atmosphere and water vapour

Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected])

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Atmosphere and Water Vapour

Earth's atmosphere consists of layers of gases surrounding our planet Earth and retained by the Earth's gravity. It contains approximately (by volume) 78% nitrogen, 20.95% oxygen, 0.93% argon, 0.038% carbon dioxide, traces of other gases, and a variable amount (1 - 3%) of water vapour. This mixture of gases is commonly known as air. The atmosphere plays an important role in protecting life on earth by absorbing harmful ultraviolet radiation from the Sun. Also, the extreme temperatures during day and night time are controlled by the presence of atmosphere. Water Vapour in Atmosphere

Water vapour is the gas phase of water. Water vapour is produced from the vaporization (evaporation or boiling) or from the sublimation of ice. Under normal atmospheric conditions, water vapour is continuously evaporating and condensing.

The volume of water vapour in atmosphere, on average, is about 1 to 3%. Therefore, normally, only about 1 to 3% of the molecules in the air are water vapour molecules. The volume of water vapour is about 4% in very warm and humid air in tropical countries (like in India). Even in tropical air, once the volume of water vapour in the atmosphere approaches 4% it will begin to condense out of the air. The condensation of water vapour prevents the percentage of water vapour in the air from further increasing. However, it is possible to have greater percentages of water vapour in the air at higher temperatures. The amount of water vapour in the air is very low in extremely arid areas and at places where the temperatures are very low. Temperature determines the maximum amount of water vapour that can exist in the air. Vapour Pressure: Vapour exerts pressure on the surface of liquid or solid from which it is created. Water vapour exerts a pressure on the water surface. If we take some liquid in a closed vessel, a part of the liquid will vaporize and the rest will remain unchanged for an unlimited time. But the process of vaporization does not really stop. Molecules continue leaving the surface of the liquid and at the same time condensation process takes place (condensation of vapour into liquid). In the closed vessel, eventually, the rate of vaporization and the rate of condensation become equal and a dynamic equilibrium is established. Thus, beginning with this instant, the amount of liquid remains unchanged in the closed vessel. The vapour is now said to be saturated.


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Saturated Vapour Pressure__________ The pressure of vapour in equilibrium with liquid or solid at any temperature is called the saturated vapour pressure of that substance at that temperature.

• The saturated vapour pressure of a liquid depends only on its chemical composition and temperature.

• In general, the higher the temperature, the higher the vapour pressure. Dew Point_________ The temperature at which the saturation vapour pressure is achieved in the air, it is said to be the dew point. Any increase of water vapour content or decrease in temperature at this point causes the condensation of water vapour. Unsaturated vapour Pressure_________ If in a situation, the rate of vaporization becomes less than the rate of condensation of vapour, the vapour is said to be unsaturated. The pressure due to unsaturated vapour is called unsaturated vapour pressure. This is possible when the liquid surface is open as the evaporation in nature occurs from the surface of sea water and from the lakes and rivers etc. Humidity The term that is used to describe the amount of water vapour in air is called humidity. The content of water vapour in air or humidity can be calculated in various ways. Absolute Humidity (AH): When the humidity is measured by the mass of water vapour in a certain volume of air or by the pressure that the water vapour exerts, it is called absolute humidity. Normally, it is measured in gm/m 3 (grams per cubic meter). For example, air at 10 C0 contains 9.4 gm/m3 of water vapour when saturated. Note the use of mixed unit rather than having C.G.S. (gm/cm 3 ) or S.I. (kg/ m 3 ) units. This has been a practice since the density of air is very low.

If all the water in one cubic meter of air could be condensed into a container, the container could be weighed to determine absolute humidity. If the mass of water vapour is wm gm and the volume of air is aV cubic meter, we can write

AH = a

w

Vm

gm/m 3 .


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The absolute humidity changes as the air pressure changes. This is because the volume of air depends on the pressure. This definition is usually not considered in case of weather forecasting. Specific Humidity (SH): Specific humidity is defined as the ratio of the mass of water vapour in the air to the total mass of the air in a particular volume of the air. If the mass of a certain volume of air is am and the mass of water vapour in that air is

wm , we can write

SH = a

w

mm

.

Relative Humidity (RH):

Air is a mixture of several gases and water vapour.

Relative humidity of air is defined as the ratio of the partial pressure of water vapour in air to the saturated vapour pressure of water at a given temperature.

