- #1

Anachronist

Gold Member

- 117

- 47

As a follow-on to this thread, which in turn followed this closed thread, I'm starting a new thread on a related real-world problem.

Say I have a 2-liter plastic soda bottle filled partway with water, and pressurized. It has been sitting inverted (with the opening pointed down, and sealed) for long enough to reach a state of equilibrium in terms of pressure, temperature, and humidity.

The cap is removed and

Assumptions:

Defining some values:

There are various answers to this problem that don't account for humidity.

Plugging in the values above into the isentropic (adiabatic) expansion formulas, we get:

$$P_F = P_0 \left ( \frac{V_0}{V_F} \right ) ^ \gamma = 448,270 \text{ Pa}\tag{1}$$

$$T_F = T_0 \left ( \frac{P_F}{P_0} \right ) ^ \frac{\gamma - 1}{\gamma} = 249\text{ K} = -44\text{°C}\tag{2}$$

This is saying that even though the bottle was only 1/3 full of water initially, the pressure dropped to almost half of the original pressure instead of 2/3 the original pressure predicted by the isothermal ideal gas law. And the temperature may well be sub-zero for dry air, but I'm skeptical that's what happens in the real world with humid air forcing out the water.

Example 2, equations [5] and [7], in https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/ gives these results, first solving for ##P_F## in [7]:

$$P_F = P_0 \left [ \frac{V_0}{V_0 + \gamma (V_F - V_0)}\right] = 465,177\text{ Pa}\tag{3}$$

$$T_F=T_0 \left [ 1-\frac{(\gamma -1)}{\gamma}\frac{(P_0-P_F)}{P_0} \right ] = 259\text{ K} = -34\text{°C}\tag{4}$$

(I may be misapplying these formulas to this problem.) The final pressure and temperature lie between the isentropic expansion and isothermal ideal gas values: ##P_F = P_0 \frac{V_0}{V_F} = 535,427\text{ Pa}## and ##T_F=T_0##.

The above results are for dry air. They aren't realistic because the air isn't dry. Final temperatures so far below 0°C are unlikely. So I'm wondering how to account for the presence of humidity in these pressure and temperature results. Is there a formula for this? I know it can get complicated, I found information on Clausius-Clapeyron equation and Kirchoff's equation, involving latent heat at the interface between liquid and gas phases.

All I really need is a reasonable approximation that gives me a number that looks agreeable to real-world observations, in which one might observe fog forming in the bottle, but it's unlikely to be ice at sub-zero temperature due to the modulating effect of the water vapor in the air.

Say I have a 2-liter plastic soda bottle filled partway with water, and pressurized. It has been sitting inverted (with the opening pointed down, and sealed) for long enough to reach a state of equilibrium in terms of pressure, temperature, and humidity.

The cap is removed and

*only the water*is allowed to escape before the cap is replaced. What is the final temperature and pressure of the air in the bottle?Assumptions:

- The bottle is rigid and insulating.
- The water escapes rapidly (not a slow leak, it's gone less than 0.1 sec),
- Ignore anything else that may have negligible effects (not sure what they would be at the moment).

Defining some values:

##V_F## = final volume of air in the bottle = 2.17 liters = 0.00217 m

##w## = volume fraction of water (1/3 full for this exercise)

##V_0 = (1-w)V_F## = initial volume of air in the bottle (then ##V_0=\frac{2}{3}0.00217## = 0.001445 m

##P_0## = initial pressure = 100 psi over 1 atm (1 atm = 101,325 Pa, therefore

##T_0## = initial temperature = 20°C (293 K)

##P_F## = final pressure in the bottle when the cap is replaced

##T_F## = final temperature of the air in the bottle when the cap is replaced

##\gamma = 1.4## = gas constant used in formulas, actually more like 1.398 for humid air according to literature I found, but we'll use 1.4.

^{3}##w## = volume fraction of water (1/3 full for this exercise)

##V_0 = (1-w)V_F## = initial volume of air in the bottle (then ##V_0=\frac{2}{3}0.00217## = 0.001445 m

^{3})##P_0## = initial pressure = 100 psi over 1 atm (1 atm = 101,325 Pa, therefore

*P*_{0}=790,801 Pa)##T_0## = initial temperature = 20°C (293 K)

##P_F## = final pressure in the bottle when the cap is replaced

##T_F## = final temperature of the air in the bottle when the cap is replaced

##\gamma = 1.4## = gas constant used in formulas, actually more like 1.398 for humid air according to literature I found, but we'll use 1.4.

There are various answers to this problem that don't account for humidity.

Plugging in the values above into the isentropic (adiabatic) expansion formulas, we get:

$$P_F = P_0 \left ( \frac{V_0}{V_F} \right ) ^ \gamma = 448,270 \text{ Pa}\tag{1}$$

$$T_F = T_0 \left ( \frac{P_F}{P_0} \right ) ^ \frac{\gamma - 1}{\gamma} = 249\text{ K} = -44\text{°C}\tag{2}$$

This is saying that even though the bottle was only 1/3 full of water initially, the pressure dropped to almost half of the original pressure instead of 2/3 the original pressure predicted by the isothermal ideal gas law. And the temperature may well be sub-zero for dry air, but I'm skeptical that's what happens in the real world with humid air forcing out the water.

Example 2, equations [5] and [7], in https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/ gives these results, first solving for ##P_F## in [7]:

$$P_F = P_0 \left [ \frac{V_0}{V_0 + \gamma (V_F - V_0)}\right] = 465,177\text{ Pa}\tag{3}$$

$$T_F=T_0 \left [ 1-\frac{(\gamma -1)}{\gamma}\frac{(P_0-P_F)}{P_0} \right ] = 259\text{ K} = -34\text{°C}\tag{4}$$

(I may be misapplying these formulas to this problem.) The final pressure and temperature lie between the isentropic expansion and isothermal ideal gas values: ##P_F = P_0 \frac{V_0}{V_F} = 535,427\text{ Pa}## and ##T_F=T_0##.

The above results are for dry air. They aren't realistic because the air isn't dry. Final temperatures so far below 0°C are unlikely. So I'm wondering how to account for the presence of humidity in these pressure and temperature results. Is there a formula for this? I know it can get complicated, I found information on Clausius-Clapeyron equation and Kirchoff's equation, involving latent heat at the interface between liquid and gas phases.

All I really need is a reasonable approximation that gives me a number that looks agreeable to real-world observations, in which one might observe fog forming in the bottle, but it's unlikely to be ice at sub-zero temperature due to the modulating effect of the water vapor in the air.

Last edited: