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How to determine internal energy without mass and pressure?

  1. Jan 18, 2014 #1
    1. The problem statement, all variables and given/known data

    The internal energy of a system is expressed by function U(S.V)=S4/3Vα, where α is a constant. The value of α is?

    3. The attempt at a solution

    Here internal energy is expressed in terms of volume and entropy but all the relations (as far as I know) are in terms of volume, entropy, mass and temperature or enthalpy. So how do you determine internal energy when pressure and mass are not given and everything else is a constant?
     
  2. jcsd
  3. Jan 18, 2014 #2

    TSny

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    How can you get the dimensions (or units) to match on both sides?

    [EDIT: Something else to think about. Is internal energy an extensive or intensive quantity?]
     
    Last edited: Jan 18, 2014
  4. Jan 18, 2014 #3
    Given that internal energy is measured in Joules, LHS becomes J and RHS is (J/k)*m3
    So to equate in terms of thermodynamic quantities we could multiply with T=1k and P=1N/m2 and divide it by enthalpy (H)=1J, but there is no relation such as this, is there?
     
  5. Jan 18, 2014 #4

    TSny

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    I don't see how to get the dimensions to work out. But I believe you can answer the question by categorizing each variable U, S, V as extensive or intensive.
     
  6. Jan 19, 2014 #5
    All of them are extensive. Still don't get how to work the problem out. Did you get the answer?
    The options are
    a) 1
    b) 1/3
    c) -1/3
    d) 3/4
     
  7. Jan 19, 2014 #6

    TSny

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    Suppose you doubled the size of the system. What would happen to each of the variables? The equation would still have to be satisfied.

    You should get one of the options.
     
  8. Jan 19, 2014 #7
    To double the size of the system we would need to provide double the internal energy so U becomes 2U, then the volume would also double and since the volume increases the density decreases by 2 and so does the pressure, that means the entropy would also decrease by 2 so I have the equation
    2U=(S/2)4/3(2V)α. Solving that would get α = 1+4/3 that means α=7/3.
    Ugh....What'd I miss?

    [EDIT: with increase in volume the entropy also increases not decreases. And bearing that in mind when I recalculated I got α=-1/3. Hoping I didn't go wrong anywhere this time. Right?]
     
    Last edited: Jan 19, 2014
  9. Jan 19, 2014 #8

    TSny

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    Looks good
     
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