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learning goals
At the end of this section you can:
- Explain the relationship between the vapor pressure of water and the ability of air to hold water vapor.
- Explain the relationship between relative humidity and the partial pressure of water vapor in the air.
- Calculate the vapor density using the vapor pressure.
- Calculate humidity and dew point.
The phrase "it's not the heat, it's the humidity" has a valid point. We stay cool in hot weather by allowing sweat to evaporate from our skin and water from our airways. Since high humidity inhibits evaporation, at a given temperature we feel warmer when humidity is high. Low humidity, on the other hand, can cause discomfort from excessive drying of the mucous membranes and lead to an increased risk of respiratory infections.
Figure \(\PageIndex{1}\):Dewdrops like this one, on a banana leaf photographed just after sunrise, form when the air temperature falls below the dew point. At the dew point, the speed at which water molecules stick together is greater than the speed at which they separate, and some of the water condenses into droplets. (Image credit: Aaron Escobar, Flickr)
When we say moisture, we really mean itRH. The relative humidity tells us how much water vapor is in the air compared to the maximum possible. At its maximum, denoted assaturation, the relative humidity is 100% and evaporation is inhibited. The amount of water vapor in the air depends on the temperature. For example, the relative humidity increases at night with falling air temperature, sometimes up to 100%dew point. insidedew pointRoom temperature, relative humidity 100% and fog can form from the condensation of water droplets if they are small enough to remain in suspension. Conversely, if you want to dry something (maybe your hair), blowing hot air on it is more effective than cold air because, among other things, the increase in temperature increases the energy of the molecules, so the rate of rise. Evaporation.
The amount of water vapor in the air depends on the vapor pressure of the water. The liquid and solid phases constantly give off vapor because some of the molecules have velocities high enough to enter the gas phase; seeFigure(To). If a lid is placed on the container, as inFigure(b), evaporation continues, increasing pressure until enough vapor accumulates for condensation to offset the evaporation. Then equilibrium is reached and the vapor pressure is equal to the partial pressure of the water in the container. Vapor pressure increases with temperature because molecular velocities increase with increasing temperature.Mesagives representative values of water vapor pressure over a temperature range.
Relative humidity is related to the partial pressure of water vapor in the air. At 100% humidity, the partial pressure is equal to the vapor pressure and no more water can get into the vapor phase. When the partial pressure is less than the vapor pressure, evaporation occurs because the humidity is less than 100%. If the partial pressure is greater than the vapor pressure, condensation occurs. In colloquial language one sometimes speaks of the ability of air to "hold" water vapour, but this is not the case. The water vapor is not retained by the air. The water content of the air is determined by the vapor pressure of the water and has nothing to do with the properties of the air.
| Temperature \(^oC\) | Vapor pressure \((Pa)\) | Saturation vapor density \(g/m^3\) |
| −50 | 4.0 | 0,039 |
| -20 | \(1.04 \times 10^2\) | 0,89 |
| -10 | \(2,60 \times 10^2\) | 2.36 |
| 0 | \(6.10 \times 10^2\) | 4.84 |
| 5 | \(8,68 \times 10^2\) | 6,80 |
| 10 | \(1.19 \times 10^3\) | 9.40 |
| 15 | \(1,69 \times 10^3\) | 12.8 |
| 20 | \(2.33 \times 10^3\) | 17.2 |
| 25 | \(3.17 \times 10^3\) | 23.0 |
| 30 | \(4.24 \times 10^3\) | 30.4 |
| 37 | \(6.31 \times 10^3\) | 44.0 |
| 40 | \(7,34 \times 10^3\) | 51.1 |
| 50 | \(1.23 \times 10^4\) | 82.4 |
| 60 | \(1,99 \times 10^4\) | 130 |
| 70 | \(3.12 \times 10^4\) | 197 |
| 80 | \(4,73 \times 10^4\) | 294 |
| 90 | \(7.01 \times 10^4\) | 418 |
| 95 | \(8,59 \times 10^4\) | 505 |
| 100 | \(1.01 \times 10^5\) | 598 |
| 120 | \(1,99 \times 10^5\) | 1095 |
| 150 | \(4,76 \times 10^5\) | 2430 |
| 200 | \(1,55 \times 10^6\) | 7090 |
| 220 | \(2.32 \times 10^6\) | 10.200 |
saturation vapor density of water
Example \(\PageIndex{1}\): Density calculation via vapor pressure
Mesagives the vapor pressure of water at .\(20.0^oC\) as \(2.33 \times 10^3 \, Pa\). Use the ideal gas law to calculate the density of water vapor in \(g/m^3\) that would produce a partial pressure equal to that vapor pressure. Compare the result with the saturation vapor density given in the table.
