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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 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:
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 m3
##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 m3)
##P_0## = initial pressure = 100 psi over 1 atm (1 atm = 101,325 Pa, therefore P0=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.
##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 m3)
##P_0## = initial pressure = 100 psi over 1 atm (1 atm = 101,325 Pa, therefore P0=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.
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