Wednesday, June 24, 2020

H C Verma solutions, SPECIFIC HEAT CAPACITIES OF GASES, EXERCISES, Q11-Q20, Chapter-27, Concepts of Physics, Part-II

Specific Heat Capacities of Gases

EXERCISES, Q11-Q20

   11. Two ideal gases have the same value of Cₚ/Cᵥ =ɣ. What will be the value of this ratio for a mixture of the two gases in the ratio 1:2?  


Answer:  Let the values for the first gas be Cₚ and Cᵥ and for the second gas Cₚ' and Cᵥ'.

Cₚ = ɣCᵥ and Cₚ' =ɣCᵥ'.

Now Cₚ -Cᵥ =R

→ɣCᵥ -Cᵥ =R

→Cᵥ =R/(ɣ-1)

Similarly Cᵥ' =R/(ɣ-1)

Suppose n mole of the first gas and 2n mole of the second gas is mixed and the temperature is raised by dT at constant volume. Now for the first gas

dU =nCᵥdT, and for the second gas,

dU' =2nCᵥ'dT.
For the mixture, dU" =dU +dU'
→3nCᵥ"dT =nCᵥdT +2nCᵥ'dT
→3Cᵥ" =Cᵥ +2Cᵥ'
→3Cᵥ" = R/(ɣ-1) +2R/(ɣ-1) =3R/(ɣ-1)
→Cᵥ" = R/(ɣ-1)

Since Cₚ" - Cᵥ" =R

→Cₚ" =R + Cᵥ" =R +R/(ɣ-1)

        ={(ɣ-1)+1}R/(ɣ-1)

        =ɣR/(ɣ-1)

Hence Cₚ"/Cᵥ" = ɣ.       

So it will be the same. 




  


    12. A mixture contains 1 mole of helium (Cₚ = 2.5R, Cᵥ = 1.5R) and 1 mole of hydrogen (Cₚ = 3.5R, Cᵥ = 2.5R). Calculate the values of Cₚ, Cᵥ and ɣ for the mixture.   


Answer:  Let the temperature is raised by dT at constant volume.

For the helium gas, dU' =nCᵥ'dT

=Cᵥ'dT 

For the hydrogen gas, dU" =nCᵥ"dT =Cᵥ"dT

For the mixture, dU =2nCᵥdT =2CᵥdT

But dU = dU' +dU"

→2CᵥdT =Cᵥ'dT +Cᵥ"dT

→Cᵥ = (Cᵥ' +Cᵥ")/2 =(1.5R +2.5R)/2 =2R

Now, Cₚ =Cᵥ +R =2R +R =3R,

and ɣ =Cₚ/Cᵥ = 3R/2R =1.5  

 



 


 

   13. Half mole of an ideal gas (ɣ = 5/3) is taken through the cycle abcda as shown in figure (27-E2). Take R = 25/3 J/mol-K (a) Find the temperature of the gas in the states a, b, c and d. (b) Find the amount of heat supplied in the processes ab and bc, (c) Find the amount of heat liberated in the process cd and da.
The figure for Q - 13
  


Answer:  (a) Given, n =1/2 mole, R =25/3 J/mol-K. From the ideal gas law,

pV =nRT

→T =pV/nR 

At the point a, 

p =100 kPa =1x10⁵ Pa, V=5000 cm³ =5x10⁻³ m³  

T =(1x10⁵)*(5x10⁻³)/{(1/2)*(25/3)} K

  2*3*5x10²/25 K =600/5 K =120 K.


At point b,

p =100 kPa =1x10⁵ Pa, V =10000 cm³ =1x10⁻² m³, so T =pV/nR,

→T =(1x10⁵)*(1x10⁻²)/{(1/2)*(25/3)} K

=3*2x10³/25 K =6000/25 K =240 K.

At point c,

p =200 kPa =2x10⁵ Pa, V =10000 cm³ =1x10⁻² m³,

So, T=pV/nR

→T =(2x10⁵)*(1x10⁻²)/{(1/2)*(25/3)} K

=2*3*2x10³/25 K =12000/25 K =480 K.


At point d,

p =200 kPa =2x10⁵ Pa, V =5000 cm³ =5x10⁻³ m³

Hence T =pV/nR

→T=(2x10⁵)*(5x10⁻³)/{(1/2)*(25/3)}

→T=2*5*2*3x10²/25 K

→T=6000/25 K =240 K.   


(b) The amount of heat supplied in process ab,

This process is at constant pressure.  

Given that ɣ =5/3, 

Cₚ =ɣR/(ɣ-1) =(5/3)*(25/3)/(5/3-1) =125/6 J/mol-K

dQ = nCₚdT =(1/2)*(125/6)*(240-120) J

=125*120/12 =1250 J


The heat supplied in process bc,

The process is at constant volume. Since the volume does not change there is no work done. The heat supplied is,

dQ = nCᵥdT

Cᵥ = R/(ɣ-1) =(25/3)/(5/3 -1) =25/2 J/mol-K

dT =480 -240 K =240 K

Hence dQ =(1/2)*(25/2)*240 J 

              =25*60 J =1500 J. 


