Wednesday, November 27, 2019

H C Verma solutions, Kinetic Theory of Gasses, EXERCISES Q1 to Q10, Chapter-24, Concepts of Physics, Part-II

Kinetic Theory of Gases

EXERCISES, Q1 to Q10

1. Calculate the volume of 1 mole of an ideal gas at STP. 


Answer:  STP means, T = 273 K and 

p = 1 atm =1.01325x10⁵ N/m². Here n = 1 mole and we know that R = 8.3 J/mol-K. Volume V =?

From the ideal gas equation,
pV =nRT
→V = nRT/p
      = 1*8.3*273/1.01325x10⁵ 
      = 2.24x10⁻² m³
      = 22.4 liters


 


2. Find the number of molecules of an ideal gas in a volume of 1.000 cm³ at STP. 


Answer:   

 Since the volume of 1 mole of an ideal gas at STP =22.4 liters =2.24x10⁻² m³, 1.000 cm³ of an ideal gas will contain 1x10⁻⁶/2.24x10⁻² =4.46x10⁻⁵ moles. Hence the number of molecules in 1.000 cm³ of an ideal gas at STP

=4.46x10⁻⁵*6.02x10²³
=2.685x10¹⁹



 3. Find the number of molecules in 1 cm³ of an ideal gas at 0°C and at a pressure of 10⁻⁵ mm of mercury.


Answer:  V = 1 cm³ =1x10⁻⁶ m³

T = 0°C = 273 K

p = 10⁻⁵ mm of mercury 

=13600*9.8*(10⁻⁵/1000) N/m²

(Since p =ρgh, ρ of mercury =13600 kg/m³) 

=1.33x10⁻³ N/m²

 R = 8.3 J/mol-K

Hence n =pV/RT

→n =1.33x10⁻³*1x10⁻⁶/(8.3*273)

      = 5.869x10⁻¹³ moles

 Hence the number of molecules,

N =n*Nₐ  

   =5.869x10⁻¹³*6.02x10²³ 

   = 3.53x10¹¹ 


 


 4. Calculate the mass of 1 cm³ of oxygen kept at STP. 


Answer:  1 cm³ = 1x10⁻⁶ m³

Since 22.4 liters i.e. 2.24x10⁻² m³ of an ideal gas contains 1 mole or 6.02x10²³ number of molecules, hence 1x10⁻⁶ m³ of gas will contain

1x10⁻⁶*1/2.24x10⁻² moles
=4.46x10⁻⁵ moles.
(The molecular weight of oxygen =2*16 =32 amu
Hence 1 mole of oxygen = 32 g)
=4.46x10⁻⁵*32 g
=1.43x10⁻³ g
=1.43 mg.

 


 5. Equal masses of air are sealed in two vessels, one of volume V₀ and other of volume 2V₀. If the first vessel is maintained at a temperature of 300 K and the other at 600 K, find the ratio of the pressures in the two vessels. 


Answer:  Here for the first vessel V = V₀, T = 300 K and let pressure =p. So,

pV₀ = nRT   ------ (i)

Similarly, for the second vessel, V =2V₀, T'=600 K and let the pressure =p'. So,
p'*2V₀ = nRT' ------(ii)
Dividing (i) by (ii),
pV₀/2p'V₀ =nRT/nRT' =T/T'
→p/2p' =300/600 =1/2
→p/p' =1
Hence the ratio of p:p' = 1 : 1.


 


 6. An electric bulb of volume 250 cc was sealed during manufacturing at a pressure of 10⁻³ mm of mercury at 27°C. Compute the number of air molecules contained in the bulb. Avogadro constant =6x10²³ per mol, the density of mercury =13600 kg/m³ and g = 10 m/s².


Answer:  V =250 cc =250/10⁶ m³ =2.5x10⁻⁴ m³. 

p= ρgh =13600*10*(10⁻³/1000) N/m²

=0.136 N/m²

R =8.3 J/mol-K
T =27+273 K =300 K
Hence from the ideal gas equation,
pV=nRT
→n = pV/RT 
      =0.136*2.5x10⁻⁴/(8.3*300) moles
      = 1.365x10⁻⁸ moles
      =1.36x10⁻⁸*6.02x10²³ molecules
      =8.1x10¹⁵ molecules



 

 7. A gas cylinder has walls that can bear a maximum pressure of 1.0x10⁶ Pa. It contains gas at 8.0x10⁵ Pa and 300 K. The cylinder is steadily heated. Neglecting any change in the volume, calculate the temperature at which the cylinder will break. 


Answer:  Since at a constant volume the pressure is directly proportional to the absolute temperature (Charles Law of pressure), p/p' = T/T'.

Here, p =8x10⁵ Pa, p' =1x10⁶ Pa, T=300 K, T'=?

So, T' =p'T/p =1x10⁶*300/8x10⁵

→T' = 3000/8 = 375 K.


 


 8. 2 g of hydrogen is sealed in a vessel of volume 0.02 m³ and is maintained at 300 K. Calculate the pressure in the vessel. 


Answer:  2 g of hydrogen = 2/2 mole =1 mole of hydrogen. So, n = 1, V=0.02 m³, T = 300 K, R=8.3 J/mol-K. p=?

From the ideal gas equation,

pV = nRT

→p = nRT/V =1*8.3*300/0.02 Pa

→p = 1.24x10⁵ Pa.

 




 9. The density of an ideal gas is 1.25x10⁻³ g/cm³ at STP. Calculate the molecular weight of the gas.


Answer:  Volume of 1 mole of an ideal gas at STP = 2.24x10⁻² m³. Hence the mass of 1 mole of the given gas at STP

 = 2.24x10⁻²*1.25x10⁻³*10⁶ g

 = 28 g

  Hence the molecular weight of the given gas = 28 g/mol.



 


 10. The temperature and pressure at Shimla are 15.0°C and 72.0 cm of mercury and at Kalka, these are 35.0°C and 76.0 cm of mercury. Find the ratio of air density at Kalka to the air density at Shimla.


Answer:   

The ideal gas equation is

pV=nRT =(m/M)RT {where m is the mass and M is molecular weight of the gas.}

→pM=(m/V)RT =ρRT
Let the air density at Shimla =ρ and at Kalka =ρ'. At Shimla p=72 cm of mercury, T=15+273=288 K
At Kalka, p' =76 cm of mercury and T'=35+273=308 K
Hence, pM/p'M =ρRT/ρ'RT'
→p/p' =ρT/ρ'T'
→ρ'/ρ =p'T/pT' 
→ρ'/ρ=76*288/72*308=0.987

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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

Click here for → Questions for Short Answer 

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)

CHAPTER- 6 - Friction

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

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


Click here for → QUESTIONS FOR SHORT ANSWER

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

Click here for "Questions for short Answers"


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"

Click here for "Questions for Short Answers"


Click here for "OBJECTIVE-II"

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