Tuesday, December 10, 2019

H C Verma solutions, Kinetic Theory of Gasses, EXERCISES Q21 to Q30, Chapter-24, Concepts of Physics, Part-II

Kinetic Theory of Gases

EXERCISES, Q21 to Q30

21. Estimate the number of collisions per second suffered by a molecule in a sample of hydrogen at STP. The mean free path (average distance covered by a molecule between successive collisions) =1.38x10⁻⁵ cm.  


Answer:  We should know the average speed of the molecules. Here at STP, T=273 K, M₀ =2 g/mol =0.002 kg/mol. R =8.3 J/mol-K. Thus the average speed, v = √(8RT)/(πM₀)

→ v =√(8*8.3*273)/(π*0.002)

→ v = 1698 m/s 

Mean free path = 1.38x10⁻⁵ cm 

=1.38x10⁻⁷ m

So the average time taken in one collision =(1.38x10⁻⁷)/1698 s

=8.13x10⁻¹¹ s

Thus collisions per second =1/(8.13x10⁻¹¹)

=1.23x10¹⁰. 



 

22. Hydrogen gas is contained in a closed vessel at 1 atm (100kPa) and 300 K. (a) Calculate the mean speed of the molecules. (b) Suppose the molecules strike the wall with this speed making an average angle of 45° with it. How many molecules strike each square meter of the wall per second?   


Answer:  (a) T = 300 K, M₀ =2 g/mol =0.002 kg/mol. R = 8.3 J/mol-K. Mean speed, v =√(8RT)/(πM₀)

=√(8*8.3*300)/(π*0.002)

=1780 m/s


(b) The momentum of one molecule =mv, where m is the mass of one molecule. Since the molecules collide at an angle of 45°, the change of momentum perpendicular to the wall =2mv*cos45°

=√2mv.

If there are n molecules striking per second per m² then the change of momentum of molecules striking per second =√2mnv = Force/m² =pressure.

Here m = (0.002/6.02x10²³) =3.32x10⁻²⁷ kg

So √2mnv =10⁵ N/m²

→n = 10⁵/(√2*3.32x10⁻²⁷*1780)

=1.2x10²⁸. 


 

  

 23. Air is pumped into an automobile tire's tube up to a pressure of 200 kPa in the morning when the air temperature is 20°C. During the day the temperature rises to 40°C and the tube expands by 2%. Calculate the pressure of the air in the tube at this temperature.   


Answer:  Let the volume of the tire in the morning =V. Temperature = 20°C =20+273 K =293 K. p =200 kPa .

     For the day time T' =273+40 K =313 K. V' =1.02V, let the pressure =p'.

Since pV/T =p'V'/T'

p' =pVT'/TV' 

   =200*V*313/(293*1.02V)
   =209 kPa.

 

 


 24. Oxygen is filled in a closed metal jar of volume 1.0x10⁻³ m³ at a pressure of 1.5x10⁵ Pa and temperature 400 K. The jar has a small leak in it. The atmospheric temperature is 300 K. Find the mass of the gas that leaks out by the time pressure and the temperature inside the jar equalize with the surrounding. 


Answer:  p =1.5x10⁵ Pa, V = 1.0x10⁻³ m³, T = 400 K, n = initial moles of molecules. At the final stage, p' = 1x10⁵ Pa, T' = 300 K, n' = final moles of molecules. V' = V. Now,

pV =nRT and p'V' =n'RT'

→n =pV/RT and n' =p'V'/RT' =p'V/RT'

The moles of molecules escaped

=n - n'

=pV/RT - p'V/RT'

=(p/T - p'/T')V/R

=(1.5x10⁵/400 - 1x10⁵/300)*1x10⁻³/8.3

=(375-333.33)*1x10⁻³/8.3

=5.02x10⁻³ moles. 

Hence the mass of oxygen gas escaped = 5.02x10⁻³*32 g

=0.16 g.  


 


 25. An air bubble of radius 2.0  mm is formed at the bottom of a 3.3 m deep river. Calculate the radius of the bubble as it comes to the surface. Atmospheric pressure = 1.0x10⁵ Pa and density of water = 1000 kg/m³.   


Answer:  Pressure due to 3.3 m deep river water = ρgh

= 1000*9.8*3.3 Pa

= 32340 Pa

Total pressure at this depth =Atmospheric pressure + Water pressure

=1.0x10⁵+32340 Pa

=1.3234x10⁵ Pa 

The radius of the air bubble, r = 2 mm
Volume, V = 4πr³/3
Near the surface,
p' = 1x10⁵ Pa
V' =4πr'³/3, where r' is the radius of the bubble near the surface. The temperature is constant. Hence from Boyle's law,
p'V' = pV
→p'(4πr'³/3) = p(4πr³/3)
→r'³ =pr³/p'
→r' =r(p/p')¹/³ =2*(1.3234)¹/³
→r' =2.2 mm.