(c) The heat liberated in the process cd,

The process cd is at constant pressure, and as calculated above, Cₚ =125/6 J/mol-K. dT =480 -240 =240 K. Hence heat liberated dQ =nCₚdT.
→dQ =(1/2)*(125/6)*240 J =125*20 J =2500 J.

The amount of heat liberated in the process da,

The process is at constant volume. As calculated Cᵥ = 25/2 J/mol-K, Change in temperature, dT =240 -120 K =120 K. The amount of heat liberated,

dQ =nCᵥdT 

     =(1/2)*(25/2)*120 J

     =25*30 J =750 J.   

 

 

 

 

   14. An ideal gas (ɣ = 1.67) is taken through the process abc shown in figure (27-E3). The temperature at the point a is 300 K. Calculate (a) the temperature at b and c, (b) the work-done in the process, (c) the amount of heat supplied in the path ab and in the path bc and (d) the change in the internal energy of the gas in the process.
The figure for Q - 14
 


Answer:  Given ɣ =1.67. At the state 'a', the pressure p =100 kPa =1x10⁵ Pa, V =100 cm³ =1x10⁻⁴ m³, Temperature T =300 K.

At the state 'b', p' =200 kPa =2x10⁵ Pa, V' =100 cm³ =1x10⁻⁴ m³, T' = ? V =V'. 

Now pV/T = p'V'/T'

→T' = p'T/p =2x10⁵*300/1x10⁵ K 

      = 600K.   


At the state 'c'

P" =200 kPa =2x10⁵ Pa, V" = 150 cm³ =1.50x10⁻⁴ m³, T" =? 

Now, pV/T = p"V"/T"

→T" =p"V"T/pV

 =2x10⁵*1.50x10⁻⁴*300/(1x10⁵*1x10⁻⁴) K

=2*1.5*300 K =900 K.


(b) Work done in the process,

In the process ab the work done is zero as it is under constant volume. The work done under the process bc will be equal to the area under the bc and the volume axis. 
Here, the work done =200 kPa*(150-100) cm³
=2x10⁵*50x10⁻⁶ J
=10 J = work done in the process abc. 
  

(c) The amount of heat supplied in the path ab,

Since no work is done in the process ab at constant volume, the amount of heat supplied, 

dQ =nCᵥdT =(pV/RT)*{R/(ɣ-1)}*(T'-T)

=(1x10⁵*1x10⁻⁴/300){1/(1.67-1)}*(600-300) J 

=10/0.67 J =14.9 J

   

The amount of heat supplied in the process bc. 

This process is at constant pressure, so

dQ =nCₚdT

Considering at b, n =p'V'/RT' 

dQ =(p'V'/RT')*{ɣR/(ɣ-1)}*(900-600) J 

=(2x10⁵*1x10⁻⁴/600)*(1.67/0.67)*300 J

= 16.7/0.67 J =24.9 J 


(d) The change in the internal energy of the gas, dU =nCᵥdT

=(pV/RT)*{R/(ɣ-1)}*(T"-T) J 

=pV(T"-T)/(ɣ-1)T  J 

=1x10⁵*1x10⁻⁴(900-300)/(0.67*300) J

=20/0.67 J 

= 29.8 J. 

Alternately,
The change in internal energy in the process abc,
dU =dQ -dW

=(14.9+24.9) -10 J

=39.8 -10 J =29.8 J  

 

 

 

 

   15. In Joly's differential steam calorimeter, 3 g of an ideal gas is contained in a rigid closed sphere at 20°C. The sphere is heated by steam at 100°C and it is found that an extra 0.095 g of steam has condensed into the water as the temperature of the gas becomes constant. Calculate the specific heat capacity of the gas in J/g-K. The latent heat of vaporization of water = 540 cal/g.  


Answer:  The amount of heat lost by steam = 0.095*540 cal =51.3 cal =51.3*4.18 J =214.4 J. The amount of heat gained by the ideal gas, 

=mcᵥdT

=3cᵥ(100-20) J

=240cᵥ J

Equating the heat lost and the heat gained,

240cᵥ =214.4

→cᵥ =214.4/240 J/g-K

= 0.90 J/g-K.    


 


 

   16. The volume of an ideal gas (ɣ = 1.5) is changed adiabatically from 4.00 liters to 3.00 liters. Find the ratio of (a) the final pressure to the initial pressure and (b) the final temperature to the initial temperature.   


Answer:  (a) V = 4.00 l, V' =3.00 l, for an adiabatic process, 

pVɣ = p'V'ɣ 

→p'/p =(V/V')ɣ

         =(4/3)1.5

          =1.54 


(b) In terms of temperature, the equation is given as,

 TVɣ-1 =T'V'ɣ-1

→T'/T =(V/V')ɣ-1

         =(4/3)1.5-1

         =√(4/3) = 1.15.  

