- #1
Telemachus
- 835
- 30
Homework Statement
Here there is another problem from callen which I couldn't finish.
It says: It is found that a certain liquid boils at a temperature of 95ºC at the top of a hill, whereas it boils at a temperature of 105º at the bottom. The latent heat is 1000cal/mole. What is the approximate height of the hill?
So this is what I did:
[tex]\frac{\Delta P}{\Delta T}=\frac{l}{T_b\Delta V_{\alpha,\beta}}\rightarrow \frac{\Delta P}{\Delta T}=\frac{l P_1}{T_b^2 R}\rightarrow \Delta P=P_1-P_2=\frac{1000\frac{cal}{mol}P_1\Delta T}{T_b^2 R}\rightarrow P_2=0.15P_1[/tex]
Where [tex]T_b[/tex] is the boiling point temperature. I've used the ideal gas equation [tex]V=\frac{RT}{P}[/tex].
l is the latent heat, which is known.
Then:
[tex]\Delta T=10K[/tex]
[tex]\Delta P= g h \rho[/tex], rho is the air density
[tex]\rho=1.22\frac{kg}{m^3}[/tex]
Then [tex]\Delta P=0.85P_b=11.956\frac{kg}{m^2s^2}h[/tex]
I need the information of the pressure at the bottom of the hill for finishing. Anyway, I'm not sure all the steps are right, and I wanted you to see this. I think I could use the pressure at sea level as an approximation.