Relative humidity is expressed as a percentage and is calculated as follows:

RH = 100×sat

part

PP

%

Where partP = the partial pressure of water vapour in the gas mixture, satP = the saturated vapour pressure. In simple terms, the definition of relative humidity is the percentage of the ratio of the observed vapour pressure to that required for saturation at the same temperature and pressure. When the unsaturated air is cooled, the relative humidity increases. Eventually it reaches a temperature when the relative humidity reaches 100%. The air is then called saturated. It is said that the dew point is reached. Further cooling leads to condensation of excess water vapour. The condensed moisture is called dew. This occurs frequently at night when the earth’s atmosphere is cooled by radiation. As the amount of water vapour does not change, the pressure due to water vapour at any temperature is equal to the saturated water vapour (in the same air) at dew point. The definition of relative humidity can be rewritten as follows:


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RH = 100×sat

dew

PP

%

Where dewP = the saturated vapour pressure at the dew point. When the air temperature is increased, the relative humidity decreases. This is because the amount of water vapour in air is now less than that could be present at this temperature.

Example:

If the amount of water vapour is 20 gm in every 1 kg of air at a certain place at 30 C0 whereas one kg of that air would be saturated with 25 gm of water vapour, the relative humidity of air would be 25/20 = 80% at 30 C0 . Note:

• The dew point has no fixed value. When air is cooled, the mass of water vapour in it remains unchanged. So the pressure due to water vapour does not change when the air is cooled. Eventually dew point is reached and the dew point (the temperature) depends on the content of water vapour in air. The dew point varies from place to place as the moisture content varies from place to place. Also, the dew point at a same location may widely vary depending on the supply of water vapour.

• The relative humidity shows the degree of saturation but it gives no idea of the actual

amount of water vapour in the air. The important points to remember: The relative humidity depends on two factors: (i) the amount of water vapour (moisture) and (ii) the temperature. As long as the moisture content in the air remains unchanged,

• Reduce the temperature, the relative humidity (RH) goes up. • Increase the temperature, the relative humidity (RH) goes down.

Measurement of Humidity: The branch of physics which deals with the determination of humidity of the atmosphere is called hygrometry. The device that is used to measure the humidity of air is called hygrometer. Working principle of a hygrometer: A hygrometer consists of two thermometers, one dry bulb or standard air temperature thermometer, and one wet bulb thermometer. The wet bulb thermometer is an ordinary


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thermometer which has the bulb covered with a muslin bag, kept moist via an absorbent wick dipped into water. Evaporation of water from the muslin lowers the temperature of the thermometer. The difference between wet and dry bulb temperatures is used to calculate the various measures of humidity.

Include a Figure like this (may not be this one)

Problems with Solutions Example 1: The temperature and dew point of air are 16 C0 and 7.6 C0 , respectively. If the atmospheric pressure is 13.5 mm Hg at 16 C0 and 7.8 mm Hg at 7.6 C0 , find the relative humidity of air. [H.S. ‘92] Solution:

Here the saturated vapour pressure at dew point i.e. at 7.6 C0 is dewP = 7.8 mm Hg and the saturated vapour pressure at 16 C0 is satP = 13.5 mm Hg.

∴The relative humidity = 100×sat

dew

PP

% = 1005.138.7× % = 57.8%.

Example 2: The room temperature and the dew point on a certain day are 18.5 C0 and 12 C0 , respectively. The saturated vapour pressure at 18 C0 , 19 C0 and 12 C0 are 15.46 mm, 15.86 mm and 10.46 mm, respectively. What is the relative humidity on that day? [H.S. ‘96] Solution: To find the relative humidity of air on that day we need to know the saturated vapour pressure on that day at 18.5 C0 and the saturated vapour pressure at dew point (12 C0 ).


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The saturated vapour pressure at dew point at 12 C0 , dewP = 10.46 mm. The saturated vapour pressure at 18 C0 and 19 C0 are 15.46 mm and 15.86 mm, respectively. ∴The rate of increase of saturated vapour pressure in this temperature range =

)46.1586.15( − = 0.4 mm/ C0 . ∴The saturated vapour pressure at room temperature at 18.5 C0 ,

satP = the saturated vapour pressure at 18 C0 + the increment of pressure due to 0.5 C0 increase in temperature = 4.05.046.15 ×+ = 15.66 mm.


dew

PP

% = 10066.1546.10

× = 66.8%.

