# 13.6 Humidity, evaporation, and boiling  (Page 3/9)

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## Percent relative humidity

We define percent relative humidity    as the ratio of vapor density to saturation vapor density, or

$\text{percent relative humidity}=\frac{\text{vapor density}}{\text{saturation vapor density}}×\text{100}$

We can use this and the data in [link] to do a variety of interesting calculations, keeping in mind that relative humidity is based on the comparison of the partial pressure of water vapor in air and ice.

## Calculating humidity and dew point

(a) Calculate the percent relative humidity on a day when the temperature is $\text{25}\text{.}0\text{º}\text{C}$ and the air contains 9.40 g of water vapor per ${\text{m}}^{3}$ . (b) At what temperature will this air reach 100% relative humidity (the saturation density)? This temperature is the dew point. (c) What is the humidity when the air temperature is $\text{25}\text{.}0\text{º}\text{C}$ and the dew point is $–\text{10}\text{.}0\text{º}\text{C}$ ?

Strategy and Solution

(a) Percent relative humidity is defined as the ratio of vapor density to saturation vapor density.

$\text{percent relative humidity}=\frac{\text{vapor density}}{\text{saturation vapor density}}×\text{100}$

The first is given to be $9\text{.}{\text{40 g/m}}^{3}$ , and the second is found in [link] to be $\text{23}\text{.}{\text{0 g/m}}^{3}$ . Thus,

$\text{percent relative humidity}=\frac{9\text{.}{\text{40 g/m}}^{3}}{\text{23}\text{.}{\text{0 g/m}}^{3}}×\text{100}=\text{40}\text{.}9\text{}\text{.%}$

(b) The air contains $9\text{.}{\text{40 g/m}}^{3}$ of water vapor. The relative humidity will be 100% at a temperature where $9\text{.}{\text{40 g/m}}^{3}$ is the saturation density. Inspection of [link] reveals this to be the case at $\text{10}\text{.}0\text{º}\text{C}$ , where the relative humidity will be 100%. That temperature is called the dew point for air with this concentration of water vapor.

(c) Here, the dew point temperature is given to be $–\text{10}\text{.}0\text{º}\text{C}$ . Using [link] , we see that the vapor density is $2\text{.}{\text{36 g/m}}^{3}$ , because this value is the saturation vapor density at $–\text{10}\text{.}0\text{º}\text{C}$ . The saturation vapor density at $\text{25}\text{.}0\text{º}\text{C}$ is seen to be $\text{23}\text{.}{\text{0 g/m}}^{3}$ . Thus, the relative humidity at $\text{25}\text{.}0\text{º}\text{C}$ is

$\text{percent relative humidity}=\frac{2\text{.}{\text{36 g/m}}^{3}}{\text{23}\text{.}{\text{0 g/m}}^{3}}×\text{100}=\text{10}\text{.}3\text{%}\text{}\text{.}$

Discussion

The importance of dew point is that air temperature cannot drop below $\text{10}\text{.}0\text{º}\text{C}$ in part (b), or $–\text{10}\text{.}0\text{º}\text{C}$ in part (c), without water vapor condensing out of the air. If condensation occurs, considerable transfer of heat occurs (discussed in Heat and Heat Transfer Methods ), which prevents the temperature from further dropping. When dew points are below $0\text{ºC}$ , freezing temperatures are a greater possibility, which explains why farmers keep track of the dew point. Low humidity in deserts means low dew-point temperatures. Thus condensation is unlikely. If the temperature drops, vapor does not condense in liquid drops. Because no heat is released into the air, the air temperature drops more rapidly compared to air with higher humidity. Likewise, at high temperatures, liquid droplets do not evaporate, so that no heat is removed from the gas to the liquid phase. This explains the large range of temperature in arid regions.

Why does water boil at $\text{100}\text{º}\text{C}$ ? You will note from [link] that the vapor pressure of water at $\text{100}\text{º}\text{C}$ is $1\text{.}\text{01}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}$ , or 1.00 atm. Thus, it can evaporate without limit at this temperature and pressure. But why does it form bubbles when it boils? This is because water ordinarily contains significant amounts of dissolved air and other impurities, which are observed as small bubbles of air in a glass of water. If a bubble starts out at the bottom of the container at $\text{20}\text{º}\text{C}$ , 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 bubble expands to keep the pressure at 1.00 atm. At $\text{100}\text{º}\text{C}$ , water vapor enters the bubble continuously since the partial pressure of water is equal to 1.00 atm in equilibrium. It cannot reach this pressure, however, since the bubble also contains air and total pressure is 1.00 atm. The bubble grows in size and thereby increases the buoyant force. The bubble breaks away and rises rapidly to the surface—we call this boiling! (See [link] .)

Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0º20.0º with the horizontal. (See [link] .) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate an
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