Saturday, July 4, 2020

H C Verma solutions, SPECIFIC HEAT CAPACITIES OF GASES, EXERCISES, Q21-Q30, Chapter-27, Concepts of Physics, Part-II

Specific Heat Capacities of Gases

EXERCISES, Q21-Q30

   21. Consider a given sample of an ideal gas (Cₚ/Cᵥ = ɣ) having initial pressure pₒ and volume Vₒ. (a) The gas is isothermally taken to a pressure pₒ/2 and from there adiabatically to a pressure pₒ/4. Find the final volume.

(b) The gas is brought back to its initial state. It is adiabatically taken to a pressure pₒ/2 and from there isothermally to a pressure pₒ/4. Find the final volume. 


Answer:  (a) After the isothermal process, pressure, p₁ = p₀/2, volume V₁ =?

Here p₀V₀ = p₁V₁

→p₀V₀ = (p₀/2)V₁

→V₁ = 2V₀

After the adiabatic process, 

pressure, p₂ =p₀/4, volume V₂ =?

For this process,

p₁V₁ˠ = p₂V₂ˠ

→(p₀/2)*(2V₀)ˠ = (p₀/4)*V₂ˠ

→V₂ˠ = 2*(2V₀)ˠ =2⁽ˠ⁺¹⁾*V₀ˠ

→V₂ = 2⁽ˠ⁺¹⁾/ˠV₀


(b) Now the first process is adiabatic, so 

p₀V₀ˠ =p₁V₁ˠ

→p₀V₀ˠ =(p₀/2)V₁ˠ

→V₁ˠ = 2V₀ˠ

→V₁ = 2¹/ˠV₀

The second process is isothermal, p₁ =p₀/2, p₂ =p₀/4, so 

p₁V₁ =p₂V₂

→V₂ = (p₁/p₂)V₁ =(2)*2¹/ˠV₀

→V₂ =2⁽ˠ⁺¹⁾/ˠV₀     




 

   22. A sample of an ideal gas (ɣ =1.5) is compressed adiabatically from a volume of 150 cm³ to 50 cm³. The initial pressure and the initial temperatures are 150 kPa and 300 K. Find (a) the number of moles of the gas in the sample, (b) the molar heat capacity at constant volume, (c) the final pressure and temperature, (d) the work done by the gas in the process and (e) the change in internal energy of the gas. 


Answer:  (a) Initial pressure pₒ =150 kPa, temperature, Tₒ =300 K, Volume Vₒ =150 cm³. V₁ =50 cm³. 

For an ideal gas, pV =nRT 

→n = pV/RT 

=(1.5x10⁵)(150x10⁻⁶)/(8.3*300)

=1.5*15/2490 mole

=0.009 mole


(b) Since Cₚ/Cᵥ = ɣ,→Cₚ =ɣCᵥ

→Cₚ - Cᵥ =R

→ɣCᵥ -Cᵥ =R

→Cᵥ =R/(ɣ-1) =R/(1.5-1) 

→Cᵥ = 2R =2*8.3 J/mol-K

       = 16.6 J/mol-K 

    

(c) If final pressure =p₁ 

→p₁V₁ᵞ = pₒVₒᵞ

→p₁ =pₒ(Vₒ/V₁)ᵞ

→p₁ =150*(150/50)¹.⁵ kPa 

→p₁ =150*31.5 kPa =780 kPa.


The final temperature =pV/nR 
=(780*1000)(50x10⁻⁶)/(0.009*8.3) K
= 522 K.


(d) Since dQ = dU +dW, here dQ = 0 for adiabatic process.  

So, the work done by the gas dW =-dU

→dW = -nCᵥ.dT

     = -0.009*16.6*(522-300) J 

     = -33 J.


(e) As we saw above that dW =-dU,

→dU = -dW = -(-33 J) = 33 J.   



 

 

   23. Three samples A, B and C of the same gas (ɣ =1.5) have equal volume and temperature. The volume of each sample is doubled, the process being isothermal for A, adiabatic for B and isobaric for C. If the final pressures are equal for the three samples, find the ratio of the initial pressures. 


Answer:  Let initial pressures be pₐ, pᵦ and p₍ and the final pressure =p.

Let initial volume V and temperature T and the final volume = 2V.    

For sample A the process is isothermal, 

So pₐV =p*2V

→pₐ =2P 

For sample B the process is adiabatic, hence pᵦVˠ = p(2V)ˠ

→pᵦ = p*2ˠ = p*21.5 = 2√2p

For sample C the process is isobaric, it means the pressure remains the same. Hence p₍ = p.

The ratio of initial pressures is

pₐ:pᵦ:p₍ = 2p:2√2p:p =2:2√2:1 


       


 

   24. Two samples A and B of the same gas have equal volumes and pressures. The gas in sample A is expanded isothermally to double its volume and the gas in B is expanded adiabatically to double its volume. If the work done by the gas is the same for the two cases, show that ɣ satisfies the equation 1-21-ɣ =(ɣ-1)ln2.


Answer:  Work-done in the isothermal process for sample A, 

=nRT*ln(Vf/Vi) =nRT*ln2

Work-done in the adiabatic process for sample B.

dW=(pₒVₒ-p₁V₁)/(ɣ-1)

=(nRT -nRT₁)/(ɣ-1)

=nR(T-T₁)/(ɣ-1)
But T₁V₁ˠ-1 =TVₒˠ-1
→T₁ =T(Vₒ/V₁)ˠ-1 =T(1/2)ˠ-1
Now dW =nRT{1-(1/2)ˠ-1}/(ɣ-1)   

Equating the two work done,

nRT{1-(1/2)ˠ-1}/(ɣ-1) =nRT*ln2

→1-1/2ˠ-1 =(ɣ-1)ln2

→1-21-ˠ = (ɣ-1)ln2     






   25. 1 liter of an ideal gas (ɣ =1.5) at 300 K is suddenly compressed to half its original volume, (a) Find the ratio of the final pressure to the initial pressure. (b) if the original pressure is 100 kPa, find the work done by the gas in the process. (c) what is the change in internal energy? (d) what is the final temperature? (e) the gas is now cooled to 300 K keeping its pressure constant. Calculate the work done during the process. (f) the gas is now expanded isothermally to achieve its original volume of 1 liter. Calculate the work done by the gas. (g) Calculate the total work done in the cycle.   


Answer:  (a) Let initial pressure =pₒ, final pressure = p, initial volume = Vₒ =1 liter, final volume =V, initial temperature = T = 300 K. 

Since the process is sudden it is adiabatic, so  

pₒVₒˠ = pVˠ

→p/pₒ  = (Vₒ/V)ˠ =2ˠ = 21.5 =2√2.


(b) Given that pₒ =100 kPa, so

p =2√2pₒ =200√2 kPa. Now work done in this process, 

=(pₒVₒ-pV)/(ɣ-1)

=(pₒVₒ-2√2pₒ*Vₒ/2)/(1.5-1)

=pₒVₒ(1-√2)*2

=(100*1000)*(1000*10⁻⁶)*(-0.41)*2 J

= -82 J.         


(c) Since dQ = dU+dW, here dQ =0, so 

dU+dW =0

→dU = -dW = -(-82 J) = 82 J.

  

(d) For an adiabatic process 

TₒVₒˠ-1 = TVˠ-1

→T = Tₒ(Vₒ/V)ˠ-1   

→T = 300*(2)1.5-1 K

→T = 300*√2 K = 424 K.

   

(e) In this process, the pressure is constant which is equal to p =200√2 kPa and the volume at the beginning of the process V =0.5 liters. Let the volume at the end = V'. Initial temperature T = 424 K {as above in (d)}, final temperature T' = 300 K.

Now for this process

V/T =V'/T'

→V' =V*(T'/T) =0.5*(300/424) 

       =0.354 liter =354 cm³

       =354x10⁻⁶ m³
       =3.54x10⁻⁴  m³.

V = 0.5 liter =5x10⁻⁴ m³

Now the work-done during the process =p*dV =p*(V' -V)

= (200√2*1000)*(3.54 -5.0)*10⁻⁴ J

= -41.3 J.


(f) For this isothermal process, the initial volume V = 3.54x10⁻⁴ m³, initial pressure p =200√2 kPa, final volume V' = 1 liter =1000 cm³ =1x10⁻³ m³. Temperature T =300 K.

Now pV =nRT and the work-done in an isothermal process is

dW =nRT*ln(V'/V)

    =pV*ln(1x10⁻³/3.54x10⁻⁴) J

    =200√2*1000*3.54x10⁻⁴*ln(10/3.54) J

=20*3.54*√2*1.03 J

=103 J.            