Example 3: The temperature of air is 26.6 C0 and the dew point is 9.5 C0 on a certain day. The maximum pressures due to water vapour at temperatures, 9 C0 , 10 C0 , 26 C0 and 27 C0 are 8.6 mm, 9.2 mm, 25.2 mm, and 26.7 mm, respectively. Determine the relative humidity of air on that day. [H.S. ‘89] Solution: To find the relative humidity of air on that day we need to know the saturated vapour pressure at dew point (9.5 C0 ) and the saturated vapour pressure at the temperature of that day (26.6 C0 ). It is given, the saturated vapour pressure at 9 C0 = 8.6 mm and the saturated vapour pressure at 10 C0 = 9.2 mm. ∴The rate of increase of saturated vapour pressure in this temperature range = )6.82.9( − = 0.6 mm/ C0 . ∴The saturated vapour pressure at the dew point at 9.5 C0 ,

dewP = the saturated vapour pressure at 9 C0 + the increment of pressure due to 0.5 C0 increase in temperature = 6.05.06.8 ×+ = 8.9 mm. The saturated vapour pressure at 26 C0 = 25.2 mm and the saturated vapor pressure at 27 C0 = 26.7 mm. ∴The rate of increase of saturated vapour pressure in this temperature range =

)2.257.26( − = 1.5 mm/ C0 . ∴The saturated vapour pressure at 26.6 C0 ,

satP = the saturated vapour pressure at 26 C0 + the increment of pressure due to 0.6 C0 increase in temperature = 5.16.02.25 ×+ = 26.1 mm.


dew

PP

% = %1001.26

9.8× = 34.1%.


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Example 4: What is the dew point when the relative humidity of air at 20 C0 is 52%? The saturated vapour pressures at 20 C0 , 10 C0 and 9 C0 are 17.5, 9.2 and 8.6 mm Hg. [J.E.E. ‘86] Solution:

The saturated vapour pressure at 20 C0 , satP = 17.5 mm. If the saturated vapour pressure at dew point be dewP , we can write the relative humidity at

20 C0 = 100×sat

dew

PP

%

∴ 100×sat

dew

PP

= 52 Or, 100

525.17 ×=dewP = 9.1 mm.

Now, for the increase of temperature from 9 C0 to 10 C0 , the pressure increases from 8.6 mm to 9.2 mm. ∴The increase in pressure by 6.0)6.82.9( =− mm corresponds to 1 C0 increase in temperature. Thus the increase in pressure by 5.0)6.81.9( =− mm will correspond to

83.06.0

5.01=

× C0 increase in temperature.

∴If the dew point pressure is 9.1 mm, the temperature = 83.9)83.09( =+ C0 . Therefore, the dew point = 9.83 C0 . Example 5: An air condition machine sucks air of 90% relative humidity at 30 C0 and cools it down to 20 C0 . In this, the relative humidity of air reduces to 50%. How much water vapour is extracted from one cubic meter of air by the machine? Neglect the volume change of air. The densities of saturated vapour at 30 C0 and 20 C0 are 30 gm/m 3 and 17 gm/m 3 . [J.E.E.] Solution:

The relative humidity at 30 C0 is the ratio of the mass of water vapour in 1 m 3 air at 30 C0 to the mass of water vapour in 1 m 3 saturated air at that temperature.

We can write, RH = 100×SatM

M %,

where =M the mass of water vapour in 1 m 3 air at 30 C0 and SatM = the mass of saturated water vapour in 1 m 3 air at 30 C0 = 30 gm. ∴According to question,

10030

90 ×=M

∴The mass of water vapour present in 1 m 3 of air, 100

3090×=M = 27 gm.

The relative humidity of air at 20 C0 becomes 50%.


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Thus in a similar way, we can write the mass of water vapour present in 1 m 3 of air at

20 C0 100

1750×= = 8.5 gm.

∴The mass of water vapour removed by the air condition machine = 5.185.827 =− gm. Example 6: The temperature of a day is 20 C0 and the relative humidity is 60%. If the temperature comes down to 5 C0 , what fraction of the water vapour will be condensed? The saturated vapour pressures at 20 C0 and 5 C0 are 17.5 mm and 6.5 mm Hg. [J.E.E.] Solution:

The temperature of air = 20 C0 . The saturated vapour pressure at 20 C0 = 17.5 mm. If now the pressure of water vapour present in the air at 20 C0 is p , we can write the relative humidity,

60 = 1005.17×

p Or, 5.10100

605.17=

×=p mm.

