Showing posts with label Chapter 27. Show all posts
Showing posts with label Chapter 27. Show all posts

Thursday, August 13, 2026

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

Specific Heat Capacities of Gases

Questions for Short Answer


  1. Does a gas have just two specific heat capacities or more than two? Is the number of specific heat capacities of a gas countable?


Answer:  A gas has more than two specific heat capacities. The specific heat capacity depends on the process involved. Since there are innumerable processes, so the number of specific heat capacities of a gas is not countable.



  2. Can we define specific heat capacity at constant temperature?


Answer:  The specific heat capacity is defined as the amount of heat supplied to unit mass per unit rise in temperature for a process. If ΔQ is the heat supplied to mass m and the rise in temperature is ΔT, then the specific heat capacity, s = ΔQ/(m.ΔT)

            If the temperature is constant, ΔT = 0. In this case, s becomes infinite. So the specific heat capacity is not defined at a constant temperature.   




  3. Can we define specific heat capacity for an adiabatic process?


Answer:  In an adiabatic process, the heat supplied is zero, i.e., ΔQ = 0. Hence, s = 0. So for an adiabatic process, the specific heat capacity has only one value equal to zero. 


  4. Does a solid also have two kinds of molar heat capacities Cₚ and Cᵥ? If yes, do we have Cₚ>Cᵥ? Cₚ-Cᵥ=R? 


Answer:  Yes, a solid also has two kinds of molar heat capacities, Cₚ and Cᵥ. But the change in volume is quite small. So still Cₚ>Cᵥ, but Cₚ-Cᵥ is very small and not equal to R.



  5. In a real gas, the internal energy depends on temperature and also on volume. The energy increases when the gas expands isothermally. Looking into the derivation of Cₚ-Cᵥ=R, find whether Cₚ-Cᵥ will be more than R, less than R, or equal to R for a real gas.


Answer:  For an ideal gas, the internal energy depends only on its temperature. The equation derived is

(dQ)ₚ = (dQ)ᵥ + nRdT

But for the real gas, the internal energy also increases when the gas expands isothermally. Let the increase due to volume expansion at constant pressure = K.

Now the derived equation changes to,

(dQ)ₚ = (dQ)ᵥ +nRdT +K

Dividing by n.dT, we get

(dQ)ₚ/ndT = (dQ)ᵥ/ndT + R + K/ndT

→Cₚ = Cᵥ + R + K/ndT
→Cₚ - Cᵥ = R + K/ndT
K/ndT is positive.
Hence, for the real gas, Cₚ - Cᵥ will be more than R. 

 



  6. Can a process on an ideal gas be both adiabatic and isothermal?


Answer:  We have ΔQ = ΔU + ΔW

 →ΔQ = nCᵥ.ΔT + ΔW

If we assume that the process is both adiabatic and isothermal, then due to the adiabatic nature, ΔQ = 0 and due to the isothermal nature, ΔT = 0. Now the equation becomes

0 = nCᵥ(0) + ΔW

→ΔW = 0

which is not true. So our assumption is also not true, and the process can not be both adiabatic and isothermal.


  


  7. Show that the slope of the p-V diagram is greater for an adiabatic process as compared to an isothermal process.


Answer:  For the isothermal process

pV = K (constant)

Differentiating with respect to V, we get

p + V*dp/dV = 0

→V*dp/dV = -p

→dp/dV = -p/V

dp/dV is the slope of the p-V diagram.

For an adiabatic process

pVɣ= K' (constant)
on differentiating w.r.t. V,
Vɣ*dp/dV + pɣV(ɣ-1) = 0
dp/dV = -pɣV⁻¹ =(-p/V)*ɣ
So the slope dp/dV is ɣ times greater in the adiabatic process. Since ɣ>1, the slope of the p-V diagram is greater for an adiabatic process as compared to an isothermal process. 



  8. Is a slow process always isothermal? Is a quick process always adiabatic?


Answer:  Theoretically, the slow process is not always isothermal.  If the walls are perfectly insulated, even slow processes will have no heat transfer. It will still be adiabatic. Practically, no walls are perfectly insulated, so a very slow process will make it nearly isothermal. So practically a slow process is always isothermal. A quick process will have no time to exchange heat with the environment. So it will always be adiabatic.



