Tuesday, December 31, 2019

H C Verma solutions, Kinetic Theory of Gasses, EXERCISES Q51 to Q62, Chapter-24, Concepts of Physics, Part-II

Kinetic Theory of Gases

EXERCISES, Q51 to Q62

  51. Using figure (24.6) of the text, find the boiling point of methyl alcohol at 1 atm (760 mm of mercury) and at 0.5 atm.   



Answer:  The boiling point of a liquid is the temperature at which the SVP equals the external pressure. We draw lines parallel to 760 mm (1 atm) and 380 mm (0.5 atm) of mercury in the graph to meet the Methyl alcohol curve. From these points draw perpendiculars to the temperature axis.
The boiling point of Methyl alcohol at 1 atm and 0.5 atm
   From the figure, we get the boiling point of methyl alcohol 65°C at 1 atm and 48°C at 0.5 atm.   


 



  52. The human body has an average temperature of 98°F. Assume that the vapor pressure of the blood in the veins behaves like that of pure water. Find the minimum atmospheric pressure which is necessary to prevent the blood from boiling. Use figure (24.6) of the text for the vapor pressures.  



Answer:  98°F = (98-32)*5/9 

=66*5/9 =330/9 =36.7°C

     Given that the vapor pressure of blood behaves like water, hence we find the pressure corresponding to 36.7°C for the water curve. It is 50 mm of mercury. Hence it is the required answer.

 



  53. A glass contains some water at room temperature 20°C. Refrigerated water is added to it slowly. When the temperature of the glass reaches 10°C, small droplets condense on the outer surface. Calculate the relative humidity in the room. The boiling point of water at a pressure of 17.5 mm of mercury is 20°C and at 8.9 mm of mercury, it is 10°C.   



Answer:  It is clear that the dew point is at 10°C. Since at 8.9 mm of mercury, the boiling point of water is 10°C, hence the SVP of water at 10°C is 8.9 mm of mercury. Similarly, the SVP at 20°C is 17.5 mm of mercury.

Relative humidity 

=SVP at Dew point/SVP at air temp.

=8.9/17.5

=0.51

=51%.

     


 


  54. 50 m³ of saturated vapor is cooled down from 30°C to 20°C. Find the mass of the water condensed. The absolute humidity of saturated water vapor is 30 g/m³ at 30°C and 16 g/m³ at 20°C.  



Answer:  At both temperatures, the vapor is saturated. Amount of water in the given volume of vapor at 30°C = 50 m³*30 g/m³ =1500 g.

Amount of water in the given volume of vapor at 20°C =50 m³*16 g/m³ =800 g.

Hence the mass of water condensed =1500-800 =700 g.    




 


  55. A barometer correctly reads the atmospheric pressure as 76 cm of mercury. Water droplets are slowly introduced into the barometer tube by a dropper. The height of the mercury column first decreases and then becomes constant. If the saturation vapor pressure at the atmospheric temperature is 0.80 cm of mercury, find the height of the mercury column when it reaches its minimum value. 



Answer:  When the mercury column becomes constant, the vapor above is saturated. The saturated vapor pressure at atmospheric pressure is given as 0.80 cm of mercury. So the vapor above the mercury column applies a pressure of 0.80 cm of mercury and the height of the column decreases by this height. Thus the final height of the mercury column =76 cm - 0.80 cm =75.2 cm.  






 

  56. 50 cc of oxygen is collected in an inverted gas jar over water. The atmospheric pressure is 99.4 kPa and the room temperature is 27°C. The water level in the jar is the same as the level outside. The saturation vapor pressure at 27°C is 3.4 kPa. Calculate the number of moles of oxygen collected in the jar.  



Answer:  Atmospheric pressure =99.4 kPa. Since the water level in the jar is the same as the level outside, the combined pressure of oxygen and the water vapor is equal to the atmospheric pressure. SVP at 27°C = 3.4 kPa, hence the pressure of oxygen gas, p =99.4-3.4 =96 kPa =9.6x10⁴ N/m².

The volume of the oxygen gas, V =50 cc =50x10⁻⁶ m³ =5x10⁻⁵ m³. R=8.3 J/mol-K, T =273+27 K =300 K.

Hence the number of moles in the oxygen gas, n =pV/RT 
→n =9.6x10⁴*5x10⁻⁵/(8.3*300)
→n =1.93x10⁻³. 

   




 

  57. A faulty barometer contains a certain amount of air and saturated water vapor. It reads 74.0 cm when the atmospheric pressure is 76.0 cm of mercury and reads 72.10 cm when the atmospheric pressure is 74 cm of mercury. Saturation vapor pressure at the air temperature = 1.0 cm of mercury. Find the length of the barometer tube above the mercury level in the reservoir.   



Answer:  Let the length of the barometer tube above the mercury level in the reservoir = L. Due to saturated vapor and air the mercury height = 74 cm. The height of the tube above the mercury column = L-74 cm. Let the air pressure in the tube = P, then P+SVP+74 cm =Atmospheric pressure in cm of Hg.

→P+1+74=76 

→P = 76-75 =1 cm.

  

When the atmospheric pressure is 74 cm, it shows 72.10 cm. If air pressure =P', So, P'+1+72.10 =74

→P' = 74-73.10 =0.90 cm

Height of tube above mercury column =L-72.1 cm

Let the area of cross-section of the tube =A cm².

Since pV =p'V'

1*(L-74)A = 0.90*(L-72.1)A

→L-74 =0.90L -0.90*72.1

→0.10L =74-64.89 =9.11

→L = 9.11/0.10 = 91.10 cm.