 

 26. Air is pumped into the tubes of a cycle rickshaw at a pressure of 2 atm. The volume of each tube at this pressure is 0.002 m³. One of the tubes gets punctured and the volume of the tube reduces to 0.0005m³. How many moles of the air has leaked out? Assume that the temperature remains constant at 300 K and that the air behaves as an ideal gas.    


Answer:  Initial pressure, p = 2 atm, volume, V = 0.002 m³. Number of moles = n, T = 300 K.

Final pressure, p' = 1 atm (atmospheric pressure), V' = 0.0005 m³, Number of moles = n'. T' = 300 K.

So the number of moles of air leaked out = n - n'

=pV/RT - p'V'/RT

=(pV - p'V')/RT

=(2x10⁵*0.002 - 1x10⁵*0.0005)/(8.3*300)

=(400-50)/2490

=350/2490

=0.14 moles. 


 


 27. 0.040 g of He is kept in a closed container initially at 100.0°C. The container is now heated. Neglecting the expansion of the container, calculate the temperature at which the internal energy is increased by 12 J.   


Answer:  Since the internal energy of a gas is the total kinetic energy of all the molecules and it is given as,

U = (3/2)nRT

n = 0.04/4 =0.01 mole

R =8.3 J/mol-K, T =273+100 =373 K.

So, U = (3/2)*0.01*8.3*373

         =46.44 J

For the required temperature T, U'=46.44+12 =58.44 J

So, 58.44 = 1.5*0.01*8.3*T

→T = 58.44/(0.1245) = 469 K =(469-273)°C = 196°C. 


 



 28. During an experiment, an ideal gas is found to obey an additional law pV² = constant. The gas is initially at a temperature T and volume V. Find the temperature when it expands to a volume 2V.   


Answer:  Since the gas is ideal hence,

pV =nRT

→p =nRT/V

Similarly for the final stage, temperature = T', pressure p' and volume V',

p' = nRT'/V'.
Now it obeys additional law
pV² = p'V'²
→(nRT/V)V² = (nRT'/V')V'²
→TV = T'V', But given that V' = 2V.
→TV = T'*2V
→T' = T/2.

 


 

 29. A vessel contains 1.60 g of oxygen and 2.80 g of nitrogen. The temperature is maintained at 300 K and the volume of the vessel is 0.166 m³. Find the pressure of the mixture.   


Answer:  V = 0.166 m³, T = 300 K.

1.60 g of oxygen = 1.60/32 moles =0.05 moles.

2.80 g of nitrogen = 2.80/28 moles =0.10 moles.

Hence total number of moles of ideal gases in the vessel, n = 0.05+0.10 =0.15

From the ideal gas equation,

pV = nRT

→p = nRT/V

→p = 0.15*8.3*300/0.166 

→p = 2250 N/m². 


  

 

30. A vertical cylinder of height 100 cm contains air at a constant temperature. The top is closed by a frictionless light piston. The atmospheric pressure is equal to 75 cm of mercury. Mercury is slowly poured over the piston. Find the maximum height of the mercury column that can be put on the piston.   


Answer:  Let the maximum height of the mercury column on the piston = h.

Diagram for Q-30

The initial pressure of air in the container, p = 75 cm of mercury = 0.75ρg N/m², where ρ = density of mercury in kg/m³.

Height of the cylinder = 100 cm = 1 m. If the area of the base of the cylinder = A m², then volume, V = A*1 =A m³.

Finally the volume of air in the cylinder, V' = A(1-h) m³. 

Pressure, p' = (0.75+h)ρg.

Since the temperature and the number of moles remain constant,

pV = p'V'

→0.75ρg*A = (0.75+h)ρg*A((1-h)

→0.75 = (0.75+h)(1-h)

→75 = (75+100h)(1-h)

→75+100h-75h-100h² =75

→100h²-25h=0

→h(100h-25)=0

Since h is not zero,

100h-25 =0

→h = 25/100 = 0.25 m = 25 cm.

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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

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

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CHAPTER- 6 - Friction

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For more practice on problems on friction solve these- "New Questions on Friction".

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CHAPTER- 5 - Newton's Laws of Motion


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Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

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CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


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Click here for "The Forces" - Exercises


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CHAPTER- 3 - Kinematics - Rest and Motion

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Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


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CHAPTER- 2 - "Physics and Mathematics"

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