 

  9. Can two states of an ideal gas be connected by an isothermal process as well as an adiabatic process?


Answer:  Let the two states of the gas have pressures and volumes as (p, V) and (p', V'), respectively. If the two states are connected by an isothermal process, then,

pV = p'V' ----------- (i)

In another case, if the two states are connected by an adiabatic process, then,

pVɣ = p'V'ɣ ---------- (ii)

Dividing (ii) by (i), we have
V(ɣ-1) = V'(ɣ-1)
→V = V'
So the two states can be connected by adiabatic as well as isothermal processes only if the initial and the final volumes are the same.


  10. The ratio Cₚ/Cᵥ for a gas is 1.29. What is the degree of freedom of the molecules of the gas?


Answer:  Cₚ/Cᵥ = 1.29 =129/100 ≈9/7

If f is the degree of freedom, then
ɣ = 1 +2/f 
→1 + 2/f = 9/7
→2/f = 9/7 - 1 =2/7
→f = 7

So the degree of freedom of molecules of the gas is 7. It is the case for diatomic gases when the molecules vibrate.      




OBJECTIVE-I


   1. Work done by a sample of an ideal gas in process A is double the work done in another process B. The temperature rises through the same amount in the two processes. If CA and CB be the molar heat capacities for the two processes,

(a) CA = CB

(b) CA < CB

(c) CA > CB

(d) CA and CB cannot be defined.   


Answer:  (c)   


EXPLANATION: From the statement of the first law of thermodynamics,

ΔQ = ΔU + ΔW.

Since the temperature rises through the same amount in both processes, ΔU is the same for both processes. 

Given (ΔW)ₐ = 2*(ΔW)ᵦ

Hence, (ΔQ)ₐ > (ΔQ)ᵦ
Molar heat capacity C ={ΔQ/n.ΔT}
n.ΔT is the same in both the processes; hence,  CA > CB. Option (c).




   2. For a solid with a small expansion coefficient,

(a) Cₚ - Cᵥ = R

(b) Cₚ = Cᵥ

(c) Cₚ is slightly greater than Cᵥ

(d) Cₚ is slightly less than Cᵥ. 


Answer:  (c)   


EXPLANATION: Since the expansion coefficient is small, there will be a change in the volume, but very small. So a very small amount of heat supplied will be used for the work done in expansion. Hence Cₚ is slightly greater than Cᵥ.

Option (c).





  3. The value of Cₚ - Cᵥ is 1.00R for a gas sample in state A and 1.08R in state B. Let pA, pB denote the pressures and TA and TB denote the temperatures of the states A and B, respectively. Most likely

(a) pA < pB and  TA > TB 

(b) pA > pB and  TA < TB 

(c) pA = pB and  TA < TB 

(d) pA > pB and  TA = TB  


Answer:  (a)   


EXPLANATION: The relation Cₚ - Cᵥ =1.00R shows that it behaves as an ideal gas at state A. But at state B, Cₚ - Cᵥ =1.08R, so in this state, the gas is not behaving as an ideal gas but as a real gas. A real gas behaves nearly as an ideal gas when it is at low pressure and high temperature. Hence, in state A, the given sample of the gas has low pressure and high temperature. So the option (a) is true.





  4. Cᵥ and Cₚ denote the molar heat capacities of an ideal gas at constant volume and constant pressure, respectively. Which of the following is a universal constant?

(a) Cₚ/Cᵥ

(b) CₚCᵥ
(c) Cₚ - Cᵥ
(d) Cₚ + Cᵥ.

  


Answer:  (c)   


EXPLANATION: The product and sum of Cₚ and Cᵥ are not constant, so the options (b) and (d) are not true. Cₚ/Cᵥ=ɣ is a constant but not a universal constant because ɣ depends on the atomicity of the gas, so the option (a) is also not true. Cₚ -Cᵥ = R, which is the universal gas constant with a value of 8.314 J/mol-K. Hence, option (c) is true.




  5. 70 calories of heat are required to raise the temperature of 2 mol of an ideal gas at constant pressure from 30°C to 35°C. The amount of heat required to raise the temperature of the same gas through the same range at constant volume is 

(a) 30 calories

(b) 50 calories

(c) 70 calories

(d) 90 calories.   


Answer:  (b)   


EXPLANATION: ΔQ = nCₚ.ΔT and

ΔQ' = nCᵥΔT. 


