Sunday, December 15, 2019

H C Verma solutions, Kinetic Theory of Gasses, EXERCISES Q31 to Q40, Chapter-24, Concepts of Physics, Part-II

Kinetic Theory of Gases

EXERCISES, Q31 to Q40

 31. Figure (24-E3) shows two vessels A and B with rigid walls containing ideal gases. The pressure, temperature and volume are pₐ, Tₐ, V in the vessel A and pᵦ, Tᵦ, V in the vessel B. The vessels are now connected through a small tube. Show that the pressure p and the temperature T satisfy

p/T =½{pₐ/Tₐ + pᵦ/Tᵦ}

when equilibrium is achieved. 
The figure for Q-31


Answer:  Let the number of moles of the ideal gas in A = n and in B = n'.

   From the ideal gas equation

pₐV = nRTₐ

→n = pₐV/RTₐ, Similarly

n' = pᵦV/RTᵦ

When the equilibrium is achieved after connecting both vessels, the total number of molecules, n" = n + n'.

In this case, the volume becomes 2V.

So the ideal gas equation for this volume is,

p(2V) = n"RT

→p/T =n"R/2V =(n+n')R/2V

→p/T = {(pₐV/RTₐ)+(pᵦV/RTᵦ)}R/2V

→p/T = ½{pₐ/Tₐ + pᵦ/Tᵦ}.

Proved.   



 32. A container of volume 50 cc contains air (mean molecular weight = 28.8 g) and is open to the atmosphere where the pressure is 100 kPa. The container is kept in a bath containing melting ice (0°C). (a) Find the mass of the air in the container when thermal equilibrium is reached. (b) The container is now placed in another bath containing boiling water (100°C). Find the mass of air in the container. (c) The container is now closed and placed in the melting-ice bath. Find the pressure of the air when thermal equilibrium is reached.   


Answer:  Number of moles of air,

n = pV/RT

(a) p = 100 kPa =1x10⁵ N/m²

V = 50 cc = 50x10⁻⁶ m³ =5x10⁻⁵ m³

R = 8.3 J/mol-K

T = 0°C = 273 K

→n = 1x10⁵*5x10⁻⁵/(8.3*273)

      = 2.207x10⁻³ moles

      = 2.207x10⁻³*28.8 g
      = 0.0635 g. 

(b) In the boiling water, T = 100°C =373 K
Now, n = pV/RT
→n = 1x10⁵*5x10⁻⁵/(8.3*373)       
      =1.615x10⁻³ moles
      =1.615x10⁻³*28.8 g
      =0.0465 g.

(c) When the container is closed, the number of moles of air in the container n=1.615x10⁻³, so the pressure in the melting ice bath,
p = nRT/V
   = 1.615x10⁻³*8.3*273/5x10⁻⁵ Pa
   = 73.2 kPa.

   



 33. A uniform tube closed at one end contains a pallet of mercury 10 cm long. When the tube is kept vertically with the closed-end upward, the length of the air column trapped is 20 cm. Find the length of the air column trapped when the tube is inverted so that the closed end goes down. Atmospheric pressure = 75 cm of mercury.  


Answer:  Let the cross-section of the uniform tube = A cm². The volume of the air column trapped = 20A cm³ =20A/10⁶ m³. The pressure = 75-10 cm =65 cm of mercury =65ρg/100 N/m² (ρ is the density of air in kg/m³). When the tube is inverted, the pressure p' = 75+10 cm = 85 cm of mercury =85ρg/100 N/m². If the length of the air column now is L cm, then its volume V'=LA/10⁶ m³. Since the temperature is constant, 

pV = p'V'

→(65ρg/100)*20A/10⁶ = (85ρg/100)*LA/10⁶

→L = 1300/85 cm = 15.3 cm.   


 

 

 34. A glass tube sealed at both ends is 100 cm long. It lies horizontally with the middle 10 cm containing mercury. The two ends of the tube contain air at 27°C and at a pressure 76 cm of mercury. The air column on one side is maintained at 0°C and the other side is maintained at 127°C. Calculate the length of the air column on the cooler side. Neglect the changes in the volume of mercury and of the glass. 


