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Defining a thermodynamic system completely

  1. Nov 26, 2013 #1
    So I read somewhere that if we know u(s,v) i.e. internal energy as a function of entropy and specific volume, we can find all other thermodynamics quantities on earth, like temperature, pressure, Gibb's free energy from it.

    However, is this true if we have u as a function of just any two thermodynamic properties?

    Also. if we know Gibb's free energy/ Helmoltz energy in terms of any 2 quantities, can we similarly find all other thermodynamic properties from it? In case there is a chemical reaction going on, then I guess we need to define 2 such parameters, right?
     
  2. jcsd
  3. Nov 26, 2013 #2

    Jano L.

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    No, knowing internal energy as a function of other set of quantities is not as informative as knowing ##U(S,V)##.

    For example, the energy of ideal gas is ##U = c_V nRT## (depends on temperature and molar number only), but from this we cannot find the equation of state ##pV = nRT##. However, from relation ##U(S,V)## we can find this equation of state. For this function, see Callen's book

    Callen H.B., Thermodynamics And An Introduction To Thermostatistics, 2ed., Wiley, 1985

    or the Wikipedia page

    https://en.wikipedia.org/wiki/Ideal_gas

    section "Thermodynamic Potentials" towards the end. The Gibbs energy provides complete thermodynamic description only if expressed as function of pressure and temperature (and molar numbers).
     
  4. Nov 26, 2013 #3
    Well can I ask a question which is indirectly related to this? (I'm just trying to correlate things I've learnt in this thermo course this semester...).
    So the question is...at the triple point for a given pure substance, where we know the temperature and pressure, how many parameters are required to know exactly the fraction of each phase that exists? For example, if you know u(s,v), it would not be enough, right? You'd require another parameter...could that be any thermodynamic parameter?
     
  5. Nov 27, 2013 #4

    Jano L.

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    Since there are three phases, we need to know two molar fractions, for example ##\frac{n_{solid}}{n}##, ##\frac{n_{liquid}}{n}##. The third molar fraction can be derived from these two, since

    $$
    n_{solid} + n_{liquid} + n_{gaseous} = n.
    $$

    (divide by ##n## and express the third fraction).

    I'm not sure what you mean. What is u(s,v)? You mean ##U(S,V)## (total energy, entropy and volume), or ##u(s,v)## (energy, entropy and volume per mole) ?
     
  6. Dec 8, 2013 #5
    Sorry for the delay in replying! I just finished my final maths exam. So I mean specific internal energy i.e. energy, entropy and volume per mole/kg. Maybe I could explain the question by going back to what you said:
    You're saying we need to know 2 mole fractions.I get that. But what if you have no way of measuring that....Wouldn't it be enough to know (as an example) the specific internal energy and specific enthalpy of the system and hence, from that knowledge (and the knowledge of the internal energy & enthalpy of each phase, from data books) determine the mole fractions?
     
  7. Dec 9, 2013 #6

    Jano L.

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    What do you mean by specific internal energy and enthalpy of the whole system? The quantities
    $$
    u = \frac{U}{N}
    $$
    and
    $$
    h = \frac{H}{N}
    $$
    where ##N## is the total molar number?

    How would you get to know these for two-phase system? Measuring molar fractions of each phase seems so much easier.

    But if we somehow knew ##u,h## for the composed system, we could solve the equations

    $$
    Nu = u_1N_1 + u_2 N_2
    $$

    $$
    Nh = h_1N_1 + h_2 N_2
    $$

    $$
    N_1 + N_2 = N
    $$

    (##1## for solid, ##2## for liquid) for ##N_1,N_2, N##, to get both molar fractions and even the total number of molecules.
     
  8. Dec 9, 2013 #7
    Yes, the definition of specific quantities is what you've mentioned at the top of your post. I'm talking about it in the perspective of problem-solving on exams...

    But you've mentioned only N1 and N2 ...we were talking about the triple point, so there will be N1, N2 and N3. Still, we have 3 equations and 3 unknowns, so I suppose its ok.
     
  9. Dec 9, 2013 #8

    Jano L.

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    Gold Member

    Hm, mathematically it seems that way. But I do not see physical motivation behind such calculation.
     
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