Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Change in Density due to Temperature and Pressure Change

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Estimate the percent change in density of iron when it is still a solid, but deep in the Earth where the temperature is [itex]2000^\circ C[/itex] and it is under [itex]5000[/itex] atm of pressure. Take into account both thermal expansion and changes due to increased outside pressure. Assume both the bulk modulus and the volume coefficient of expansion do not vary with temperature and are the same as at normal room temperature. The bulk modulus for iron is about [itex]90\times10^9N/m^2[/itex].

    We also know that the volumetric expansion coefficient of iron is [itex]35\times10^{-6}(^{\circ}C)^{-1}[/itex].

    Also, [itex]1atm=1.103\times10^5N/m^2[/itex].

    2. Relevant equations
    [itex]\frac{\Delta V}{V_0}=-\frac{1}{B}\Delta P[/itex] where B is the bulk modulus.
    [itex]\Delta V=V_0\beta\Delta T[/itex] where [itex]\beta[/itex] is the volumetric expansion coefficient.

    3. The attempt at a solution
    I'm assuming the change in volumes adds up so we have using the two equations successively, we have
    [itex]\frac{\Delta V}{V_0}=\left(-\frac{1}{90\times10^9}\right)\left(5000atm\right) \left(\frac{1.103\times10^5N/m^2}{1atm}\right)=-0.00563[/itex]
    [itex]\frac{\Delta V}{V_0}=(4975^\circ C)(35\times10^-6\left(^\circ C\right)^{-1}=0.17413[/itex]

    So the total change in volume is [itex]0.17413-0.00563=0.168495[/itex].

    Then we could easily get the percent change in density. However, I am unsure if I am applying this correctly. Am I allowed to add the change in volumes?
  2. jcsd
  3. Feb 16, 2012 #2


    User Avatar
    Homework Helper

    The volume becomes (1+ΔV/V0) times the initial one in both processes, and you need to multiply them when there is change of volume both because of pressure and temperature. When ΔV/V0<<1, you can simply add the relative changes (with the proper signs) but 0.17 is not very small compared to unity.

  4. Feb 16, 2012 #3
    Wait, so it should be (0.17413)(-0.00563)=-0.0009803519? That doesn't make intuitive sense to me. Why do we multiply the change in volumes?
  5. Feb 16, 2012 #4


    User Avatar
    Homework Helper

    NO. We multiply the volume ratios. V/V0=1.17413*(1-0.00563)=1.16752. ΔV/V0=0.16752.

  6. Feb 16, 2012 #5
    Oh, I see now. Could you explain why it must be multiplied? I don't understand.
  7. Feb 17, 2012 #6


    User Avatar
    Homework Helper

    The formulae for volume change due to pressure and temperature refer to cases when the other quantity is constant. Pretend you put the piece of iron at high pressure first, when the volume decreases by Vo(ΔP/B). The new volume, V1=Vo(1-ΔP/B) is warmed up, and the new volume serves as "Vo" in thermal expansion. But the results are almost the same, and the the formulae are approximations anyway. So both ways should be accepted for the relative change of the volume, which is 0.168 with 3 significant digits by both methods.

  8. Feb 17, 2012 #7
    Oh, thanks you so much. I understand now. :smile:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook