Monday, June 15, 2020

H C Verma solutions, SPECIFIC HEAT CAPACITIES OF GASES, EXERCISES, Q1-Q10, Chapter-27, Concepts of Physics, Part-II

Specific Heat Capacities of Gases

EXERCISES, Q1-Q10

   1. A vessel containing one mole of a monoatomic ideal gas (molecular weight = 20 g/mol) is moving on a floor at a speed of 50 m/s. The vessel is stopped suddenly. Assuming that the mechanical energy lost has gone into the internal energy of the gas, find the rise in its temperature.


Answer:  Kinetic energy of the gas 

E =½mv².

Here v = 50 m/s and mass m = 20 g =0.020 kg.

So E = ½*0.020*50² J =25 J

This mechanical energy has gone into the internal energy of the gas. The change in the internal energy of the gas is given as (from equipartition of energy) 

dU = ½nd*R*dT,

here n = 1, dU = 25 J, R = 8.3 J/mol-K, degree of freedom, d = 3 for a monoatomic gas. So,

dT = 25*2/(3*8.3) = 2.0 K.

The rise in temperature will be 2.0 K.  


 

   2. 5 g of gas is contained in a rigid container and is heated from 15°C to 25°C. The specific heat capacity of the gas at constant volume is 0.172 cal/g-°C and the mechanical equivalent of heat is 4.2 J/cal. Calculate the change in the internal energy of the gas. 


Answer:  Given, m = 5 g,

cᵥ = 0.172 cal/g-°C, 

ΔT = 25°C -15°C =10°C.

From the first law of thermodynamics,

ΔQ =ΔU +ΔW

Since the volume is constant, work done is zero, i.e. ΔW = 0. So,

ΔU =ΔQ = mcᵥΔT

Hence the change in internal energy,

ΔU = mcᵥΔT 

=(5 g)*(0.172 cal/g-°C)*(10°C)

=8.6 cal =8.6*42 J =36 J.



 

   3. Figure (27-E1) shows a cylindrical container containing oxygen (ɣ=1.4) and closed by a 50 kg frictionless piston. The area of cross-section is 100 cm², atmospheric pressure is 100 kPa and g is 10 m/s². The cylinder is slowly heated for some time. Find the amount of heat supplied to the gas if the piston moves out through a distance of 20 cm.
The figure for Q - 3
   

Answer:  The piston exerts constant pressure on the gas which is equal to the atmospheric pressure and the weight of the piston.  Here piston weight = mg = 50*10 =500 N, Area of the piston, A = 100 cm² =0.01 m². Pressure by piston weight = 500/0.01 Pa =50000 Pa = 50 kPa. The total pressure on the gas,

p =100 kPa +50 kPa =150 kPa.

The piston moves by a distance, d of 20 cm =0.20 m. The increase in volume =A*d=0.01*0.20 m³ =0.002 m³.

Hence the work done by the gas, 

ΔW =p*ΔV 

=(150*1000 Pa)*(0.002 m³)

=300 J

Also p*ΔV =nRΔT
→nΔT = 300/R

ΔU = nCᵥΔT 

From the first law of thermodynamics,

ΔQ =ΔU+ΔW 

= nCᵥΔT + 300 --------- (i)

→300*Cᵥ/R +300 J

Since Cₚ-Cᵥ =R and Cₚ/Cᵥ =ɣ
→ɣCᵥ-Cᵥ =R
→Cᵥ/R =1/(ɣ-1) =1/0.4 =2.5  

Hence from (i),
ΔQ = 300*2.5 + 300 J
=1050 J. 

  



   4. The specific heat capacities of hydrogen at constant volume and at constant pressure are 2.4 cal/g-°C and 3.4 cal/g-°C respectively. The molecular weight of hydrogen is 2 g/mol and the gas constant R =8.3x10⁷ erg/mol-°C. Calculate the value of J.