Here ΔQ = 70 cal =70*4.2 J, n = 2 mol, ΔT = 5°K, so Cₚ = 70*4.2/2*5 =7*4.2 J/mol-K.
Now Cᵥ = Cₚ - R =7*4.2 -8.3 =29.4-8.3 =21.1 J/mol-K.
Now ΔQ' =nCᵥ*ΔT =2*21.1*5 =211 J =211/4.2 cal ≈50 cal. Hence option (b).

Note: You may not convert cal into J and write the unit of R as cal/mol-K.




  6. Figure (27-Q1) shows a process on a gas in which pressure and volume both change. The molar heat capacity for this process is C.
Figure for Q - 6

(a) C = 0

(b) C = Cᵥ

(c) C > Cᵥ

(d) C < Cᵥ    




Answer:  (c)   


EXPLANATION: Suppose the given process takes the state of the gas from A to B; the difference in temperature is dT and the heat involved is dQ. The difference in internal energy = dU.

dQ = nC.dT

If the work done in this process = dW, then from the first law of thermodynamics, dQ =dU + dW ---- (i) 

       Now consider a process on the same amount of gas at a constant volume that takes the temperature of the gas through the same difference. Hence the heat supplied in this process =dQ' =nCᵥ.dT

The work done in this process is zero due to constant volume. Hence dU = dQ'.

From (i), dQ = dU + dW

→dQ =dQ' +dW

→nC.dT = nCᵥdT +dW

→C = Cᵥ +dW/ndT

The area under p-V diagram in the given figure is positive hence, C > Cᵥ. Option (c) is true.



 

  7. The molar heat capacity for the process shown in figure (27-Q2) is
Figure for Q - 7

(a) C = Cₚ

(b) C = Cᵥ 

(c) C > Cᵥ

(d) C = 0.   


Answer:  (d)   


EXPLANATION: In the given figure

p =K/Vɣ

→pVɣ = K, which shows that it is an adiabatic process in which the heat supplied = zero. The molar heat capacity is defined as the amount of heat required per mol of the gas to raise the temperature through 1K.     Since no heat is being given here to raise the temperature, the molar heat capacity is zero. Option (d). 





  8. In an isothermal process on an ideal gas, the pressure increases by 0.5%. The volume decreases by about

(a) 0.25%

(b) 0.5%

(c) 0.7%

(d) 1%   


Answer:  (b)   


EXPLANATION: In an isothermal process pV = constant. Hence, if p increases by 0.5%, V will decrease by 0.5% to keep the product of the new values of p and V constant. Hence, the option (b).



 


  9. In an adiabatic process on a gas with ɣ =1.4, the pressure is increased by 0.5%. The volume decreases by about

(a) 0.36%

(b) 0.5%

(c) 0.7%

(d) 1%.   


Answer:  (a)   


EXPLANATION: In an adiabatic process, pVˠ = K. So if p is increased by 0.5%, Vˠ should be decreased by 0.5% to keep the product constant. Hence now, V'ˠ =Vˠ/1.005

→V' = V/(1.005)(1/ˠ) 
       =V/(1.005)(1/1.4)
       =V/1.0036 

Which means the volume decreases by 0.36%. Option (a) is true. 




 

  10. Two samples A and B are initially kept in the same state. Sample A is expanded through an adiabatic process and sample B through an isothermal process. The final volumes of the samples are the same. The final pressures in A and B are pA and pB, respectively.  

(a)  pA > pB 

(b)  pA = pB 

(c)  pA< pB  

(d) The relation between pA and pB cannot be deduced.   

 


Answer:  (c)   


EXPLANATION: In the adiabatic process pVˠ = K and in the isothermal process pV =K'.

Now pVˠ = pA V'ˠ and pV =  pB V'

Dividing we get, V(ˠ-1) =(pA/pB)V'(ˠ-1)

Hence pA/pB =(V/V')(ˠ-1) 

The gas is expanded, so V' > V, so

pA/pB < 1

→pA < pB.

Option (c) is true. 



   


  11. Let Tₐ and Tᵦ be the final temperatures of samples A and B, respectively, in the previous question.

(a) Tₐ < Tᵦ

(b) Tₐ = Tᵦ

(c) Tₐ > Tᵦ

(d) The relation between Tₐ and Tᵦ cannot be deduced.  


Answer:  (a)   


EXPLANATION: Since sample B goes through the isothermal process, its initial and final temperatures will be the same and equal to Tᵦ. Initially, both the samples are at the same temperature; hence, the initial temperature of sample A = Tᵦ. This sample expands adiabatically; hence, the work done by the gas is at the expense of its internal energy. Hence the final temperature of this sample will decrease, so Tₐ < Tᵦ. Option (a) is true.





