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JSGandora
JSGandora is offline
#1
Feb15-12, 07:47 PM
P: 92
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]
and
[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?
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