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Homework Help: [Thermo.] thermal expansion and isothermal compressibility coefficients.

  1. Jun 23, 2010 #1
    1. The problem statement, all variables and given/known data

    A certain metal whose thermal expansion coefficient [tex]\beta[/tex] is 5,0 × 10^-5 °C^-1 and whose isothermal compressibility [tex]\kappa_T[/tex] is 1,2 × 10^-6 atm^-1 is at an initial pressure of 1 atm and an initial temperature of 20°C. A thick layer of Invar is thermally insulating the system. The Invar's coefficients are not to be taken into consideration.

    a) What would be the final pressure if the temperature raised to 32°C?

    b) If the layer could resist to a pressure up to 1200 atm, what would be the highest possible temperature achievable?

    2. Relevant equations

    [tex]\kappa_T = \frac{-1}{V} (\frac{\partial V}{\partial P})_T [/tex]

    [tex]\beta = \frac{1}{V} (\frac{\partial V}{\partial T})_P [/tex]

    3. The attempt at a solution

    We know from the statement that:

    [tex]\kappa_T = \frac{-1}{V} (\frac{\partial V}{\partial P})_T = 1,2 \times 10^{-6}[/tex]

    [tex]\beta = \frac{1}{V} (\frac{\partial V}{\partial T})_P = 5,0 \times 10^{-5}[/tex]


    Working out both expressions to a common equation gives us

    [tex]- 1,2 \times 10^{-6} dP = \frac{dV}{V} [/tex]

    [tex]5,0 \times 10^{-5} dT = \frac{dV}{V} [/tex]


    [tex]- 1,2 \times 10^{-6} dP = 5,0 \times 10^{-5} dT [/tex]


    [tex] - 1,2 \times 10^{-6} \int_{P_i}^{P_f}\ dP = 5,0 \times 10^{-5} \int_{T_i}^{T_f}\ dT [/tex]

    [tex] - 1,2 \times 10^{-6} (P_f - 1) = 5,0 \times 10^{-5} (32 - 20) [/tex]

    [tex]P_f = 1 - 500 = -499\ atm [/tex]

    Is this correct? A negative pressure?


    [tex]- 1,2 \times 10^{-6} (1200 - 1) = 5,0 \times 10^{-5} (T_f - 20) [/tex]

    [tex]T_f = -8.776\ degrees\ Celsius[/tex]

    So what am I doing wrong here?


    EDIT: Is it possible that [tex]\kappa_T = - 1,2 \times 10^{-6} [/tex] instead of [tex]+ 1,2 \times 10^{-6} [/tex] ?

    That would work out, it seems.
    Last edited: Jun 23, 2010
  2. jcsd
  3. Jun 24, 2010 #2
    Because the invar has remarkably low thermal expansion coefficient and it covers the metal, the metal barely expands: [tex]dV = 0[/tex] (this is the important thing you missed!).
    One thing to notice: [tex]dV = \frac{\partial V}{\partial P}dP + \frac{\partial V}{\partial T}dT[/tex] (this is where you made the mistake!).
    You can work things out from here, can't you? :smile:

    P.S: [tex]K_T[/tex] cannot be negative, because the larger the pressure is, the more the metal is compressed, the smaller the volume becomes.
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