  12. Let ΔWₐ and ΔWᵦ be the work done by systems A and B, respectively, in the previous question.

(a) ΔWₐ > ΔWᵦ

(b) ΔWₐ = ΔWᵦ

(c) ΔWₐ < ΔWᵦ

(d) The relation between ΔWₐ and ΔWᵦ cannot be deduced.   


Answer:  (c)   


EXPLANATION: For sample A, which goes through an adiabatic process, ΔQ = 0. Since ΔQ = ΔU + ΔW, here,

0 = ΔU + ΔWₐ  

→ΔWₐ = -ΔU

For sample B, ΔQ = ΔU' + ΔWᵦ

Since the process on B is isothermal, the temperature does not change. Thus there is no change in the internal energy. ΔU' = 0. So ΔWᵦ = ΔQ.

ΔWₐ has a negative value while ΔWᵦ has a positive value. Clearly, ΔWₐ < ΔWᵦ. Hence, option (c) is true.

 

  13. The molar heat capacity of oxygen gas at STP is nearly 2.5R. As the temperature is increased, it gradually increases and approaches 3.5R. The most appropriate reason for this behavior is that at high temperatures

(a) oxygen does not behave as an ideal gas

(b) oxygen molecules dissociate in atoms

(c) the molecules collide more frequently

(d) molecular vibrations gradually become effective.   


Answer:  (d)   


EXPLANATION: As the temperature goes high, the molecular vibrations gradually become effective. The molecules start vibrating about their mean positions. So a part of the heat given is utilized for this vibration, and more heat is required to raise the temperature through the same degree than at the STP. Hence, the molar heat capacity is more at high temperatures. Option (d) is true.    




OBJECTIVE-II


   1. A Gas kept in a container of finite conductivity is suddenly compressed. The process 

(a) must be very nearly adiabatic 

(b) must be very nearly isothermal

(c) may be very nearly adiabatic

(d) may be very nearly isothermal.

 


Answer:  (c), (d).   


EXPLANATION: Since it is not clear how much the conductivity of the container is and how suddenly the gas is compressed, there may be two conditions.

First, the thermal conductivity is very low, and the compression is very fast. In this situation, there will be negligible heat transfer through the container. The process may be very nearly adiabatic. Option (c).

Second, the conductivity is very high, and the sudden compression is not very fast. In this case, the temperature approximately remains constant. The process is very nearly isothermal. Option (d).  



 


   2. Let Q and W denote the amount of heat given to an ideal gas and the work done by it in an isothermal process,

(a) Q = 0,

(b) W = 0,

(c) Q ≠ W,

(d) Q =W. 


Answer:  (d).   


EXPLANATION: From the first law of thermodynamics, dQ = dU + dW.

In an isothermal process, the temperature remains constant and so does the internal energy U. So dU = 0. Thus, dQ = dW. Hence, for the question, Q = W. Option (d) is correct.  





   3. Let Q and W denote the amount of heat given to an ideal gas and the work done by it in an adiabatic process,

(a) Q = 0

(b) W = 0,

(c) Q = W,

(d) Q ≠ W. 

 


Answer:  (a), (d).   


EXPLANATION: For an adiabatic process, there is no heat transfer; hence Q = 0. Thus the first law of thermodynamics for this process is,

 0 = U + W

→W = -U. W is not zero.
Hence the options (b) and (c) are not true. The options (a) and (d) are true.





   4. Consider the processes A and B, shown in Figure (27-Q3). Is it possible that
The figure for Q - 4

(a) both the processes are isothermal

(b) both the processes are adiabatic

(c) A is isothermal and B is adiabatic

(d) A is adiabatic and B is isothermal.  


Answer:  (c)   


EXPLANATION: Both the processes start from the same state; hence, both of them cannot be either isothermal or adiabatic. In that case, both will traverse the same path. So the options (a) and (b) are not true.

        Since one of them is isothermal and the other adiabatic, the steeper slope path B will be adiabatic, and the other A will be isothermal. Hence, option (c) is true.






   5. Three identical adiabatic containers A, B and C contain helium, neon and oxygen respectively at equal pressure. The gases are pushed to half their initial volumes.

(a) The final temperatures in the three containers will be the same. 

