hmm... Well the ratio is much greater than 10X--Vtank/Vwater--the volume of the tank is about 1.5*.08*2.5 m^3 = .3m^3, and the volume of the water is about 100ml at most, so Vwater is about .0001 m^3, so the ratio is about 3000. The valve is high enough so that no water will flow to the other tank. I do appreciate your response, and I wasn't sure how to incorporate the vapor pressure into the equilibrium condition, so you just add the pressures?
I've rethought the problem some more, and I think that I may have formulated the problem incorrectly to begin with; the scenario is very similar though. To start, the point of this setup is to detect very small leaks in a test tank (tank B), by measuring airflow from a non-leaking tank (tank A).
The problem is that there is water in tank A, and this tank is connected to a sensor that will test for air flow between the tank and another tank (tank B) at twice the volume of tank A. There is no water in tank B. Initially, the pressure is equalized between the tanks by a bypass valve, which is then shut off when data taking begins. Now, suppose both tanks don't leak. What effect does the water in tank A have if the temperature changes? Will there be an airflow between the tanks? The sensor is very sensitive, and can detect as small as a cubic millimeter of airflow. Now, if airflow is detected due to the water changing phases, this could be a false positive for a leak detection. So, I think this is the main problem, and I hope this is a much more clear problem to work through. I suppose the best way to approach this is to figure dP/dT in tank A, and compare with dP/dT in tank B.
Initial pressure of tanks = 20 psi.
water in tank A = .0001 m^3.
Volume of tank B = 2*tank A.
Max temperature change = 18 to 30 degrees Celsius.
I was thinking of using the Clausius-Clapeyron equation
<br />
\frac{dP}{dT} = \frac{L}{T \Delta V} <br />
where \Delta V = V_{gas} - V_{liquid} and L is the latent heat of vaporization for water.
for calculating a change in pressure in tank A, and then comparing that with the change in pressure in tank B due to temperature changes. So, I think I'd rather do things in integral form... first off, we can make the approximation that V_{gas} \gg V_{liquid}
dP = \frac{L}{T V_{gas}} dT
Then using the equation of state
PV = NRT
NR \frac{dP}{P} = \frac{L}{T^2} dT \rightarrow NR ln \left( \frac{P_2}{P_1} \right) = L \left( \frac{1}{T_1} - \frac{1}{T_2} \right)
\rightarrow P_{2_A} = P_1 exp \left( \frac{L}{N_w R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \right)
Where P_{2_A} is the final pressure of tank A and N_w is the number of moles of water. So, if V_{w} = 100 cm^3 and \rho _{w} = .997 \frac{g}{cm^3}, then m_{w} = 99.7 g. The molar weight of water is \simeq 18 \frac{g}{mol} = M_{w}, so N_{w} \simeq \frac{m_w}{M_w} = \frac{99.7}{18} = 5.54 mol. Given that L \simeq 42 \frac{kJ}{mol}, and R = 8.31 \frac{J}{mol K}, we get
P_{2_A} = 20 psi exp \left( \frac{42000}{5.54*8.31} \left( \frac{1}{291} - \frac{1}{303} \right) \right) = 22.644 psi
Now, we need to compare that with the pressure change which occurs in tank B for the same temperature change. Using the equation of state, and noting that V and N are constant in this tank, we get
\frac{P_1}{T_1} = \frac{P_2}{T_2} \rightarrow P_{2_B} = \frac{T_2}{T_1} P_1 \rightarrow P_{2_B} = \frac{303}{291} 20 psi = 20.82 psi
so P_{2_A} - P_{2_B} = 1.82 psi. Am I doing this right?