(g) The total work done in the cycle 

 =dWₐ +dWᵦ +dW₍

= -82 J + (-41.3 J) + 103 J 

= -20.3 J   




  

 

   26. Figure (27-E4) shows a cylindrical tube with adiabatic walls and fitted with an adiabatic separator. The separator can be slid into the tube by an external mechanism. An ideal gas (ɣ =1.5) is injected in the two sides at equal pressures and temperatures. The separators remain in equilibrium in the middle. It is now slid to a position where it divides the tube in the ratio 1:3. Find the ratio of the temperatures in the two parts of the vessel. 
The figure for Q - 26
   

Answer:  Let the volume of the tube = V. So the initial volume of each side Vₒ =V/2. Assume the initial temperature =T. 

      After the separator is slid to a new position let the temperature of the right part =T' and that of left part =T". Now the volume of the right part V' =V/4 and that of the left part V" = 3V/4.   

    Since the process in both the parts is adiabatic, For the right part 

T'V'ˠ-1 = TVₒˠ-1 

      And for the left part,

T"V"ˠ-1 = TVₒˠ-1

Hence T'V'ˠ-1 =T"V"ˠ-1

→T'/T" = (V"/V')ˠ-1 

        ={(3V/4)/(V/4)}1.5-1

        = √3

So, T':T" = √3:1 




    

   27. Figure (27-E5) shows two rigid vessels A and B, each of volume 200 cm³ containing an ideal gas (Cᵥ =12.5 J/mol-K). The vessels are connected to a manometer tube containing mercury. The pressure in both the vessels is 75 cm of mercury and the temperature is 300 K. (a) Find the number of moles of the gas in each vessel. (b) 5.0 J of heat is supplied to the gas in the vessel A and 10 J to the gas in the vessel b. Assuming no appreciable transfer of heat from A to B calculate the difference in the heights of mercury in the two sides of the manometer. Gas constant R = 8.3 J/mol-K.
The figure for Q - 27


Answer:  Volume of each vessel, V =200 cm³ =200x10⁻⁶ m³ =2x10⁻⁴ m³.

Pressure in each vessel, p= 75 cm of mercury =ρhg

=(13600 kg/m³)*(0.75 m)(9.8 m/s²) Pa

=99960 Pa 

T = 300 K

Cᵥ = 12.5 J/mol-K, R =8.3 J/mol-K


(a) Number of moles of each gas 

n =pV/RT =99960*2x10⁻⁴/(8.3*300) mol

=0.008.       


(b) Let the difference of height of mercury in the manometer = h cm =h/100 m.

Pressure due to this height,

pₙ =ρhg/100 Pa.

The equilibrium of pressures is,

pₐ +pₙ = pᵦ --------------- (i)

Now for vessel A, dQ = 5 J 

dQ = nCᵥdT 

→dT =5/(0.008*12.5) = 50 K

Thus the temperature of A, T' = 300+50 =350 K. 

For vessel B, dQ = 10 J 

→dT =10/(0.008*12.5) =100 K.

Thus the temperature of B, T"  =300+100 =400 K.      

From equation (i) 

nRT'/V + ρgh/100= nRT"/V

→ρgh =100*nR(T" -T')/V 

→ρgh =100*0.008*8.3*(50)/2x10⁻⁴ cm

→h =332/(2x10⁻⁴*13600*9.8)

      =12.5 cm.   

  



     

   28. Figure (27-E6) shows two vessels with adiabatic walls, one containing 0.1 g of helium (ɣ = 1.67, M = 4 g/mol) and the other containing some amount of hydrogen (ɣ = 1.4, M = 2 g/mol). Initially, the temperatures of the two gases are equal. The gases are electrically heated for some time during which equal amounts of heat are given to two gases. It is found that the temperatures rise through the same amount in the two vessels. Calculate the mass of hydrogen.
The figure for Q-28


Answer:  Let the mass of hydrogen = m, so the number of moles of hydrogen n= m/2.

The number of moles of helium, n' =0.1/4 =0.025.

Since Cᵥ =R/(ˠ-1), 

For helium Cᵥ' =R/(ˠ'-1)    

Now dQ =n'Cᵥ'dT

For hydrogen, Cᵥ =R/(ˠ-1)  

Here dQ =nCᵥdT.

dQ and dT are same for the two gases.

So, n'Cᵥ'dT =nCᵥdT  

→n =n'Cᵥ'/Cᵥ

→m/2 =0.025*{R/(ˠ'-1)}/{R/(ˠ-1)}

→m =2*0.025*(ˠ-1)/(ˠ'-1)

       =0.05*(1.4-1)/(1.67-1)

       =0.05*0.40/0.67 

       = 0.03 g.            




 

   29. Two vessels A and B of equal volume Vₒ are connected by a valve. The vessels are fitted with pistons which can be moved to change the volumes. Initially, the valve is open and the vessels contain an ideal gas (Cₚ/Cᵥ =ɣ) at atmospheric pressure pₒ and atmospheric temperature Tₒ. The walls of the vessel A is diathermic and the those of B are adiabatic. The valve is now closed and the pistons are slowly pulled out to increase the volumes of the vessels to double the original value. (a) Find the temperatures and pressures in the two vessels. (b) The valve is now opened for sufficient time so that the gases acquire a common temperature and pressure. Find the new values of temperature and pressure.

    

Answer:  (a) For the vessel A, the temperature will remain the same  Tₒ because of the diathermic wall.  

And pV = pₒVₒ

→p*2Vₒ = pₒVₒ    

→p = pₒ/2. 


For vessel B:-

It is an adiabatic process. Hence,

pVˠ = pₒVₒˠ 

→p*(2Vₒ)ˠ = pₒVₒˠ

→p = pₒ(Vₒ/2Vₒ)ˠ =pₒ/2ˠ. 

If T is final temperature, 

TVˠ-1 = TₒVₒˠ-1

→T = Tₒ(Vₒ/2Vₒ)ˠ-1 =Tₒ/2ˠ-1.       

 

(b) Since the valve is opened for sufficient time, the temperature will be that of the atmosphere equal to Tₒ due to the diathermic wall in vessel A. Considering both the vessel together, original volume =2Vₒ, and the final volume =4Vₒ. If the final pressure = p' then,

p'*4Vₒ = pₒ*2Vₒ

→p' = pₒ/2.      

  

 


  

   30. Figure (27-E7) shows an adiabatic cylindrical tube of volume Vₒ divided into two parts by a frictionless adiabatic separator. Initially, the separator is kept in the middle, an ideal gas at pressure p₁ and temperature T₁ is injected into the left part and another ideal gas at pressure p₂ and temperature T₂ is injected into the right part. Cₚ/Cᵥ =ɣ is the same for both gases. The separator is slid slowly and released at a position where it can stay in equilibrium. Find (a) the volumes of the two parts, (b) the heat given to the gas in the left part, and (C) the final common pressure of the gases.
The figure for Q-30


Answer:  (a) Initial volume of each part =V₀/2. Pressures and temperatures of left and right parts are p₁, T₁ and p₂, T₂ respectively. 

Let the final volume of the left part = V' and that of right part = V". Assume final common pressure = p. It is an adiabatic process hence for the left part, 

p₁(Vₒ/2)ˠ = pV'ˠ

→p = p₁(Vₒ/2V')ˠ ---------- (i)

Since V" +V' = Vₒ → V" =Vₒ - V'

For the right part,

pV"ˠ = p₂(Vₒ/2)ˠ

→p =p₂(Vₒ/2V")ˠ

Equating the two values of p,

p₁(Vₒ/2V')ˠ = p₂(Vₒ/2V")ˠ

→(1/2V')ˠ/(1/2V")ˠ = p₂/p₁

→(V"/V')ˠ = p₂/p₁

→{(Vₒ -V')/V'}ˠ = p₂/p₁

→Vₒ/V' -1 =(p₂/p₁)1/ˠ 

→Vₒ/V' = 1 + (p₂/p₁)1/ˠ 

→V' = Vₒ/{1+(p₂/p₁)1/ˠ}

      = Vₒ*p₁1/ˠ/(p₁1/ˠ +p₂1/ˠ) 

      = p₁¹/ˠVₒ/A

Where  A = p₁¹/ˠ + p₂¹/ˠ  

Now V" = Vₒ - V' =Vₒ -p₁1/ˠVₒ/A

            =(A-p₁1/ˠ)Vₒ/A   

            = p₂¹/ˠVₒ/A


(b) Since the walls of the cylinder and the separator are adiabatic, there will be no heat transfer. Hence the heat transferred to the left part = zero.


(c) From (i) the common pressure,

p = p₁(Vₒ/2V')ˠ, put the value of V'

   =p₁(VₒA/2p₁1/ˠVₒ)ˠ

   =Aˠ*p₁/p₁2ˠ  

   = (A/2)ˠ, where A = p₁¹/ˠ + p₂¹/ˠ  

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CHAPTER- 20 - Dispersion and Spectra


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CHAPTER- 14 - Fluid Mechanics



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