(b) The final pressures in the three containers will be the same. 

(c) The pressures of helium and neon will be the same but that of oxygen will be different. 

(d) The temperatures of helium and neon will be the same, but that of the oxygen will be different. 


Answer:  (c), (d).   


EXPLANATION: Helium and Neon are monatomic gases for which Cₚ/Cᵥ =ɣ =1.67, but oxygen is a diatomic gas for which ɣ = 1.4.

     For an adiabatic process, pVɣ = constant or TV(ɣ-1) =constant.

      Hence the changes for helium and neon will be the same, but for oxygen, it will be different. So the options (c) and (d) are true.





   6. A rigid container of negligible heat capacity contains one mole of an ideal gas. The temperature of the gas increases by 1°C if 3.0 cal of heat is added to it. The gas may be

(a) helium

(b) argon

(c) oxygen

(d) carbon dioxide. 


Answer:  (a), (b).   


EXPLANATION: From the given situation, at constant volume, one mole of the ideal gas requires 3.0 cal of heat to increase its temperature by 1°C, which is by definition, is the molar heat capacity of the gas at constant volume, Cᵥ. So, Cᵥ = 3.0 cal/mol-K =3.0*4.18 J/mol-K = 12.5 J/mol-K.

    We know that Cₚ/Cᵥ =ɣ, 
→Cₚ =ɣCᵥ 
Also, Cₚ - Cᵥ = R
→ɣCᵥ - Cᵥ = R
→Cᵥ = R/(ɣ-1)
For a monoatomic gas ɣ = 1.67, thus
Cᵥ = R/0.67 =1.5R =1.5*8.314 J/mol-K =12.5 J/mol-K.
So the given gas is monatomic. Out of the four options, only helium and argon are monatomic because they are inert gases. Hence the options (a) and (b) are true.





   7. Four cylinders contain equal numbers of moles of argon, hydrogen, nitrogen, and carbon dioxide at the same temperature. The energy is minimum in

(a) argon

(b) hydrogen

(c) nitrogen

(d) carbon dioxide. 


Answer:  (a).   


EXPLANATION: Taking the energy to be zero at T = 0, the energy of the gas at temperature T is given as

U = nCᵥT

Since in the given four cylinders, n and T are the same for all the gases, the energy will be minimum for the gas for which Cᵥ is minimum. 

    Since Cᵥ =R/(ɣ-1), it will be minimum for which ɣ is maximum. We know that ɣ is maximum for a monoatomic gas and decreases as the atomicity increases. Out of the four options, only argon is monoatomic, and the others are either diatomic or triatomic. So Cᵥ is a minimum for argon. Thus, option (a) is true.     



EXERCISES


   1. A vessel containing one mole of a monatomic 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

Work 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 monatomic ideal gas in a process in which the gas performs a work Q/2 on its surroundings. 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.  



   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 moles of the first gas and 2n moles of the second gas are 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 a 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 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 surroundings. (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 the 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.    



    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 temperature 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 an 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, 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 the initial pressures be pₐ, pᵦ and p₍ and the final pressure =p.

Let the initial volume be 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, the 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 the 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 in 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 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 the 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 vessels together, the 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 the right part = V". Assume the 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     



   31. An adiabatic cylinder tube of cross-sectional area 1 cm² is closed at one end and fitted with a piston at the other end. The tube contains 0.03 g of an ideal gas. At 1 atm pressure and at the temperature of the surroundings, the length of the gas column is 40 cm. The piston is suddenly pulled out to double the length of the column. The pressure of the gas falls to 0.355 atm. Find the speed of sound in the gas at atmospheric temperature.      


Answer:  (a) m = 0.03 g =3x10⁻⁵ kg, 

Area of the piston, A = 1 cm² =1x10⁻⁴ m², 

Length, L = 40 cm =0.40 m. 

Initial volume, V =AL

       =1x10⁻⁴*0.40 =4x10⁻⁵ m³.

Initial density, ρ =m/V

=3x10⁻⁵/4x10⁻⁵ kg/m³

=0.75 kg/m³  

Initial pressure, p =1atm =1x10⁵ Pa.

