Friday, December 20, 2019

H C Verma solutions, Kinetic Theory of Gasses, EXERCISES Q41 to Q50, Chapter-24, Concepts of Physics, Part-II

Kinetic Theory of Gases

EXERCISES, Q41 to Q50

  41. Figure (24-E6) shows a cylindrical tube of cross-sectional area A fitted with two frictionless pistons. The pistons are connected to each other by a metallic wire. Initially, the temperature of the gas is T₀ and its pressure is p₀ which equals the atmospheric pressure. (a) What is the tension in the wire? (b) What will be the tension if the temperature is increased to 2T₀?
The figure for Q-41
 


Answer:  (a) Since the pressure on both sides of the pistons is the same, there is no net pressure on the pistons hence no net force on the pistons. Thus the tension in the wire is zero.


(b) Since the volume is constant,

p/T = p'/T'

Here p = p₀, T = T₀ and T' = 2T₀, so

p₀/T₀ = p'/2T₀ 

→p' = 2p₀

So the pressure inside the tube is 2p₀ and ouside it is p₀. So net pressure on a piston = 2p₀ - p₀ =p₀. Area of a piston = A. The force on each of the piston = tension in the wire = p₀A. 


 

   42. Figure (24-E7) shows a large closed cylindrical tank containing water. Initially, the air trapped above the water surface has a height h₀ and pressure 2p₀ where p₀ is the atmospheric pressure. There is a hole in the wall of the tank at a depth h₁ below the top from which water comes out. A long vertical tube is connected to as shown. (a) Find the height h₂ of the water in the long tube above the top initially. (b) Find the speed with which water comes out of the hole. (c) Find the height of the water in the long tube above the top when the water stops coming out of the hole. 
The figure for Q-42
 

Answer:  (a) Let us consider the pressure at the water level in the tank. It is given as 2p₀ but through the vertical tube water height, it comes to 

p₀+(h₀+h2)ρg. Equating these two we get,

p₀+(h₀+h₂)ρg = 2p₀

→(h₀+h₂)ρg = p₀ 

→h₀+h₂ =p₀/ρg

→h₂ = p₀/ρg - h₀


(b) Consider the exit level of the water as the reference line. If the velocity of outgoing water is v and the velocity inside the tank at this level =v' then according to the Bernoulli's theorem sum of three heads inside and outside the exit should be constant,
p₀ +ρgh +½ρv² = p + ρgh' +½ρv'²   

      (outside ↑)           (inside ↑)

Since the reference line is at the exit level, h = h' = 0. Velocity inside the tank at this level is negligible, so v' = 0. The pressure at this level inside,
 p = 2p₀ + ρg(h₁-h₀), putting these values we get,
p₀+½ρv² = 2p₀ +ρg(h₁-h₀)
→½ρv² = p₀ + ρg(h₁-h₀)
→ρv² = 2{p₀+ρg(h₁-h₀)}
→v² = 2{p₀+ρg(h₁-h₀)}/ρ
→v = √[2{p₀+ρg(h₁-h₀)}/ρ]       

(c) Let the level of water in the tube above the top of the tank = x when water stops to come out. Then pressure at the exit level = (x+h₁)ρg. But this should be zero because water stops to come out, hence
(x+h₁)ρg = 0
→x+h₁ = 0
→x = -h₁  



    43. An ideal gas is kept in a long cylindrical vessel fitted with a frictionless piston of a cross-sectional area 10 cm² and weight 1 kg (figure 24-E8). The vessel itself is kept in a big chamber containing air at atmospheric pressure 100 kPa. The length of the gas column is 20 cm. If the chamber is now completely evacuated by an exhaust pump, what will be the length of the gas column? Assume the temperature remains constant throughout the process. 
The figure for Q-43


Answer:  Pressure, p on the gas column, will be equal to atmospheric pressure plus the pressure due to the weight of the piston. So,

p = p₀+mg/A

m = 1 kg, A =10 cm² =1x10⁻³ m², p₀ =100 kPa = 1x10⁵ N/m².

p = 1x10⁵ + 1x10/(1x10⁻³) Pa
   = 1x10⁵ + 1x10⁴ Pa
   = 1.1x10⁵ Pa
Volume V = (20/100)A =0.20*1x10⁻³ =2x10⁻⁴ m³ 
After the chamber is completely evacuated the pressure on the gas column is only due to the weight of the piston.
p' = mg/A =1x10/10⁻³ =1x10⁴ Pa
If the length of the gas column now = X cm, then volume V' = (X/100)*(10⁻³) =X*10⁻⁵ m³
Since temperature is constant,
pV = p'V'
→(1.1x10⁵)*(2x10⁻⁴) =(1x10⁴)*(X*10⁻⁵)
→X = 2.2x10/10⁻¹ =2.2x100 cm
→X = 2.2 m.   

 


 

    44. An ideal gas is kept in a long cylindrical vessel fitted with a frictionless piston of a cross-sectional area 10 cm² and weight 1 kg. The length of the gas column in the vessel is 20 cm. The atmospheric pressure is 100 kPa. The vessel is now taken into a spaceship revolving around the earth as a satellite. The air pressure in the spaceship is maintained at 100 kPa. Find the length of the gas column in the cylinder. 


