# Thermodynamics question, need help

1. Aug 29, 2007

### Jonny_trigonometry

How does an amount of water inside a sealed tank affect a change in pressure of the tank if you open a valve that connects the tank to another tank at a lower pressure and twice the volume? We have to consider the problem with and without water in the tank. We have to consider the vapor pressure of the water, and the fact that when pressure is reduced in the tank, more water will turn to vapor in order to reach the vapor pressure. The initial difference in pressure between the two tanks is lower than 10 psi at max. The temperature of the two tanks may change a little, but we can assume that it is constant over a long period of time.

I'm trying to figure out how to formulate the question and then answer it. Does anyone know how to work this problem? Is anyone familiar with accounting for vapor pressure in problems like these? Ultimately, I'd like to figure out the difference in final equilibrium pressures between the cases of (1) an amount of water in the tank and (2) no water in the tank.

2. Sep 1, 2007

### Gokul43201

Staff Emeritus
The lowest pressure that you can have above the water at any temperature is the equilibrium vapor pressure of water at that temperature (look up vapor tables for water). Unless the volume of the tanks is large (> 10X) compared to the volume of water, you can ignore the loss of liquid water to vapor. Do you have a rough value for this ratio of volumes? Also, is the valve connecting the tanks located at the bottom (i.e., only permits water flow, but not vapor flow)?

The equilibrium condition is that the total (gauge + vapor) pressure at the valve be the same from both tanks. In the limit of negligible liquid loss (to vapor), you can write the gauge (liquid column) pressures as a function of the level, and you can also calculate the vapor pressures simply from the ideal gas relation, using the initial pressures. There is no need to worry about partial pressures of water vapor.

3. Sep 3, 2007

### Jonny_trigonometry

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

$$\frac{dP}{dT} = \frac{L}{T \Delta V}$$

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?

Last edited: Sep 4, 2007
4. Sep 4, 2007

### Jonny_trigonometry

hmm... The units aren't right in the exponential argument, I think that N should actually be the number of water particles, so instead of 5.54, it should be 5.54*6.02*20^23. If this is the case, then the change in pressure due to water is negligible.

5. Sep 4, 2007

6. Sep 5, 2007

### Jonny_trigonometry

thanks, any help is much appreciated. I still haven't been able to satisfy myself with the answer. I don't understand how to calculate the fraction of water molecules in a gaseous state to those in a liquid state under 20 psi of regular (dry) air.