# 2.2 Pressure, temperature, and rms speed  (Page 7/18)

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Another important application of partial pressure is vapor pressure    , which is the partial pressure of a vapor at which it is in equilibrium with the liquid (or solid, in the case of sublimation) phase of the same substance. At any temperature, the partial pressure of the water in the air cannot exceed the vapor pressure of the water at that temperature, because whenever the partial pressure reaches the vapor pressure, water condenses out of the air. Dew is an example of this condensation. The temperature at which condensation occurs for a sample of air is called the dew point . It is easily measured by slowly cooling a metal ball; the dew point is the temperature at which condensation first appears on the ball.

The vapor pressures of water at some temperatures of interest for meteorology are given in [link] .

Vapor pressure of water at various temperatures
T $\left(\text{°}\text{C}\right)$ Vapor Pressure (Pa)
0 610.5
3 757.9
5 872.3
8 1073
10 1228
13 1497
15 1705
18 2063
20 2338
23 2809
25 3167
30 4243
35 5623
40 7376

The relative humidity (R.H.) at a temperature T is defined by

$\text{R.H.}=\frac{\text{Partial pressure of water vapor at}\phantom{\rule{0.2em}{0ex}}T}{\text{Vapor pressure of water at}\phantom{\rule{0.2em}{0ex}}T}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%.$

A relative humidity of $100%$ means that the partial pressure of water is equal to the vapor pressure; in other words, the air is saturated with water.

## Calculating relative humidity

What is the relative humidity when the air temperature is $25\phantom{\rule{0.2em}{0ex}}\text{ºC}$ and the dew point is $15\phantom{\rule{0.2em}{0ex}}\text{ºC}$ ?

## Strategy

We simply look up the vapor pressure at the given temperature and that at the dew point and find the ratio.

## Solution

$\text{R.H.}=\frac{\text{Partial pressure of water vapor at}\phantom{\rule{0.2em}{0ex}}15\phantom{\rule{0.2em}{0ex}}\text{°C}}{\text{Partial pressure of water vapor at}\phantom{\rule{0.2em}{0ex}}25\phantom{\rule{0.2em}{0ex}}\text{°C}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%=\frac{1705\phantom{\rule{0.2em}{0ex}}\text{Pa}}{3167\phantom{\rule{0.2em}{0ex}}\text{Pa}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%=53.8%.$

## Significance

R.H. is important to our comfort. The value of $53.8%$ is within the range of $40%\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}60%$ recommended for comfort indoors.

As noted in the chapter on temperature and heat, the temperature seldom falls below the dew point, because when it reaches the dew point or frost point, water condenses and releases a relatively large amount of latent heat of vaporization.

## Mean free path and mean free time

We now consider collisions explicitly. The usual first step (which is all we’ll take) is to calculate the mean free path    , $\lambda ,$ the average distance a molecule travels between collisions with other molecules, and the mean free time $\tau$ , the average time between the collisions of a molecule. If we assume all the molecules are spheres with a radius r , then a molecule will collide with another if their centers are within a distance 2 r of each other. For a given particle, we say that the area of a circle with that radius, $4\pi {r}^{2}$ , is the “cross-section” for collisions. As the particle moves, it traces a cylinder with that cross-sectional area. The mean free path is the length $\lambda$ such that the expected number of other molecules in a cylinder of length $\lambda$ and cross-section $4\pi {r}^{2}$ is 1. If we temporarily ignore the motion of the molecules other than the one we’re looking at, the expected number is the number density of molecules, N / V , times the volume, and the volume is $4\pi {r}^{2}\lambda$ , so we have $\left(N\text{/}V\right)4\pi {r}^{2}\lambda =1,$ or

$\lambda =\frac{V}{4\pi {r}^{2}N}.$

Taking the motion of all the molecules into account makes the calculation much harder, but the only change is a factor of $\sqrt{2}.$ The result is

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