Answer:  Given, Cₚ =2.4 cal/g-°C, Cᵥ =3.4 cal/g-°C, Since Cₚ-Cᵥ = R,

→R = 3.4 -2.4 = 1 cal/g-°C.

(Since it is given that the molecular weight of hydrogen = 2 g/mol)
→R = 2 cal/mol-°C. 

But also given that R = 8.3x10⁷ erg/mol-°C.

Let us equate these two values of R.
J*2 cal/mol-°C =8.3x10⁷ erg/mol-°C
→J = 4.15x10⁷ erg/cal. 
   



 

   5. The ratio of the molecular heat capacities of an ideal gas is Cₚ/Cᵥ = 7/6. Calculate the change in internal energy of 1.0 mole of the gas when its temperature is raised by 50 K (a)  keeping the pressure constant, (b) keeping the volume constant and adiabatically. 


Answer:  (a) When pressure is kept constant. ΔQ =nCₚΔT

Let the change in volume = ΔV

Hence the work-done, ΔW =p.ΔV

From the ideal gas law,

p.ΔV =nRΔT =ΔW,

From the first law of thermodynamics,

ΔQ =ΔU +ΔW

→ΔU =ΔQ -ΔW =nCₚΔT -nRΔT .....(i)

Now Cₚ-Cᵥ =R, Cₚ/Cᵥ =7/6, →Cᵥ=6Cₚ/7

→Cₚ -6Cₚ/7 =R

→Cₚ =7R

From (i).

ΔU =nΔT(Cₚ-R) =nΔT(7R-R) 

=nΔT*(6R)

=1*50*6*8.3 J

=2490 J. 



 (b). When the volume is kept constant the work done by the gas is zero and the heat-given is used to increase the internal energy of the gas.

ΔU =ΔQ =nCᵥΔT.

n =1, ΔT = 50 K. Given Cₚ/Cᵥ= 7/6.

→Cₚ = 7Cᵥ/6. But Cₚ-Cᵥ=R, so,

(7/6 -1)Cᵥ =R

→Cᵥ = 6R

Now ΔU = 1*6R*50 =300 R

=300*8.3 J=2490 J


(c) When the temperature is increased adiabatically.

ΔQ =0. From the first law of thermodynamics,

ΔQ =ΔU +ΔW

→ΔU = -ΔW

Wok-done by the gas in an adiabatic process =nRΔT/(ɣ-1)

Here n = 1 mol, R=8.3 J/mol-K, ΔT=50 K, ɣ =7/6, Hence

ΔW = -1*8.3*50/(7/6 -1)

     = -8.3*50*6 J =-2490 J

It is negative because the work is done on the gas, not by the gas.
So, ΔU =-(-2490 J)

           = 2490 J. 



 

   6. A sample of air weighing 1.18 g occupies 1.0 x 10³ cm³ when kept at 300 K and 1.0 x 10⁵ Pa. When 2.0 cal of heat is added to it at constant volume, its temperature increases by 1°C. Calculate the amount of heat needed to increase the temperature of air by 1°C at constant pressure if the mechanical equivalent of heat is 4.2x10⁷ erg/cal. Assume that air behaves as an ideal gas.


Answer:  m =1.18 g, V =1000 cm³ =0.001 m³, p=1x10⁵ Pa, T =300 K, ΔQ=2 cal, ΔT=1°C. Hence from,

pV =nRT
→n =pV/RT =1x10⁵*(0.001/8.3*300)
      =0.04 mol

Now, ΔQ = nCᵥΔT

→2 =0.04*Cᵥ*1 

→Cᵥ =2/0.04 =50 J/mol-°C

       =50*4.2x10⁷ erg/mol-°C

       =210x10⁷ erg/mol-°C

Now Cₚ =Cᵥ +R 

          =210x10⁷+8.3x10⁷ erg/mol-°C

          =218.3x10⁷ erg/mol-°C.

