A cloud chamber contains water vapor at its equilibrium...

A cloud chamber contains water vapor at its equilibrium vapor pressure $P_{\infty}\left(T_0\right)$ corresponding to an absolute temperature $T_0$. Assume that (i) the water vapor may be treated as an ideal gas; (ii) the specific volume of water may be neglected compared to that of the vapor; (iii) the latent heat $l$ of condensation and $\gamma=c_P / c_V$ may be taken to be constants: $l=540 \mathrm{cal} / \mathrm{g}, \gamma=\frac{1}{2}$. (a) Calculate the equilibrium vapor pressure $P_{\infty}(T)$ as a function of the absolute temperature $T$. (b) The water vapor is expanded adiabatically until the temperature is $T, T<T_0$. Assume the vapor is now supersaturated. If a small number of droplets of water is formed (catalyzed, e.g., by the presence of ions produced by the passage of an $\alpha$ particle), what is the equilibrium radius of these droplets? (c) In the approximations considered, does adiabatic expansion always lead to supersaturation?

A cloud chamber contains water vapor at its equilibrium vapor pressure $P_{\infty}\left(T_0\right)$ corresponding to an absolute temperature $T_0$. Assume that
(i) the water vapor may be treated as an ideal gas;
(ii) the specific volume of water may be neglected compared to that of the vapor;
(iii) the latent heat $l$ of condensation and $\gamma=c_P / c_V$ may be taken to be constants: $l=540 \mathrm{cal} / \mathrm{g}, \gamma=\frac{1}{2}$.
(a) Calculate the equilibrium vapor pressure $P_{\infty}(T)$ as a function of the absolute temperature $T$.
(b) The water vapor is expanded adiabatically until the temperature is $T, T<T_0$. Assume the vapor is now supersaturated. If a small number of droplets of water is formed (catalyzed, e.g., by the presence of ions produced by the passage of an $\alpha$ particle), what is the equilibrium radius of these droplets?
(c) In the approximations considered, does adiabatic expansion always lead to supersaturation?

Key Concepts

Latent Heat

Latent heat is the energy absorbed or released during a phase transition at constant temperature. In the context of vaporization or condensation, it is a key parameter in determining how much energy is required to change phases, and it plays an integral role in the Clausius-Clapeyron equation.

Ideal Gas Law

The ideal gas law relates the pressure, volume, and temperature of a gas under the assumption that the gas molecules do not interact. This simplification is useful in deriving relationships like the Clausius-Clapeyron equation when modeling water vapor behavior before condensation occurs.

Nucleation Theory

Nucleation theory describes the initial formation of small particles (or droplets) in a supersaturated vapor. It involves analyzing the balance between the reduction in free energy due to phase change and the cost associated with creating a new interface, thereby determining the critical radius for stable droplet growth.

Clausius-Clapeyron Equation

The Clausius-Clapeyron equation provides a relation between the change in vapor pressure and temperature for a phase transition. It uses latent heat and the difference in specific volumes between the phases to quantify how the equilibrium vapor pressure changes, making it particularly useful when the vapor is treated as an ideal gas.

Adiabatic Process

An adiabatic process is one in which no heat is exchanged with the surroundings. In the context of expanding water vapor, adiabatic expansion leads to a temperature drop that affects vapor pressure according to the specific heat ratio (?). This process is essential for understanding how supersaturation can occur during rapid expansion.

Supersaturation

Supersaturation occurs when the actual vapor pressure exceeds the equilibrium vapor pressure at a given temperature. This metastable condition is crucial in cloud formation and droplet nucleation, as it provides the necessary conditions for condensation to be energetically favorable.

Transcript

00:03 All right, so this question is a little bit complicated, just trying to figure out exactly what's going on when we first started.

00:11 But we've got these three flasks that are connected, but the valves are closed currently.

00:17 If we open the valves, it'll be connected.

00:19 In this first flask on the left, flask a, we have a mixture of water vapor, nitrogen, and carbon dioxide.

00:30 And then the other two, b and c, are empty, but they are temperature jacketed.

00:39 So what that means is that each of these is going to maintain a constant temperature, but the three of them are not the same temperature.

00:49 And we're told to assume that the volume of the tubing connecting them is zero.

00:55 So the first thing that's going to happen is we're going to open this first valve, between a and b, and the pressure will drop, and we'll have to, and we can make some calculations on that.

01:09 And then the second thing is we're going to open the other valve, and the pressure will drop again.

01:14 So let's talk about what's happening physically first.

01:18 So when we open this first valve between a and b, what's going to happen is the gas in flask a is going to spread out between a and b, and it's going to spread out until the pressure equalizes.

01:31 But one of the things you'll notice is that the temperature of b is below the freezing point of water.

01:40 So that water vapor that makes it into flask b is going to freeze.

01:48 Now we're going to make a couple of assumptions here.

01:51 The first assumption is that the frozen water has no volume.

01:56 We're going to make that assumption.

01:58 It's usually an assumption that you make in gas lot problems is that solids and liquids don't have any volume.

02:03 Just gases that have volume.

02:04 So we're going to assume that anything that solidifies isn't taking up any volume.

02:10 The second thing that we're going to assume is that there is no equilibrium between the gas phase and the solid phase.

02:19 So once it's frozen, it's going to stay frozen.

02:22 And we can imagine that's a pretty good assumption, because if any of you live in parts of the world that get cold during the winter, you know that it gets pretty dry.

02:33 There's not a lot of moisture in the air.

02:35 You might notice that your throat starts to feel dry or there's more static electricity, but it gets pretty dry when it gets below the freezing point of water.

02:45 So we can assume that there's not a lot of water in the air.

02:48 So those two things combined, really the second one more than more than anything, we can assume that when we open that first valve, we're going to freeze all of the water.

03:00 Because what's going to happen is if we imagine that, we imagine that.

03:04 We're going to at first we have some water vapor in flask a and some in flask b, and they're at equilibrium between the two of them.

03:11 And that's fine.

03:12 But then the water vapor in flask b condenses and freezes, and now there's no water vapor in flask b anymore.

03:19 So they're going to equalize again.

03:21 Some of the a is going to come into b.

03:23 Some of the water vapor in a is going to come into b, and it's going to condense and it's going to freeze.

03:27 And eventually we're going to get to a point where all of the water vapor has randomly migrated into flask b and frozen and is stuck there.

03:36 And when we open the third valve, the same thing is going to happen to the carbon dioxide.

03:41 It's going to deposit on the sea flask.

03:46 So that's the physical principle of what's going on.

03:50 It's going to allow us to solve this problem.

03:53 So let's get to the math.

03:58 So the first thing that we want to calculate is how much? how many moles of gas do we have in flaskay? so we know that our volume of flaskay is 1 .000 liters.

04:14 And our temperature is 25 celsius, which is 298 kelvin.

04:23 And we also know that our pressure to start off with is 564 millimeters of mercury.

04:36 We're trying to solve for is n.

04:37 So we're going to use our ideal gas law.

04:41 Pv equals n rt.

04:44 We're going to rearrange it to get to n by itself.

04:47 So pv over rt equals n.

04:53 And now let's go ahead and plug in some numbers.

04:56 So we have our 564 times our 1.

05:00 That's times 1, not 0 .1.

05:04 The r value for millimeters of mercury is 62 .4, and the temperature is to 98.

05:13 So this is the first thing we're going to do is we're going to calculate what is the total number of moles of gas that we have in this system.