Strategy
To solve this problem, we need to divide it into two steps. The partial pressure follows the ideal gas law,
\[VP = nRT,\]
where \(n\) is the number of moles. Let's solve this equation for \(n/V\)
To calculate the number of moles per cubic yard, we can convert that amount to grams per cubic yard if desired. To do this, we need to use the molecular mass of water, which is given on the periodic table.
Solution
1. Identify the known ones and convert them to the appropriate units:
- Temperature \(T = 20^oC = 203 \, K\)
- the vapor pressure \(P\) of water at \(20^oC\) is \(2.33 \times 10^3 \, Pa\)
- the molar mass of water is \(18.0 \, g/m\)
2. Solve the ideal gas law according to \(n?V\).
\[\dfrac{n}{V} = \dfrac{P}{RT}\]
3. Plug the known values into the equation and solve for \(n/V\).
\[\dfrac{n}{V} = \dfrac{P}{RT} = \dfrac{2.33\times 10^3\,Pa}{(8.31\,J/mol\cdotK)(2 \,K ) } = 0.957\,mol/m^3\]
4. Convert the density in moles per cubic meter to grams per cubic meter.
\[\rho = \left(0,957\dfrac{mol}{m^3}\right)\left(\dfrac{18,0 \, g}{mol}\right) = 17,2 \, g/m ^3\]
discussion
The density is obtained assuming a pressure equal to the vapor pressure of water in \(20.0^oC\). The density found is identical to the value inMesa, which means that a vapor density of \(17.2 \, g/m^3\) at \(20.9^oC\) has a partial pressure of \(2.33 \times 10^3 \, Pa\) generated, equal to the vapor pressure of water at that temperature. If the partial pressure is equal to the vapor pressure, the liquid and vapor phases are in equilibrium and the relative humidity is 100%. Therefore at \(20.0^oC\) there can be no more than 17.2 g of water vapor per \(m^3\), so this value is the saturation vapor density at that temperature. This example shows how water vapor behaves like an ideal gas: the pressure and density agree with the ideal gas law (assuming the density in the table is correct). The saturation vapor densities listed inMesaare the maximum amounts of water vapor that air can hold at different temperatures.
relative humidity percentage
We definerelative humidity percentageas the ratio of vapor density to saturation vapor density, or
\[Percent \,relative\,humidity = \dfrac{steam\,density}{saturation \,steam\,density} \times 100\]
We may use this and the data it containsMesato do a variety of interesting calculations, bearing in mind that relative humidity is based on a comparison of the partial pressure of water vapor in air and ice.
We may use this and the data it containsMesato do a variety of interesting calculations, bearing in mind that relative humidity is based on a comparison of the partial pressure of water vapor in air and ice.
Example \(\PageIndex{2}\): Calculation of humidity and dew point
(a) Calculate the percentage of relative humidity on a day when the temperature is \(25.0^oC\) and the air contains 9.40 g of vapor per \(m^3\). (b) At what temperature does this air reach 100% relative humidity (the saturation density)? This temperature is the dew point. (c) What is the humidity at an air temperature of \(25.0^oC\) and a dew point of \(-10.0^oC\)?
strategy and solution
(a) Percent relative humidity is defined as the ratio of vapor density to saturation vapor density.
\[Percent \,relative\,humidity = \dfrac{steam\,density}{saturation \,steam\,density} \times 100\]
The first is given as \(9.40 \, g/m^3\), the second inMesalength \(23.0\, g/m^3\). Daher,
\[Percent\, relative\, humidity = \dfrac{9.40\,g/m^3}{23.0\,g/m^3}\times 100 = 40.9\%\]
(b) Air contains \(9.40 \, g/m^3\) water vapor. Relative humidity is 100% at a temperature where \(9.40 \, g/m^3\) is the saturation density. inspection ofMesashows that this is the case at \(10.0^oC\),
where the relative humidity is 100%. This temperature is called the dew point for air with this water vapor concentration.