Answer:     

  Let the pressure on the both sides in the second case = p. If the length of the air column on the cooler side = L cm, then the length of the air column on the hotter side =90-L cm. The volume of air on the cooler side V =LA/10⁶ m³ and temperature T = 273 K.
So, pV =nRT
→p(LA/10⁶) =nR*273 -------- (i)
On the other side, pressure = p, Volume, V' =(90-L)A/10⁶ m³, temperature T' = 127+273 =400 K
So, pV' =nRT'
→.p(90-L)A/10⁶ =nR*400 -----(ii)
Dividing (i) by (ii)
L/(90-L) = 273/400
→400L =273*90 - 273L
→(400+273)L =273*90
→673L = 273*90
→L = 273*90/673 = 36.5 cm.

 


 35. An ideal gas is trapped between a mercury column and the closed end of a narrow vertical tube of a uniform base containing the column. The upper end of the tube is open to the atmosphere. The atmospheric pressure equals 76 cm of mercury. The lengths of the mercury column and the trapped air column are 20 cm and 43 cm respectively. What will be the length of the air column when the tube is tilted slowly in a vertical plane through an angle of 60°? Assume the temperature to remain constant. 


Answer:  The pressure of the trapped air column in the vertical tube, p = 76+20 =96 cm of mercury =96ρg/100 Pa. If the cross-section area of the tube is A cm², volume V = 43A/10⁶ m³.

Diagram for Q-35

 When the tube is tilted through 60°, the pressure of the air column, p' = 76+20*cos 60° cm =76+10 cm =86 cm of mercury = 86ρg/100 Pa. If the length of the air column now is = L cm then volume, V' = LA/10⁶ m³.

  Since the amount of air and the temperature is the same,

pV = p'V'

→(96ρg/100)*43A/10⁶ = (86ρg/100)*LA/10⁶

→96*43 = 86*L

→L = 96*43/86 = 48 cm.



 

 36. Figure (24-E4) shows a cylindrical tube of length 30 cm which is partitioned by a tight-fitting separator. The separator is very weakly conducting and can freely slide along the tube. Ideal gases are filled in the two parts of the vessel. In the beginning, the temperatures in the parts A and B are 400 K and 100 K respectively. The separator slides to a momentary equilibrium position shown in the figure. Find the final equilibrium position of the separator, reached after a long time. 
The figure for Q-36
 


Answer:  Since the separator is weakly conducting, after a long time the temperature on both sides will be the same, sat T. Also the separator can freely slide, hence the pressure on both the side will be equal. Take initial pressure =p and final pressure = p'.

   If the area of cross-section of the tube is S cm², the volume of the part A, V = 20*S/10⁶ m³ =20F m³ (Where F =S/10⁶). And of part B, V' =10F m³.

  Assume that after a long time the separator's position is x cm from the left. Now the volume of part A, Vₐ' =xF m³ and that of part B, Vᵦ' =(30-x)F m³.
     Now the ideal gas equation for part A, 
pV/400 = p'Vₐ'/T
→p*20F/400 = p'*xF/T
→p/p' = 20x/T  --------------- (i)

Similarly for part B,
pV'/100 = p'Vᵦ'/T
→p*10F/100 = p'*(30-x)F/T
→p/p' = 10(30-x)/T ----------(ii)

Equating these two values of p/p' we get

20x/T = 10(30-x)/T
→20x = 300 - 10x
→30x = 300
→x = 10 cm from the left end.      




 

 37. A vessel of volume V₀ contains an ideal gas at pressure p₀ and temperature T. Gas is continuously pumped out of this vessel at a constant volume-rate dV/dt=r keeping the temperature constant. The pressure of the gas being taken out equals the pressure inside the vessel. Find the (a) the pressure of the gas as a function of time, (b) the time taken before half the original gas is pumped out. 


