Latent heat and Clapeyron equation

[Telemachus]

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:
\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
Where T_b is the boiling point temperature. I've used the ideal gas equation V=\frac{RT}{P}.
l is the latent heat, which is known.
Then:
\Delta T=10K
\Delta P= g h \rho, rho is the air density
\rho=1.22\frac{kg}{m^3}

Then \Delta P=0.85P_b=11.956\frac{kg}{m^2s^2}h

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.

 

[yapboz]

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:
\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
Where T_b is the boiling point temperature. I've used the ideal gas equation V=\frac{RT}{P}.
l is the latent heat, which is known.
Then:
\Delta T=10K
\Delta P= g h \rho, rho is the air density
\rho=1.22\frac{kg}{m^3}

Then \Delta P=0.85P_b=11.956\frac{kg}{m^2s^2}h

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.

is there anyone who could solve this problem?

 

[Chestermiller]

I haven't gone over the rest of your calculations in detail, but you should be using the pressure at sea level (1Bar) as the pressure at the bottom of the hill. There is a more accurate way of doing this problem by taking into account the change in density with altitude, but this probably isn't needed here. In any event, after you complete the calculation, you can recalculate the density at the top and bottom of the hill to see if you judge that it really matters.