(c) Here the dew point temperature is \(10.0^oC\). wear and tearMesa, we see that the vapor density is \(2.36 \, g/m^3\) because this value is the saturation vapor density at \(-10.0^oC\). It can be seen that the saturation vapor density at \(25.0^oC\) is \(23.0 \,g/m^3\). Therefore, the relative humidity is \(25.0^oC\).
\[Percent\, relative\, humidity = \dfrac{2.36\,g/m^3}{23.0\,g/m^3}\times 100 = 10.3\%\]
discussion
The importance of the dew point is that the air temperature cannot drop below \(10.0^oC\) in part (b) or \(-10.0^oC\) in part (c) without water vapor condensing out of the air . When condensation occurs, considerable heat transfer takes place (discussed inheat and heat transfer processes), which prevents the temperature from falling further. When dew points are below \(0^oC\), temperatures below freezing are more likely, which explains why farmers keep dew point records. Low humidity in deserts means low dew point temperatures. Condensation is therefore unlikely. When the temperature drops, the vapor does not condense into liquid droplets. Since no heat is released into the air, the air temperature drops faster compared to air with higher humidity. In addition, liquid droplets do not vaporize at high temperatures, so no heat is transferred from the gas phase to the liquid phase. This explains the large temperature differences in arid regions.
Why does water boil at \(100^oC\)? You will noticeMesathat the vapor pressure of water is at \(100^oC\) \(1.01 \times 10^5 \, Pa\) or 1.00 atm. Therefore, at that temperature and pressure, it can vaporize indefinitely. But why does it bubble when it boils? This is because water typically contains significant amounts of dissolved air and other impurities, which can be seen in a glass of water as tiny air bubbles. If a bubble starts at the bottom of the container at \(20^oC\), it contains water vapor (about 2.30%). The pressure inside the bubble is fixed at 1.00 atm (we ignore the slight pressure exerted by the water around it). As the temperature rises, the amount of air in the bubble stays the same, but the water vapor increases; the bladder expands to maintain the pressure at 1.00 atm. At \(100^oC\) water vapor continuously enters the bubble since the partial pressure of water in equilibrium is 1.00 atm. However, this pressure cannot be reached because the bladder also contains air and the total pressure is 1.00 atm. The bubble gets bigger and the buoyancy increases. The bubble ruptures and quickly rises to the surface. We call that cooking! (SeeFigure.)
check your understanding
- Freeze drying is a process in which substances such as food are dried by placing them in a vacuum chamber and reducing the atmospheric pressure around them. How does reduced atmospheric pressure speed up the drying process and why does it lower the temperature of the food?
[hide solution]
The decrease in atmospheric pressure leads to a decrease in the partial pressure of water and thus to lower humidity. For example, the evaporation of water from food is improved. The water molecules most likely to be released from food are those with the highest speeds. The remaining ones therefore have a lower average speed and a lower temperature. This can (and does) cause the food to freeze and dry out; Hence the process is aptly named freeze drying.
PHET EXPLORATIONS: STATES OF MATTER
Check out how different types of molecules form a solid, liquid or gas. Add or remove heat and watch the phase change. Change the temperature or volume of a container and see how a real-time pressure-temperature graph responds. Link the interaction potential to the forces between molecules.
Summary
- Relative humidity is the proportion of water vapor in a gas compared to the saturation value.
- The saturation vapor density can be determined from the vapor pressure at a given temperature.
- Percent relative humidity is defined as \[percent \,relative\,humidity = \dfrac{vapour\,density}{saturation \,vapour\,density} \times 100\%\]
- The dew point is the temperature at which the air reaches 100% relative humidity.
glossary
- dew point
- the temperature at which the relative humidity is 100%; the temperature at which water begins to condense from the air
- saturation
- the state of 100% relative humidity
- relative humidity percentage
- the ratio of vapor density to saturation vapor density
- RH
- the amount of water in the air in relation to the maximum amount that the air can hold