Answer:  Let at time t the pressure of the gas in the vessel = p, and volume given Vₒ. If there is n number of moles in the vessel and in a time dt the mole of gas pumped out = dn the decrease in pressure of the vessel = dV. From the ideal gas equation,

(p-dp)Vₒ = (n-dn)RT

→pVₒ - Vₒdp = nRT - dn*RT

But pVₒ = nRT, so
Vₒdp = dn*RT --------------- (i)

Now the pressure of the gas taken out is equal to the internal pressure hence

(p-dp)*dV = dnRT

→pdV - dp*dV = dnRT

→pdV = dnRT, -------------- (ii)

(since dp*dV is negligible.)

Equating (i) and (ii)

Vₒdp = pdV   

→dp/p = dV/Vₒ

→dp/p = rdt/Vₒ  
(since dV/dt = r, dV =-rdt, since volume decreases with time)
integrating between pₒ to p and t = 0 to t,
→∫dp/p = -∫rdt/Vₒ
→[ln p] = -r[t]/Vₒ
→(ln p- ln pₒ) = -rt/Vₒ
→ln (p/pₒ) = -rt/Vₒ
→p/pₒ = e-(rt/V0)   
→p = pₒe-(rt/V0)


(b) Half of the gas has been pumped out, so the number of mole inside the vessel = n/2. If the pressure now is p', then
p'Vₒ =(n/2)RT = nRT/2 = pVₒ/2
→p' = p/2, so the pressure now is half the original pressure. 
From the equation derived in part (a).
e-(rt/V0) = 1/2
→-(rt/Vₒ) = ln(1/2) = -ln 2
→t = (Vₒ.ln 2)/r       

 

 38. One mole of an ideal gas undergoes a process 

p = p₀/{1+(V/V₀)²}

where p₀ and V₀ are constants. Find the temperature of the gas when V = V₀. 


Answer:  Given n = 1 mole. So 

pV = nRT, and for n =1 and V = Vₒ

→pVₒ = RT    -------- (i)

But for the given process,

p = pₒ/{1+(V/Vₒ)²}

putting V = Vₒ

→p = pₒ/2  

From (i) 

(pₒ/2)Vₒ = RT

→pₒVₒ = 2RT

→T = pₒVₒ/2R mol⁻¹     




 

 39. Show that the internal energy of the air (treated as an ideal gas) contained in a room remains constant as the temperature changes between day and night. Assume that the atmospheric pressure around remains constant and the air in the room maintains this pressure by communicating with the surrounding through the windows etc.  


Answer:  Internal energy U of an ideal gas is equal to the total translational kinetic energy of all the molecules and it is given as 

U = (3/2)nRT

But from the ideal gas equation

pV = nRT, hence

U = (3/2)pV    

Since p and V are constant, U is also constant. 




 40. Figure (24-E5) shows a cylindrical tube of radius 5 cm and length 20 cm. It is closed by a tight-fitting cork. The friction coefficient between the cork and the tube is 0.20. The tube contains an ideal gas at a pressure of 1 atm and a temperature of 300 k. The tube is slowly heated and it is found that the cork pops out when the temperature reaches 600 K. Let dN denote the magnitude of the normal contact force exerted by a small length dl of the cork along the periphery (see the figure). Assuming that the temperature of the gas is uniform at any instant, calculate dN/dl.
The figure for Q-40
  


Answer:  For the constant volume

p/T = p'/T'

here T = 300 K, T' = 600 K, p = 1 atm.

→p' = (600/300)*1 atm = 2 atm.

The radius r = 5 cm = 0.05 m

 Circumference of the cork C =2πr

→C = 2π(0.05) =0.10π

 The normal force on length dl = dN

The force of friction on dl =µdN

Hence the total force of friction between cork and tube, F = C.µdN/dl

→F = 0.10π*0.20*dN/dl, (since µ = 0.20)

→F = 0.02π dN/dl

When the cork pops out the net pressure on the cork area =p' - p =1 atm, because outside pressre is still p.

Area of the cork, A = πr² =π(0.05)²

The force on the cork, F = (p'-p)A

→F = (1x10⁵ Pa)*{π(0.05)² m²}

Equating the force of friction and the force by gas on the cork we get, 

0.02πdN/dl = (1x10⁵){π(0.05)²}

→dN/dl = (1x10⁵)*0.0025/0.02

→dN/dl = 0.125x10⁵ = 1.25x10⁴ N/m

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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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