Thermodynamic state entirely determined by only two quantities?

In summary: So knowing ##H## and ##p## allows us to calculate ##M## and ##\langle M \rangle##. However, if we only knew ##p## we would be unable to calculate ##M## or ##\langle M \rangle##. This is why we need ##n+1## variables in order to determine all thermodynamic properties of a system.
  • #1
mSSM
33
1
I am trying to understand why I can specify the specific volume [itex]v[/itex] of a fluid element as a function of the equilibrium pressure, [itex]p[/itex], and the equilibrium entropy, [itex]s[/itex]. This is for example done in this article http://www.sno.phy.queensu.ca/sno/str/SNO-STR-95-051.pdf , on this website: http://wind.mit.edu/~emanuel/geosys/node3.html , and in Landau & Lifshitz fluid mechanics, page 7.

I have spent a while looking some source explaining this, and remembered from statistical physics (quote from Landau & Lifshitz, Statistical Physics, Page 42):
Thus the macroscopic state of a body at rest in equilibrium is entirely determined by only two quantities, for example the volume and the energy. All other thermodynamic quantities can be expressed as functions of these two.

I am trying to understand how the knowledge of [itex]p[/itex] and [itex]s[/itex] would help me to, for example, obtain [itex]T[/itex]? How can I get access to other quantities knowing only those two variables?
 
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  • #2
The statement in Landau is only true if there is only one generalized force associated with the system or equivalently if there is only one generalized coordinate characterizing the system; in the case of Landau's discussion the generalized coordinate is the volume and the conjugate force is the pressure. In more general cases, wherein one has ##n## generalized coordinates or generalized forces, one needs ##n+1## variables in order to completely determine the thermodynamic properties of the system.

With that being said, the enthalpy ##H## of a system is entirely characterized by ##p## and ##S## i.e. ##H = H(S,p)##. You can therefore determine the temperature using ##T = \frac{\partial H}{\partial S}|_p## so that ##T = T(S,p)##. The enthalpy also allows you to determine the volume as a function of these coordinates i.e. ##V = V(S,p)## using ##V = \frac{\partial H}{\partial p}|_S##. All you need to do is determine the enthalpy of the system as a function of the pressure and entropy.

This can always be done in principle by putting the system in thermal and mechanical contact with a heat bath and using the Gibbs canonical ensemble to calculate the Gibbs partition function ##\mathcal{Z}## from which we get the enthalpy ##H = -\frac{\partial}{\partial \beta}\ln \mathcal{Z}##. This will tell us ##H(T,p)## since ##\mathcal{Z} = \mathcal{Z}(T,p)## on account of the system being in both thermal and mechanical contact with the heat bath. We can also get ##S = S(T,p)## from ##\mathcal{Z}## and in principle we can invert this to get ##T = T(S,p)## so that ##H = H(S,p)##.
 
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  • #3
Alright, thanks for the explanation. This leaves me with two questions:
  • [itex]T[/itex] and [itex]S[/itex] are not viewed as a pair of generalized forces/coordinates in this context?
  • What is the argument for needing [itex]n+1[/itex] variables in order to completely determine all thermodynamic properties?
 
  • #4
mSSM said:
[itex]T[/itex] and [itex]S[/itex] are not viewed as a pair of generalized forces/coordinates in this context?

This is really a matter of taste. Some authors will consider ##T## and ##S## as a pair of generalized forces/coordinates whereas others will reserve these qualifications only for coordinates and forces that are related through some form of work ##\delta W## and not through spontaneous heat flow ##\delta Q## as ##T## and ##S## are. I prefer the latter.

mSSM said:
What is the argument for needing [itex]n+1[/itex] variables in order to completely determine all thermodynamic properties?

Say we have ##n## generalized coordinates for a system. This can be for example volume ##V## and an external magnetic field ##B## if we have a sample of ferromagnetic material and wish to calculate its average magnetization and pressure. The Gibbs partition function will have dependence ##\mathcal{Z} = \mathcal{Z}(T,x_1,...,x_n)## i.e. it will depend on ##n+1## coordinates. As you know, all thermodynamic properties of a system can be determined entirely from ##\mathcal{Z}##. For example, going back to the example of the ferrmagnetic, we have ##p = \frac{1}{\beta}\partial_V \ln \mathcal{Z}## and ##\langle M \rangle = \frac{1}{\beta}\partial_B \ln\mathcal{Z}## as well as ##H = -\partial_{\beta}\ln\mathcal{Z}##.
 
  • #5


I can provide a response to this question by explaining the concept of thermodynamic state and its determination by two quantities.

Thermodynamic state refers to the specific conditions of a system at a given point in time. It is described by various thermodynamic variables such as pressure, temperature, volume, and entropy. These variables are interrelated and can be used to fully characterize the state of a system.

In thermodynamics, there are two fundamental laws that govern the behavior of a system: the first law, which states that energy cannot be created or destroyed, only transferred or converted between different forms, and the second law, which states that the entropy of an isolated system always increases or remains constant.

The relationship between these two laws and the two quantities, pressure and entropy, can be explained through the concept of thermodynamic potentials. These potentials, such as internal energy, enthalpy, and Helmholtz free energy, are functions of the state variables and provide a way to understand the behavior of a system.

In particular, the Helmholtz free energy, denoted by F, is a function of temperature, volume, and entropy. Using the fundamental relation of thermodynamics, dF = -SdT - pdV, we can see that at constant temperature and volume, the change in Helmholtz free energy is directly related to the change in entropy. This means that at a given temperature and volume, the value of the Helmholtz free energy is determined by the value of entropy.

Similarly, at constant temperature and pressure, the Gibbs free energy, denoted by G, is a function of temperature, pressure, and entropy. Using the fundamental relation, dG = -SdT + Vdp, we can see that at a given temperature and pressure, the value of the Gibbs free energy is determined by the value of entropy.

Therefore, by knowing the values of pressure and entropy, we can determine the values of temperature and other thermodynamic quantities through the use of these potentials. This is why the knowledge of pressure and entropy is sufficient to fully determine the thermodynamic state of a system.

In conclusion, the understanding of thermodynamic state and its determination by two quantities, pressure and entropy, is a fundamental concept in thermodynamics. These two quantities provide a way to access and understand the behavior of a system and its various thermodynamic properties.
 

1. What are the two quantities that determine the thermodynamic state of a system?

The two quantities that determine the thermodynamic state of a system are temperature and pressure. These two parameters are known as state variables and are used to describe the physical state of a system.

2. Why are temperature and pressure considered the most important quantities in determining thermodynamic state?

Temperature and pressure are considered the most important quantities in determining thermodynamic state because they directly affect the physical properties of a system. Temperature determines the kinetic energy of the particles in a system, while pressure determines the force exerted by the particles on the walls of the container.

3. Can other quantities, such as volume or density, determine the thermodynamic state of a system?

While other quantities such as volume or density can provide additional information about a system, they cannot fully determine the thermodynamic state. Temperature and pressure are the only two quantities that are required to completely describe the state of a system.

4. How do changes in temperature and pressure affect the thermodynamic state of a system?

Changes in temperature and pressure can cause a system to undergo phase changes or change its physical properties. For example, increasing pressure can cause a gas to condense into a liquid, while increasing temperature can cause a solid to melt into a liquid.

5. Are there any limitations to using temperature and pressure to determine thermodynamic state?

While temperature and pressure are important in determining thermodynamic state, they are not the only factors that can affect a system. Other factors such as chemical composition, external forces, and time can also play a role in determining the state of a system. Additionally, temperature and pressure are not always independent of each other, which can complicate their use in determining thermodynamic state.

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