As the pressure of water vapour in the air should be proportional to the mass of water vapour in the air, we can write the mass of water vapour, 5.101 ×= km gm, where k is a constant. When the temperature comes down to 5 C0 , some water vapour would be condensed and the air would be saturated by the remaining vapour. The saturated vapour pressure at 5 C0 = 6.5 mm. ∴The mass of the water vapour present in air at 5 C0 , 5.62 ×= km gm. ∴The mass of condensed water vapour due to reduction of temperature from 20 C0 to 5 C0 = kkk 4)5.65.10( =− gm.

∴The fraction of condensed water vapour = 218

5.104

=k

k .

Example 7: The volume of an air tight and dry room is 76 m 3 and its temperature is 17 C0 . A pot of water is now kept in that room. After how much evaporation of water the rest of the water and the vapour will be in equilibrium? The saturated vapour pressure of water at 17 C0 is 15 mm Hg. [I.I.T.] Solution:

At equilibrium, the air inside the room at 17 C0 is saturated by water vapour where the saturated vapour pressure is 15 mm Hg. We have, the pressure 1P = 15 mm Hg, the temperature 290172731 =+=T K and the volume =1V 76 m 3 = 61076× c.c. Let the volume of the same water vapour at N.T.P. is 2V c.c. whereas the temperature and pressure are 2732 =T K and =2P 760 mm Hg.

Assuming the vapour follows the ideal gas law, we have2

22

1

11

TVP

TVP

=


9

Or, 273

760290

107615 26 V×=

×× Or, 52 1015

290273

××=V = 51012.14 ×

∴The volume of the same saturated water vapour at N.T.P. is 51012.14 × c.c whereas the mass should remain the same. The mass of 1 Mole of water vapour (H 2 O) = 1612 +× = 18 gm and the volume is 22.4 litre at N.T.P.

∴The density of water vapour at N.T.P. = 3104.2218×

gm/cc.

∴The mass of water vapour = 35

104.22181012.14×

×× = 310135.1 × gm = 1.135 kg.

Example 8: Determine the mass of 1 litre moist air when the air temperature is 27 C0 and the barometer reading is 753.6 mm. The dew point is 16.1 C0 . The saturated vapour pressure at 16.1 C0 is 13.6 mm Hg; the density of air at N.T.P. is 0.001293 gm/cc and the density of saturated water vapour is 0.000808 gm//cc. [I.I.T.] Solution: The moist air contains dry air and water vapour. ∴The pressure of moist air = the pressure of dry air + the pressure of water vapour present in the air. The pressure of water vapour in the air at 27 C0 = the pressure of water vapour in the same air at dew point i.e. at 16.1 C0 = 13.6 mm. ∴The pressure due to dry air present in the moist air at 27 C0 = 7406.136.753 =− mm.

Let us consider the air and water vapour follow ideal gas law, 2

22

1

11

TVP

TVP

= .

For dry air, 7401 =P mm, =1V 1 litre = 310 c.c., CT 01 27= = 27273 + = 300 K .

If the volume of the same air at N.T.P. is 2V , where 7602 =P mm and 2732 =T K , we can write

273

760300

10740 23 V×=

× Or, 760300

10273740 3

2 ×××

=V = 886.05 c.c.

∴ The mass of dry air present in the moist air at 27 C0 = the mass of the same dry air at N.T.P. = 001293.005.886 × = 1.1457 gm. For the water vapour present in the air at 27 C0 , =1P 13.6 mm, =1V 310 c.c., 310 c.c.,

310 c.c., =1T 300 K . If the volume at N.T.P. is ′2V , we can write

273

760300

106.13 23 ′×=

× V Or, ′2V =

300760102736.13 3

××× = 16.28 c.c.


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∴ The mass of water vapour present in the moist air at 27 C0 = the mass of the same water vapour at N.T.P. = 000808.028.16 × = 0.0131 gm. ∴The mass of 1 litre moist air = the mass of dry air + the mass of water vapour = 1.1457 + 0.0131 = 1.1588 gm. Example 9: A closed room measures 5 m×5 m×4 m. The room is filled with air at 27 C0 . If the relative humidity inside the room reduces from 90% to 40%, determine the amount of moisture that is condensed. The pressure of saturated water vapour at 27 C0 = 26.7 mm Hg; the gas constant 7103.8 ×=R erg/(mole C0 ); the atomic weight of water = 18; =g 980 cm/sec 2 . [H.S. ’02; I.I.T.] Solution:

Let the mass of water vapour required to saturate the air in the room at 27 C0 be Sm gm. As the mass of 1 gm-mole water vapour = 18 gm, the number of moles in the water

vapour, 18

Smn = .