  


 

   17. An ideal gas at pressure 2.5x10⁵ Pa and temperature 300 K occupies 100 cc. It is adiabatically compressed to half of its original volume. Calculate (a) the final pressure, (b) the final temperature and (c) the work done by the gas in the process. Take ɣ = 1.5. 


Answer:  (a) p =2.5x10⁵ Pa, ɣ =1.5, 

T = 300 K, V =100 cc.

V' =V/2, p' =? 

For an adiabatic process,

pVɣ = p'V'ɣ

→p' =p(V/V')ɣ =2.5x10⁵*(2)1.5 Pa

      = 7.1x10⁵ Pa.


(b) Also for the adiabatic process,

TVɣ-1 =T'V'ɣ-1

→T' =T(V/V')ɣ-1 

      =300*(2)1.5-1 

      =300√2 

      =424 K.  


(c) The work done by the gas in an adiabatic process =(pV-p'V')/(ɣ-1)

=(2.5x10⁵*V -7.1x10⁵*V/2)/(1.5-1)

=(-2.1x10⁵/2)V/0.5 J

=-2.1x10⁵*100x10⁻⁶ J
=-21 J. 

   


 

   18. Air (ɣ = 1.4) is pumped at 2 atm pressure in a motor tire at 20°C. If the tire suddenly bursts, what would be the temperature of the air coming out of the tire? Neglect any mixing with the atmospheric air.  


Answer:  p = 2 atm, T =20°C =293 K,

ɣ =1.4, p' =1 atm, T' =?

Since the process is sudden it will be an adiabatic process. The relation between pressure and temperature in an adiabatic process is given as, 

Tɣ/pɣ-1 = T'ɣ/p'ɣ-1 

→T'ɣ = Tɣ*(p'/p)ɣ-1 

→T' = T*(p'/p)ɣ-1/ɣ =293*(1/2)0.4/1.4

→T' =293*0.82 =240 K. 



      

 

   19. A gas is enclosed in a cylindrical can fitted with a piston. The walls of the can and the piston are adiabatic. The initial pressure, volume and temperature of the gas are 100 kPa, 400 cm³ and 300 K respectively. The ratio of the specific heat capacities of the gas is Cₚ/Cᵥ =1.5. Find the pressure and the temperature of the gas if it is (a) suddenly compressed (b) slowly compressed to 100 cm³.   


Answer:  (a) Given that,  

p =100 kPa = 1x10⁵ Pa,

V =400 cm³, T = 300 K, ɣ = 1.5, 

V' = 100 cm³. 

When suddenly pressed the adiabatic relationship is 

pVɣ = p'V'ɣ 

→p' = p*(V/V')ɣ

      = 100*(400/100)1.5 kPa 

      = 800 kPa.  


The adiabatic relation between temperature and volume is given as, 

TVɣ-1 = T'V'ɣ-1 

→T' = T*(V/V')ɣ-1

      = 300*(400/100)0.5 K

      = 300*√4 K

      = 600 K.  


(b) Though the process is slow the walls of the can and the piston are adiabatic, so still, there will be no heat transfer and the process will be adiabatic. Since the final volume is the same as in the first process, the pressure and temperature will be the same as in the first process.

       





 

   20. The initial pressure and volume of a given mass of gas (Cₚ/Cᵥ = ɣ) are pₒ and vₒ. The gas can exchange heat with the surrounding. (a) It is slowly compressed to a volume vₒ/2 and then suddenly compressed to vₒ/4. Find the final pressure. (b) If the gas is suddenly compressed from the volume vₒ to vₒ/2 and then slowly compressed to vₒ/4, what will be the final pressure?  


Answer:  (a) When the gas is slowly compressed the process is isothermal because of heat exchange with outside. For an isothermal process,

pV = p'V',    

Here, p = pₒ, V = vₒ, V' = vₒ/2, hence,

p' = p(V/V') = pₒ*2 =2pₒ 

The next process is adiabatic because it is sudden. Here the relation between p and V is given as,   

pVɣ = p'V'ɣ 

Here, p =2pₒ, V =vₒ/2, V' = vₒ/4, hence the final pressure p' =p(V/V')ɣ 

→p' = 2pₒ*(2)ɣ = 2ɣ+1pₒ.

              

(b) This time the first process is adiabatic in which, p = pₒ, V = vₒ, V' =vₒ/2, the pressure at the end of the adiabatic process,  

p' =pₒ(V/V')ɣ = pₒ*2ɣ 

Now the second process is isothermal,

 so pV = p'V', and p' =p(V/V') 

Here p =pₒ*2ɣ, V= vₒ/2. V' = vₒ/4 

Now the final pressure.

 p' = pₒ*2ɣ*(2) =2ɣ+1pₒ.  

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CHAPTER- 20 - Dispersion and Spectra


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CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








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