Answers to Problems on Specific Heat Capacities of Gases HC Verma's Questions for Short Answer

Q#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.

Q#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 infinity. So the specific heat capacity is not defined at a constant temperature.   

Q#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. 

Q#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.

Q#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. 

Q#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.

Q#7

Show that the slope of 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 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 more in the adiabatic process. Since ɣ>1, so the slope of p-V diagram is greater for an adiabatic process as compared to an isothermal process. 

Q#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 be still 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 be always adiabatic.

Q#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'        (1)

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

pVɣ = p'V'ɣ    (2)

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.

Q#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.