If we assume the water vapour to follow ideal gas law , we have nRTPV = Here, =P the pressure of the saturated water vapour = 2.67 mm Hg = 9806.1367.2 ×× dyne/cm 2 , =V the volume of the room = 455 ×× m 3 = 610100× cm 3 = 810 c.c., =T the temperature of water vapour = 27327 + = 300 K , 7103.8 ×=R erg/(mole C0 )

∴ 300103.818

109806.1367.2 78 ×××=××× Sm

Or, 303.8

189806.1367.2×

×××=Sm = 2572.46 gm

The relative humidity, RH = 100×Sm

m %, where m gm is the mass of water vapour

present in the air at 27 C0 .

Now, when the relative humidity is 90%, we write 10046.2572

90 ×=m Or,

21.231590.046.2572 =×=m . ∴The mass of water vapour present in the air when the relative humidity is 90% = 2315.21 gm. Similarly, the mass of water vapour present in the air when the relative humidity is 40% = 98.102840.046.2572 =× gm. Therefore, the amount of water vapour condensed due to the reduction of relative humidity = 23.128698.102821.2315 =− gm.


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Example 10: A flask contains 1 gm saturated steam at 100 C0 . How much steam is condensed when the temperature is brought down to 25 C0 ? The pressure of water vapour at 25 C0 = 30 mm Hg. [J.E.E.] Solution: For a certain fixed volume of air, the mass of water vapour in it is proportional to its pressure.

∴We can write, 100

25

100

25

pp

mm

=

Here, 25m = mass of water vapour at 25 C0 , =100m mass of water vapour at 100 C0 = 1 gm, =25p pressure of water vapour at 25 C0 = 30 mm Hg, =100p pressure of water vapour at 100 C0 = 760 mm Hg (normal pressure)

∴ 176030

25 ×=m = 763 gm.

∴The mass of condensed steam (water vapour) as the temperature is brought down from

100 C0 to 25 C0 = 7673

7631 =− = 0.96 gm.

Example 11: In a closed room, the air temperature is 15 C0 and the dew point is found to be 8 C0 . If the temperature is brought down to 10 C0 , will there be any change in dew point? The pressure of water vapour at 7 C0 is 7.49 mm and at 8 C0 it is 8.02 mm. [J.E.E. ‘97] Solution: We know that the dew point is the temperature at which the water vapour present in the volume of air becomes saturated. As long as the air contains the same amount of water vapour, the dew point remains unchanged. If we assume that in a closed room, the amount of water remains the same, the dew point is also set to be the same. Note that the other sets of information, as given in the question, are unnecessary. Example 12: The temperature of air on some day is 23 C0 and the relative humidity is 60%. Now if the air temperature comes down to 10 C0 , what fraction of water vapour present in the air will condense? The saturated vapour pressure at 23 C0 and 10 C0 are 21.1 mm and 9.2 mm, respectively. [J.E.E. ‘95] Solution: The air temperature is 23 C0 and the saturated vapour pressure at this temperature = 21.1 mm. If the vapour pressure of air at 23 C0 is p mm, we can write the relative humidity,

1001.21

60 ×=p Or, 1.216.0 ×=p = 12.66 mm.


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As the mass of water vapour present in the air is proportional to the vapour pressure, we can write the mass of water vapour = 66.12×k gm, where k is a constant. When the temperature comes down to 10 C0 , some of the water vapour gets condensed and the air remains saturated by the remaining water vapour. The saturated water vapour at 10 C0 = 9.2 mm ∴The mass of water vapour present in the air at 10 C0 = 2.9×k gm. ∴When the temperature reduces from 23 C0 to 10 C0 , the mass of condensed vapour =

kkk 46.32.966.12 =− gm.

∴The fraction of condensed water vapour = 27.066.1246.3

=kk .

