Sunday, December 1, 2019

H C Verma solutions, Kinetic Theory of Gasses, EXERCISES Q11 to Q20, Chapter-24, Concepts of Physics, Part-II

Kinetic Theory of Gases

EXERCISES, Q11 to Q20

11. Figure (24-E1) shows a cylindrical tube with adiabatic walls and filled with a diathermic separator. The separator can be slid in the tube by an external mechanism. An ideal gas is injected in the two sides at equal pressures and equal temperatures. The separator remains in equilibrium in the middle. It is now slid to a position where it divides the tube in the ratio of 1:3. Find the ratio of the pressures in the two parts of the vessel. 
The figure for Q-11
 

Answer:  Since n = pV/RT. Here when the separator is in the middle V is the same for both sides and p and T are also the same (given). Hence n is also the same for both sides.

         The total volume of both sides = 2V. When the separator divides the tube in 1:3, the volume of one side V' = 2V/4 and of the other side V" = 6V/4. Since the separator is diathermic final temperature will remain T.

Now for the one side, p' =nRT/V', and for the other side p" =nRT/V". Hence the ratio

p'/p" = V"/V' =(6V/4)/(2V/4) = 3:1     



      


12. Find the rms speed of hydrogen molecules in a sample of hydrogen gas at 300 K. Find the temperature at which rms speed is double the speed calculated in the previous part. 




Answer:  The rms speed is given as

v = √(3RT/M₀)
Here for hydrogen, M₀ = 2 g/mol =0.002 kg/mol, T = 300K and R = 8.3 J/mol-k, so
v = √(3*8.3*300/0.002)
   = 1933 m/s

Since the square of the rms speed is proportional to the absolute temperature,

v²/v'² = T/T'
Here, v' = 2v, T = 300 K, T' = ?
so, v²/4v² = 300/T'
→T' = 300*4 = 1200 K

 rms speed, v = √(3pV/M) 





 13. A sample of 0.177 g of an ideal gas occupies 1000 cm³ at STP. Calculate the rms speed of the gas molecules. 


Answer:  Given that M = 0.177 g =1.77x10⁻⁴ kg, V = 1000 cm³ =1x10⁻³ m³, T = 273 K, p = 1 atm =1x10⁵ N/m². Hence the rms speed 

v = √(3pV/M) 

=√(3*1x10⁵*1x10⁻³/1.77x10⁻⁴) 

=√(1.69x10⁶)

=1.30x10³ m/s

=1300 m/s

 



14. The average translational kinetic energy of air molecules is 0.040 eV (1 eV = 1.6x10⁻¹⁹ J). Calculate the temperature of the air. Boltzmann constant k = 1.38x10⁻²³ J/K.  


Answer:  Since the average translational kinetic energy is given as

K =3kT/2 

Here, K = 0.04 eV =0.04*1.6x10⁻¹⁹ J. T =?

From this, T = 2K/3k

= 2*0.04*1.6x10⁻¹⁹/(3*1.38x10⁻²³)

= 309 K 

 


  

 

15. Consider a sample of oxygen at 300 K. Find the average time taken by a molecule to travel a distance equal to the diameter of the earth. 


Answer:  Let the average speed of oxygen molecules = v. Then

v =√(8RT/πM₀)

Here T = 300 K, for oxygen M₀ =32 g/mol =0.032 kg/mol and R=8.3 J/mol-K. Hence

v = √(8*8.3*300/0.032π) = 445 m/s

The earth's diameter is approximately 12800 km =1.28x10⁷ m. So average time taken by an oxygen molecule to travel a distance equal to earth's diameter = 1.28x10⁷/445 =28764 s =28764/3600 hr =8.0 hr 

  

 

16. Find the average magnitude of the linear momentum of the helium molecule in a sample of helium gas at 0°C. Mass of a helium molecule = 6.64x10⁻²⁷ kg and Boltzmann constant = 1.38x10⁻²³ J/K. 


Answer:  Given m = 6.64x10⁻²⁷ kg, T =273 K, k =1.38x10⁻²³ J/K. The average speed is given as

v = √(8kT/πm) 

Hence the average momentum 

mv =√(8kTm/π)

=√(8*1.38x10⁻²³*273*6.64x10⁻²⁷/π)

=80x10⁻²⁵ kg-m/s

=8.0x10²⁴ kg-m/s   



  

 

17. The mean speed of the molecules of a hydrogen sample equals the mean speed of the molecules of a helium sample. Calculate the ratio of the temperature of the hydrogen sample to the temperature of the helium sample. 


Answer:  Mean speed or average speed  of hydrogen molecules is 

v =√(8RT/πM₀) 

and for the helium molecules

v' =√(8RT'/πM₀')

Hence v/v' =√(TM₀'/T'M₀)

But here, v = v', M₀' =4 g/mol and M₀ =2 g/mol. Hence

1 = √(4T/2T') =√(2T/T')

→2T/T' = 1

→T/T' =1/2

→T:T' = 1:2  



  

18. At what temperature the mean speed of the molecules of hydrogen gas equals the escape speed from the earth?  


Answer:  Escape speed from the earth =√2gr

The mean speed of the molecules of hydrogen gas =√(8RT/πM₀).

M₀ =0.002 kg/mol
r =6.4x10⁶ m
g =9.8 m/s²
R = 8.3 J/mol-K

for the given condition

8RT/πM₀ = 2gr

→T = πgrM₀/4R

=3.14*9.8*6.4x10⁶*0.002/(4*8.3) 

≈ 11800 K


  

 

19. Find the ratio of the mean speed of hydrogen molecules to the mean speed of nitrogen molecules in a sample containing a mixture of the two gases. 


Answer:  For hydrogen M₀ = 2 g/mol, For nitrogen M₀' =28 g/mol. Since both are in a mixture T is same for both. Mean speed of Hydrogen

v =√(8RT/πM₀), and mean speed of nitrogen molecules,

v' =√(8RT/πM₀')

Hence the ratio

v/v' = √(M₀'/M₀) =√(28/2) =√14 =3.74 



20. Figure (24-E2) shows a vessel partitioned by a fixed diathermic separator. Different ideal gases are filled in the two parts. The rms speed of the molecules in the left part equals the mean speed of the molecules in the right part. Calculate the ratio of the mass of a molecule in the left part to the mass of a molecule in the right part. 
The figure for Q-20


Answer:  Due to the diathermic separator the temperature in both the part is the same as T. The rms speed in the left part v = √(3kT/m). The mean speed in the right part v' =√(8kT/πm'), where m and m' are masses of molecules of gases in the left and right part respectively. Here given that

v = v'

→√(3kT/m) =√(8kT/πm')

→3/m = 8/πm'

→m/m' = 3π/8 = 1.18. 

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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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Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)

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Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

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Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

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Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

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"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


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CHAPTER- 3 - Kinematics - Rest and Motion

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Click here for "OBJECTIVE-I"


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)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Physics and Mathematics"

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