     




 

  58. On a winter day, the outside temperature is 0°C and the relative humidity 40%. The air from outside comes into a room and is heated to 20°C. What is the relative humidity in the room? The saturation vapor pressure at 0°C is 4.6 mm of mercury and at 20°C, it is 18 mm of mercury.   



Answer:  For the outside air, 

Relative Humidity =VP/SVP

→40% =VP/(4.6 mm of mercury)

→VP = 0.40*4.6 mm of mercury

       =1.84 mm of mercury.

Temperature, T =273 K.

Now in the room at 20°C, SVP =18 mm of mercury.

Temperature, T' =273+20 =293 K. Since the air comes into the room and heated to 20°C, it must expand because the air pressure is the same. Consider a volume V outside which changes to volume V' inside. At constant pressure,  
V/T = V'/T'
→V/273 = V'/293
→V'/V = 293/273 =1.07.
Hence 1 m³ of air expands to 1.07 m³ in the room. The amount of vapor present in 1 m³ outside is now present in 1.07 m³ of air inside. Thus in 1 m³ of air inside the amount of vapor is 1/1.07 of the amount in 1 m³ outside. Since the vapor pressure is proportional to the amount of vapor present in the same volume of the gas, the VP in-room =(1/1.07) of 1.84 mm of mercury. 
=1.72 mm of mercury. 
Hence the relative humidity in the room =VP/SVP
=1.72/18 =0.095 =9.5%.

    


  

 

  59. The temperature and humidity of air are 27°C and 50% on a particular day. Calculate the amount of vapor that should be added to 1 cubic meter of air to saturate it. The saturation vapor pressure at 27°C = 3600 Pa.    



Answer:  RH =VP/SVP

→0.50 = VP/3600

→VP = 1800 Pa

Since the saturation vapor pressure is 3600 Pa, the amount of water vapor (m) should be added to increase the vapor pressure further 1800 Pa. So,

p =1800 Pa =1800 N/m²

V =1 m³

T =273+27 =300 K

R = 8.3 J/mol-K

n = m/M =m/18, {For water, M=18}

From the equation

pV =(m/M)RT

→m = pVM/RT

      =1800*1*18/(8.3*300)

      =13 g.   





 

  60. The temperature and humidity in a room are 300K and 20% respectively. The volume of the room is 50 m³. The saturation vapor pressure at 300 K is 3.3 kPa. Calculate the mass of the water vapor present in the room.   



Answer:  Humidity =VP/SVP 

→0.20 =VP/3.3

→VP =0.20*3.3 =0.66 kPa =660 N/m² =p

Volume, V =50 m³

T =300 K, m = mass of vapor.

pV =(m/M)RT

→m = pVM/RT

      =660*50*18/(8.3*300)

      = 238 g.   




 


  61. The temperature and relative humidity are 300K and 20% in a room of volume 50 m³. The floor is washed with water, 500 g of water sticking on the floor. Assuming no communication with the surrounding, find the relative humidity when the floor dries. The changes in temperature and pressure may be neglected. Saturation vapor pressure at 300 K = 3.3 kPa.  



Answer:  RH = 20% =0.20, 

SVP = 3.3 kPa.

RH = VP/SVP

→0.20 = VP/3.3

→VP = 0.66 kPa =660 Pa.

When 500 g of water is absorbed in the air, it will add to the vapor pressure in the air. Let VP corresponding to this amount =p'. Amount of water, m =500 g. The molecular weight of water, M =18 g. V = 50 m³. T =300 K. From,

pV =(m/M)RT

→p'*50 = (500/18)*8.3*300

→p' =1383 Pa.

So the vapor pressure after the floor dries =Previous VP +p'

=660+1383 Pa

=2043 Pa. 

Hence the relative humidity =VP/SVP

=2043/3300

=0.62

=62%.      





 

  62. A bucket full of water is placed in a room at 15°C with initial relative humidity 40%. The volume of the room is 50 m³. (a) How much water will evaporate? (b) If the room temperature is increased by 5°C how much more water will evaporate? The saturation vapor pressure of water at 15°C and 20°C are 1.6 kPa and 2.4 kPa respectively.    


Answer:  (a) RH at 15°C =40% =0.40 

SVP =1.6 kPa, VP = p =?

RH =VP/SVP 

→0.40 =p/1.6

→p =0.40*1.6 kPa =0.64 kPa =640 Pa.

The vapor will evaporate till the air is saturated i.e. vapor pressure is = SVP =1.6 kPa =1600 Pa.

So extra vapor pressure, p' =1600 -640 Pa =960 Pa.

The amount of water evaporated from the bucket = m, the molecular weight of water, M =18. T =273+15 K =288 K.  V =50 m³. From,

p'V =(m/M)RT

→960*50 =(m/18)*8.3*288

→m =960*50*18/(8.3*288)

→m =361 g.


(b) When the temperature is increased by 5°C, the new temperature is T =273+20 K =293 K. Now the increased vapor pressure, p =2.4-1.6 kPa =0.8 kPa =800 Pa. V =50 m³. M =18 g, from

pV=(m/M)RT

→800*50 =(m/18)*8.3*293

→m =800*50*18/(8.3*293)

→m =296 g.

So, 296 g more water will evaporate in this case.         

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Links to the Chapters



CHAPTER- 20 - Dispersion and Spectra


CHAPTER- 19 - Optical Instruments

CHAPTER- 18 - Geometrical Optics



CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








CHAPTER- 11 - Gravitation




CHAPTER- 10 - Rotational Mechanics




CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

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CHAPTER- 4 - The Forces

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CHAPTER- 3 - Kinematics - Rest and Motion

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