Discussions of a few Questions

Q.1 At which condition the temperature of a room and the dew point are equal? [H.S. ‘98] Ans. The temperature at which the air is saturated by the water vapour present in it is called the dew point. If the temperature of the room has to be equal to the dew point then the air in the room has to be saturated by the water vapour at this temperature. Q.2 When will there be no dew point? Ans. When the air is completely dry i.e. there is no water vapour present in the air, there will not be any dew point. Q.3 What will be the situation of the surroundings when the room temperature and the dew point become equal? [H.S.] Ans. When the room temperature and the dew point are equal, the air inside the room will be saturated by the water vapour present in it. Therefore, the relative humidity will

Remember the following points: • Saturation or 100% relative humidity can be achieved either by

introducing more water vapour or by lowering the temperature. • The temperature at which the air is saturated by the water vapour in it

is called the dew point. • Evaporation is faster when the vapour pressure is low which means

the amount of water vapour is less in the air and that means the relative humidity is low.


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be 100%. The air can not contain any more water vapour; any more addition of water vapour will result in condensation. Q.4 If the room temperature is increased, what will be its effect on the dew point and the relative humidity? [H.S. ’96; J.E.E. ’99, ‘85] Ans. The dew point is the temperature at which the air gets saturated by the water vapour present in it. Therefore, as long as the amount of water vapour in the air is not changed, the dew point also does not change. Thus the increase of room temperature has no effect on the dew point. The relative humidity at a certain temperature is the ratio of the mass of the water vapour present in the air to the mass of water vapour required to saturate the air at that temperature. As the temperature is increased, more water vapour is now required to saturate the air. However, the amount of water vapour in the air remains the same which makes the ratio to decrease. Thus the relative humidity reduces due to the increase in air temperature. Q.5 If some water is sprinkled in a room, how does it affect (i) the dew point and (ii) the relative humidity? [H.S.] Ans. (i) The air in the room contains a certain amount of water vapour. If now the temperature of the room is reduced, eventually a temperature is reached when the air becomes saturated by the water vapour in it and thus the dew point is reached. Now the amount of water vapour in the air is increased as the water is sprinkled in the room. Thus we will not have to reduce the temperature that much; the saturation will be reached at a temperature above the earlier dew point. Therefore, the dew point will increase. (ii) We know that the relative humidity at a certain temperature is the ratio of the mass of the water vapour present in the air to the mass of water vapour required to saturate the air at that temperature. We assume that the temperature of the air in the room does not change due to the sprinkling of water. This means that the mass of water vapour required to saturate the air remains the same. On the other hand, the air in the room now contains more water vapour at the same temperature. Thus the ratio increase which means the relative humidity increases. Q.6 Two rooms have the same temperature. The relative humidity of air in one room is more than that of the other. Which room will be more comfortable to you? Ans. We control our body temperature in summer by sweating. The evaporation of perspiration from our skin cools our body. But the evaporation depends on the relative humidity of air surrounding us. Higher the relative humidity, the less effective the evaporation is. Therefore, we feel hotter in the room which has higher relative humidity. Q.7 In some day, the dew point inside and outside the room had been same but the relative humidity was different. Explain this. Ans. The dew point is the temperature at which the air gets saturated with the amount of water vapour in it. If the air inside and outside the room contain the same proportion of water vapour, the dew point may be the same at two places. But the relative humidity depends on the air temperature apart from the amount of vapour present in the air. If the


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temperatures inside and outside the room are different, the relative humidity will be different. Q.8 Why does the dry hot air is more comfortable than less hot and humid air? [H.S.] Ans. When the air is hot, our body tries to control its temperature by cooling through evaporation of sweat from our skin. Thus we feel comfort. The evaporation depends on the relative humidity of the air. If the air is dry, the humidity is low which makes the evaporation of sweat easier. Thus our body can cool down even if the air temperature is high. On the other hand, if the relative humidity is high the evaporation can not happen at as efficiently as before. Thus we can not cool down our body by sweating even if the air temperature is not as high as before. Q.9 The critical temperature of CO 2 is 31.4 C0 . Would you call it a gas or vapour if its temperature is 25 C0 ? [J.E.E. ‘97] Ans. If the temperature of a gaseous substance is above its critical temperature, it should be called a gas and if it is below the critical temperature it should be termed as vapour. In this case since CO 2 is at 25 C0 (below the critical point), it should be called vapour. Q.10 Can there be a dew point below 0 C0 ? Ans. If the air can be saturated by water vapour at a temperature below 0 C0 , then that temperature will be a dew point. The saturation vapour pressure decreases monotonically with temperature and which becomes as low as around 4.6 mm Hg at 0 C0 . For the air to be saturated even at a temperature 0 C0 , the vapour pressure has to be equal to the above value. But below 0 C0 , the water vapour directly becomes ice rather than reaching at the saturation point. Thus the dew point is never reached below 0 C0 . Q.11 What do you mean by the dew point of air to be 15 C0 ? [H.S. ‘02] Ans. This means that the air would become saturated at 15 C0 by the amount of water vapour present in it. Q.12 Why does the saturated vapour pressure of a liquid at boiling point become equal to the pressure of the gas above the liquid surface? [H.S. ’02; J.E.E. ‘98] Ans. A liquid boils when its saturated vapour pressure becomes equal to the external pressure on the liquid. During boiling bubbles of vapour form throughout the liquid and they rise above the liquid surface and escape. If the external pressure is higher than the saturated vapour pressure, these bubbles are prevented from forming.