Hence the amount of heat needed to increase the temperature by 1°C at constant pressure =nCₚ*ΔT

=0.04*218.3x10⁷*1 erg

=8.732x10⁷ erg

=8.732x10⁷/4.2x10⁷ cal

=2.08 cal. 

      



 

   7. An ideal gas expands from 100 cm³ to 200 cm³ at a constant pressure of 2.0x10⁵ Pa when 50 J of heat is supplied to it. Calculate (a) the change in internal energy of the gas, (b) the number of moles in the gas if the initial temperature is 300K, (c) the molar heat capacity Cₚ at constant pressure and (d) the molar heat capacity Cᵥ at constant volume.    


Answer:  (a) Change in volume, ΔV =200-100 =100 cm³ =1x10⁻⁴ m³, Pressure, p =2x10⁵ Pa, So the work-done by the gas =ΔW =p*ΔV 

= 2x10⁵*1x10⁻⁴ J =20 J.

Given, ΔQ = 50 J.

From the first law of thermodynamics,

ΔQ =ΔU +ΔW

→ΔU =ΔQ -ΔW =50 -20 =30 J.


(b) The initial volume, V = 100 cm³ 

=1x10⁻⁴ m³.

The initial pressure, p=2x10⁵ Pa, and Temperature, T = 300 K. If n is the number of moles then,

pV =nRT

→n =pV/RT

      =2x10⁵*1x10⁻⁴/(8.3*300)

      =0.008

 

(c) pΔV =nRΔT
→nΔT =pΔV/R =ΔW/R
Since ΔU =nCᵥΔT
→Cᵥ =ΔU/nΔT
       =ΔU*R/ΔW
       =30*8.3/20 =12.45 J/mol-K  
Since Cₚ =Cᵥ +R =12.45+8.3 J/mol-K
=20.75 J/mol-K


(d) The molar heat capacity at constant volume Cᵥ =12.45 J/mol-K
{As derived in (c) above}



 

  8. An amount Q of heat is added to a monoatomic ideal gas in a process in which the gas performs a work Q/2 on its surrounding. Find the molar heat capacity for the process.  


Answer:  From the first law of thermodynamics, 

ΔQ =ΔU +ΔW. Here,

Q =ΔU +Q/2

→ΔU = Q/2

But for a monoatomic gas,

 ΔU=3nRΔT/2

→Q/2 =3nRΔT/2

→Q =3nRΔT  

For the heat at constant pressure,

Q =nCₚΔT

Equating, nCₚΔT = 3nRΔT

→Cₚ = 3R 

   



   9. An ideal gas is taken through a process in which the pressure and the volume are changed according to the equation p = kV. Show that the molar heat capacity of the gas for the process is given by C = Cᵥ + R/2.


Answer:  Since for an ideal gas,

pV =nRT

→kV² =nRT ,  

Differentiating, 2kVdV =nRdT

→dV =nRdT/2kV 

Since, dQ =dU +dW

nCdT =nCᵥdT +pdV

(Where C is the molar heat capacity of the gas)

→nCdT =nCᵥdT +kV*nRdT/2kV

→C = Cᵥ + R/2, Proved.


 


   10. An ideal gas (Cₚ/Cᵥ = ɣ) is taken through a process in which the pressure and the volume vary as p = aVb. Find the value of b for which the specific heat capacity in the process is zero.  


Answer:  We know that for a process the heat given, dQ =nCdT

Where C = specific heat capacity. Given that, C = 0. Hence dQ = 0.

This is the condition for an adiabatic process. For an adiabatic process 

pVɣ = Constant.

But given that the process follows

p =aVb

→p/Vb = a

→pV-b =a =constant

Comparing the two equations,

-b = ɣ

→b = -ɣ.  

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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

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CHAPTER- 7 - Circular Motion

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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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CHAPTER- 4 - The Forces

The Forces-

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CHAPTER- 3 - Kinematics - Rest and Motion

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CHAPTER- 2 - "Physics and Mathematics"

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