Final volume, V' = 2V

Final pressure, p' = 0.355 atm

If Cₚ/Cᵥ =ˠ for the gas, in the above adiabatic process,

p'V'ˠ = pVˠ

→(2V)ˠ = (p/p')Vˠ

→2ˠ = (1/0.355) = 2.817


Taking log to both sides,
ˠln2 =ln2.817
→ˠ =ln2.817/ln2 = 1.5


The speed of sound in a gas is given as,

v = √(ˠp/ρ)

   =√{1.5*1x10⁵/0.75}

   =√2x10⁵

   =447 m/s. 

 



 

    32. The speed of sound in hydrogen at 0°C is 1280 m/s. The density of hydrogen at STP is 0.089 kg/ m³. Calculate the molar heat capacities Cₚ and Cᵥ of hydrogen.   


Answer:  The density of hydrogen at STP,

 ρ =0.089 kg/m³.

     Speed of sound at this temperature,

v =1280 m/s, but,

v =√(ˠp/ρ)

→ˠ =ρv²/p =0.089*1280²/10⁵ =1.46


Since R = 8.3 J/mol-K,

Cᵥ = R/(ˠ-1) =8.3/0.46 

    = 18.0 J/mol-K.


And Cₚ = ˠCᵥ =1.46*18.0 
           = 26.3 J/mol-K. 


 

 

   33. 4.0 g of helium occupies 22400 cm³ at STP. The specific heat capacity of helium at constant pressure is 5.0 cal/mol-K. Calculate the speed of sound in helium at STP.   


Answer:  Density of helium at STP,

ρ = 4*10⁶/22400*1000

   = 0.179 kg/m³ 


Cₚ =5.0 cal/mol-K

   =5.0*4.2 J/mol-K

   = 21.0 J/mol-K. 


Since Cₚ =ˠR/(ˠ-1)

→21 = ˠ*8.3/(ˠ-1)

→21ˠ -21 = 8.3ˠ

→12.7ˠ =21

→ˠ = 21/12.7 =1.65


At STP, the pressure of helium,

p = 10⁵ Pa.


Speed of sound in helium at STP,

v = √(ˠp/ρ)

 =√(1.65*10⁵/0.179)

 =960 m/s.   

 

 

   34. An ideal gas having density 1.7x10⁻³ g/cm³ at a pressure 1.5x10⁵ Pa is filled in a Kundt's tube. When the gas is resonated at a frequency of 3.0 kHz, nodes are formed at a separation of 6.0 cm. Calculate the molar heat capacities Cₚ and Cᵥ of the gas.   


Answer:  The resonating frequency,

f = 3.0 kHz =3000 Hz.


Pressure, p = 1.5x10⁵ Pa


Since the nodes are formed at a separation of 6.0 cm, it means,

λ/2 = 6.0 cm = 0.06 m, where λ is the wavelength of sound.

→λ = 2*0.06 =0.12 m. 


Hence speed of sound, v =fλ

→v = 3000*0.12 m/s

     = 360 m/s.


Given ρ =1.7x10⁻³ g/cm³

 =1.7x10⁻³*10⁶/1000 kg/m³

 =1.7 kg/m³

 

Now the speed of sound in gas is given as, v =√(ˠp/ρ)

→ˠ = ρv²/p

     = 1.7*360²/1.5x10⁵

     = 1.47  


Now Cᵥ = R/(ˠ-1) =8.3/0.47

           =17.7 J/mol-K.


And Cₚ =ˠ*Cᵥ = 1.47*17.7

          =26.0 J/mol-K.


 

 

   35. Standing waves of frequency 5.0 kHz are produced in a tube filled with oxygen at 300 K. The separation between the consecutive nodes is 3.3 cm. Calculate the specific heat capacities Cₚ and Cᵥ of the gas.   


Answer:  The frequency of sound waves, f = 5.0 kHz = 5000 Hz.

   The separation between the consecutive nodes = 3.3 cm = 3.3x10⁻² m. If λ is the wavelength of the sound waves, then 

λ/2 = 3.3x10⁻² m

→λ = 6.6x10⁻² m.


Hence the speed of sound in the given sample of oxygen, v = fλ.

→v =5000*6.6x10⁻² m/s 

      = 330 m/s.


Given T = 300 K.


Now, ρ = m/V           


Since v =√(ˠp/ρ)

→v² = ˠ*(nRT/V)/(m/V)

       = ˠRT(n/m)

       = ˠRT(m/Mm)

       = ˠRT/M

→ˠ = Mv²/RT

 {For oxygen, M = 32 g =32x10⁻³ kg} 

     = 32x10⁻³*330²/(8.3*300)

     = 1.4    


Now Cₚ = ˠR/(ˠ-1)

    = 1.4*8.3/(1.4-1)

    = 29.0 J/mol-K.