Questionnaire

Very Short Questions: Mark: 1


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(Answer in one or two words) 1. What is it called when the temperature of the atmosphere keeps reducing and that the air gets saturated by the water vapour in it at a certain temperature? [dew point] 2. How much is the relative humidity of air at the dew point? [100%] 3. At a time when the air temperature is 20 C0 , the dew point is 15 C0 . What will be the dew point when the air temperature rises to 25 C0 ? [15 C0 ]

(Fill in the Blanks) 1. If the air temperature is 10 C0 and the dew point is also10 C0 , the relative humidity is --------%. [100] 2. In summer, Delhi is more comfortable than Kolkata because the relative humidity of Delhi is ---------. [less] 3. If the air temperature is reduced, the relative humidity -------------. [goes up] 4. If the air temperature increases, the relative humidity -------------. [goes down]

(Multiple Choice type) 1. A container contains saturated water vapour. If now the temperature is increased keeping the volume unchanged, (a) the saturated vapour will be unsaturated (b) the vapour will reach dew point (c) some water vapour will be condensed (d) nothing will happen. [(a)] 2. A container contains saturated water vapour. If now the volume is decreased keeping the temperature unchanged, (a) the saturated vapour will be unsaturated (b) the vapour will reach dew point (c) some water vapour will be condensed (d) nothing will happen. [(c)] 3. The science of measuring moisture in the atmosphere is called (a) thermometry (b) calorimetry (c) hygrometry (d) thermodynamics [(c)] 4. When the temperature is increased, the saturated vapour pressure (a) increases linearly with absolute temperature (b) increases, but not linearly with absolute temperature (c) decreases with absolute temperature (d) decreases, but not linearly with absolute temperature. [(b)] 5. The pressure due to saturated vapour at 100 C0 is (a) zero (b) one atmospheric pressure (c) less than one atmospheric pressure (d) greater than one atmospheric pressure. [(b)] 6. The dew point of air is 10 C0 at a time. The moisture in air will start becoming unsaturated at a temperature of


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(a) 10 C0 (b) less than 10 C0 (c) more than 10 C0 (d) 0 C0 [(a)] 7. If the air temperature is 10 C0 and the dew point is also 10 C0 , then the relative humidity is (a) 0% (b) 200% (c) 100% (d) not sure [(c)] 8. If the air is dry, then (a) there is no dew point (b) the dew point is 0 C0 (c) the dew point is 100 C0 (d) the dew point becomes equal to the temperature of the air. [(a)] 9. Two rooms are at different temperatures but the dew points in them are the same. What will be the relation between the relative humidity in the two rooms? (a) They are equal (b) The relative humidity in the warmer room is greater than the other (c) The relative humidity in the warmer room is lower than the other (d) The relative humidity will depend on the size of the room. [(c)] 10. It is not possible to get the wet cloth dried when (a) the relative humidity in the air is 100% (b) the relative humidity in the air is 0% (c) the air pressure is low (d) the dew point is high. [(a)] Short Questions: Marks: 2 1. What is vapour pressure? [H.S. ‘06] 2. What do you mean by saturated vapour pressure? [H.S. ‘06] 3. What is unsaturated vapour pressure? [H.S. ‘06] 4. Compare the saturated and the unsaturated vapour. [H.S. ‘03] 5. What is the effect of temperature on the saturated vapour pressure of water? 6. What is the difference between gas and vapour? 7. What is dew point? [H.S. ’94, ‘92] 8. Give the definitions of absolute humidity and relative humidity. [H.S. ‘92]

atmosphere and water vapour

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