systems involving these concept.
See for example figure attached. I say PVn=const. is not ideal anymore on the actual conditions. Where n≠0(no such actual process as -isobaric), n≠1(no such thing as isothermal), n≠k (no such thing as isenthalpic) and n>∞ isochoric process.

Actual values might be close to this curves(sets of state points) but, real thing, there is no such thing as ideal gas and reversible process, thus n deviates from ideal condition on actual measurement of PVT properties.

864PPFig1.gif
 
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  • #8
I think what you are trying to say is that a reversible path is an idealization that an actual process path can be made to approach as the driving force for the change is caused to approach zero. In my judgement, this does not relate materially to the OP's original question. The OP confirmed that his question was fully addressed by post #3. Therefore, I am closing this thread.

Chet
 
1.

What is the first law of thermodynamics?

The first law of thermodynamics states that energy cannot be created or destroyed, it can only be transferred or converted from one form to another.

2.

What is the significance of equation 6.1 in Smith, Van Ness, and Abbott's thermodynamics textbook?

Equation 6.1 is a mathematical representation of the first law of thermodynamics, also known as the energy balance equation. It describes the relationship between the change in internal energy, heat transfer, and work done in a thermodynamic system.

3.

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

The first law of thermodynamics is applied in various fields, including engineering, physics, and chemistry. It is used to analyze and understand energy conversion and transfer processes, such as in power plants, engines, and refrigeration systems.

4.

What are some common misconceptions about the first law of thermodynamics?

One common misconception is that the first law only applies to closed systems, when in fact it also applies to open systems. Another is that the first law only applies to thermal energy, when it also applies to other forms of energy, such as mechanical and chemical.

5.

Can the first law of thermodynamics be violated?

No, the first law of thermodynamics is a fundamental law of nature and cannot be violated. However, it can be difficult to accurately measure all the energy transfers and conversions in a system, leading to discrepancies in energy balance calculations.

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