Answer:  In the laboratory at the earth,

p =1.1x10⁵ Pa, Volume V = 2x10⁻⁴ m³.

In the spaceship the value of g = 0, hence the piston is weightless and it has no pressure on the gas column. Hence the pressure now, p' = 1x10⁵ Pa. If the length of the gas column = x cm, then volume V' = (X/100)*(10⁻³) m³ =10⁻⁵*X m³. Since the temperature is constant,

pV = p'V' 

→1.1x10⁵*2x10⁻⁴ =1x10⁵*10⁻⁵*X

→X = 2.2*10/1 =22 cm.

    

  


 

    45. Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at 0°C at a pressure of 76 cm of mercury. One of the bulbs is then placed in melting ice and the other is placed in a water bath maintained at 62°C. What is the new value of the pressure inside the bulb? The volume of the connecting tube is negligible.  


Answer:  Let the volume of each bulb = V. Initial pressure = p and temperature, T = 273 K. If each bulb has n moles of gas initially then the ideal gas equation for each of the bulbs is,

pV =nRT

→n = pV/RT.

In the second stage let the number of moles in the bulb in melting ice = n' and in the other = n". The pressure in each of the bulbs = p'. 

So for the first bulb in melting ice

p'V =n'RT, →n' = p'V/RT

and for the second bulb at T' = 62°C i.e. 335 K,

p'V = n"RT', →n" = p'V/RT'

Since the total number of moles in both bulbs do not change,

n'+n" = 2n 

→p'V/RT + p'V/RT' = 2pV/RT

→p'(1/T + 1/T') = 2p/T

{Here T = 273 K, T' =335 K and p =76ρg/100 Pa.}

→p'(1/273 + 1/335) = 2*(76ρg/100)/273

→p'(335+273)/(273*335)=152ρg/100*273

→p' =152ρg*335/(100*608)

      =0.84ρg Pa

      =0.84 m of mercury column

      =84 cm of mercury column.  



 

    46. The weather report reads, "Temperature 20°C: Relative humidity 100%" What is the dew point?  


Answer:  100% Relative humidity means that the air is fully saturated with the water vapor. It can not hold more water vapor at this temperature of 20°C. If the temperature is reduced further some of the water vapor will condense. Hence the dew point here is 20°C. 



 

    47. The condition of air in a closed room is described as follows. Temperature = 25°C, relative humidity = 60°C, pressure = 104 kPa. If all the water vapor is removed from the room without changing the temperature, what will be the new pressure? The saturation vapor pressure at 25°C = 3.2 kPa.   


Answer:  

Relative humidity =Vapor pressure of air/Saturation                     vapor pressure at the same temperature 

Here RH =60% i.e. 0.6

Saturation vapor pressure at 25°C =3.2 kPa. So,
VP/SVP = 0.6
→VP =0.6*SVP =0.6*3.2 =1.92 kPa ≈2 kPa
When all the water vapor is removed the vapor pressure in the air becomes zero. Hence the pressure of air in the room = 104 kpa - 2 kPa = 102 kPa. 
  



 

    48. The temperature and the dew point in an open room are 20°C and 10°C. If the room temperature drops to 15°C, what will be the new dew point? 


Answer:  If the dew point is 10°C, it means the air has saturation vapor pressure at this temperature. Even if the room temperature drops, the saturation vapor pressure for the given air will not change and the dew point will remain 10°C. 




 

    49. Pure water vapor is trapped in a vessel of volume 10 cm³. The relative humidity is 40%. The vapor is compressed slowly and isothermally. Find the volume of the vapor at which it will start condensing. 


Answer:  When vapor pressure = Saturation vapor pressure, the vapor starts condensing. Given that relative humidity = 40%.

But, RH = VP/SVP

→0.40 = VP/SVP

→VP = 0.40*SVP = p (say) 

V = 10 cm³. 

Let the volume of vapor at which it starts condensing = V'. Pressure here = SVP = p' (say). Since the process is isothermal,

p.V = p'V'

→(0.40*SVP)*10 = (SVP)*V' 

→V' = 0.40*10 cm³ = 4.0 cm³.    

  


 

    50. A barometer tube is 80 cm long (above the mercury reservoir). It reads 76 cm on a particular day. A small amount of water is introduced in the tube and the reading drops to 75.4 cm. Find the relative humidity in the space above the mercury column if the saturation vapor pressure at the room temperature is 1.0 cm. 


Answer:  The atmospheric pressure = 76 cm of mercury column.    

Diagram for Q-50

When a small amount of water is introduced in the tube it turns in to water vapor above the mercury column due to low pressure. This water vapor exerts pressure above the mercury level in the tube and pushes it down. Since the mercury column is pushed down by 76-75.4 = 0.6 cm by the vapor, the present vapor pressure, VP = 0.6 cm of mercury. 

Given that SVP = 1.0 cm, hence the relative humidity

RH = VP/SVP = 0.6/1.0 = 0.60 = 60%. 

 

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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




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CHAPTER- 8 - Work and Energy

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