And Cᵥ = Cₚ/ˠ =29.0/1.4

          = 20.7 J/mol-K.        


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Links to the Chapters



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CHAPTER-6 - Friction

Questions for Short Answers

Objective - I

Objective - II

Exercises - Q 1 to Q 10

Exercises - Q 11 to Q 20

Exercises - Q 21 to Q 31

CHAPTER-7 - Circular Motion

Questions for Short Answers

Objective - I

Objective - II

Exercises - Q 1 to Q 10

Exercises - Q 11 to Q 20

Exercises - Q 21 to Q 30

CHAPTER-8 - Work and Energy

Questions for Short Answers

Objective - I

Objective - II

Exercises - Q 1 to Q 10

Exercises - Q 11 to Q 20

Exercises - Q 21 to Q 30

Exercises - Q 31 to Q 42

Exercises - Q 43 to Q 54

Exercises - Q 55 to Q 64


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CHAPTER- 9 - Center of Mass, Linear Momentum, Collision

Questions for Short Answers

Objective - I

Objective - II

EXERCISES Q-1 TO Q-10

EXERCISES Q-11 TO Q-20

EXERCISES Q-21 TO Q-30

EXERCISES Q-31 TO Q-42

EXERCISES Q-43 TO Q-54

EXERCISES Q-55 TO Q-64

CHAPTER- 10 - Rotational Mechanics

Questions for Short Answers

OBJECTIVE - I

OBJECTIVE - II

EXERCISES Q-01 TO Q-15

EXERCISES Q-16 TO Q-30

EXERCISES Q-31 TO Q-45

EXERCISES Q-46 TO Q-60

EXERCISES Q-61 TO Q-75

EXERCISES Q-76 TO Q-86


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CHAPTER- 11 - Gravitation


Questions for Short Answers

OBJECTIVE - I

OBJECTIVE - II

EXERCISES Q-01 TO Q-10

EXERCISES Q-11 TO Q-20

EXERCISES Q-21 TO Q-30

EXERCISES Q-31 TO Q-39 (With Extra 40th problem)

CHAPTER- 12 - Simple Harmonic Motion 


Questions for Short Answers 


OBJECTIVE - I

OBJECTIVE-II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-30

EXERCISES - Q-31 TO Q-40

EXERCISES - Q-41 TO Q-50

EXERCISES - Q-51 TO Q-58 with EXTRA QUESTIONS Q-59 and Q-60 


CHAPTER- 13 - Fluid Mechanics 


Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-30

EXERCISES - Q-31 TO Q-35



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CHAPTER- 14 - Some Mechanical Properties of Matter 


Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q -11 TO Q -20

EXERCISES - Q -21 TO Q -32

CHAPTER- 15 - Wave Motion and Waves on a String

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-30

EXERCISES - Q-31 TO Q-40

EXERCISES - Q-41 TO Q-50

EXERCISES - Q-51 TO Q-57

CHAPTER- 16 - Sound Waves

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-30

EXERCISES - Q-31 TO Q-40

EXERCISES - Q-41 TO Q-50

EXERCISES - Q-51 TO Q-60

EXERCISES - Q-61 TO Q-70 

EXERCISES - Q-71 TO Q-80

EXERCISES - Q-81 TO Q-89

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CHAPTER- 17 - Light Waves


CHAPTER- 18 - Geometrical Optics

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Part-II

Solutions - "Concepts of Physics" Part-II, by H C Verma

CHAPTER- 23 - Heat and Temperature

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-34


CHAPTER- 24 - Kinetic Theory of Gases

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 to Q10

EXERCISES - Q-11 to Q-20

EXERCISES - Q-21 to Q-30


EXERCISES - Q-31 to Q-40

EXERCISES - Q-41 to Q-50

EXERCISES - Q-51 to Q-62




CHAPTER- 25 - Calorimetry

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q-1 to Q-10

EXERCISES - Q11 to Q-18 




CHAPTER- 26 - Laws of Thermodynamics

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q-1 to Q-10

EXERCISES - Q-11 to Q-22




CHAPTER- 27 - Specific Heat Capacities of Gases

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q-1 to Q-10 




CHAPTER- 28 - Heat Transfer


Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q-1 to Q-10











  

















    



























































































 





































































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