What is the Ratio Between Water and Gas in a Geyser?

[mike_shaulov]

Please help:

Homework Statement

A simple model of a Geyser is an underground huge lake connected to the surface
by a small tube.
let the depth (and the tube length) be 90m.
proove that the ratio between the mass of the water in the lake and the gas which sparks from the geyser is:

\frac{{\Delta M}}{M} = \frac{c}{L}(T_{10} - T_1 )

where:
T1, T10 - are boiling tempertures at 1 atm. pressure, and 10 atm. pressure.
c - water heat capacity.
L - Latent heat.

Homework Equations

Clasius Claperon equation (Coexistence line):
P = P_0 \times Exp[\frac{{ - L}}{\tau } + \frac{L}{{\tau _0 }}]

 

[BigJDubs]

where: P - pressureL - Latent heatT - tempertureThe Attempt at a SolutionLet us assume that the temperature and pressure of the water in the geyser remain constant with depth.Using the Clausius-Clapeyron equation, we can calculate the ratio between the mass of the water in the lake and the gas which sparks from the geyser.The Clausius-Clapeyron equation states that the pressure of a gas is related to its temperature by:P = P_0 \times Exp[\frac{{ - L}}{\tau } + \frac{L}{{\tau _0 }}]where: P - pressureL - Latent heatT - tempertureSince the pressure and temperature remain constant with depth, we can calculate the ratio between the mass of the water in the lake and the gas which sparks from the geyser as:\frac{{\Delta M}}{M} = \frac{c}{L}(T_{10} - T_1 )where:T1, T10 - are boiling tempertures at 1 atm. pressure, and 10 atm. pressure.c - water heat capacity.L - Latent heat.Therefore, the ratio between the mass of the water in the lake and the gas which sparks from the geyser is:\frac{{\Delta M}}{M} = \frac{c}{L}(T_{10} - T_1 )