Specific Heat Capacities of Gases HC Verma Solutions of Que for Short Ans Ch-27 Vol-2
Specific Heat Capacities of Gases HC Verma Solutions of Que for Short Ans Ch-27 Vol-2 Concept of Physics. Step by Step Solution of Questions for short answer of Ch-27 Specific Heat Capacities of Gases HC Verma Question of Bharti Bhawan Publishers . Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Specific Heat Capacities of Gases HC Verma Solutions of Que for Short Ans Ch-27 Vol-2
| Board | ISC and other board |
| Publications | Bharti Bhawan Publishers |
| Ch-27 |
Specific Heat Capacities of Gases |
| Class | 12 |
| Vol | 2nd |
| writer | H C Verma |
| Book Name | Concept of Physics |
| Topics | Solutions of Question for Short Answer |
| Page-Number | 76, |
-: Select Topics :-
Ques for Short Ans
Objective-I
Objective-II
Exercise
Specific Heat Capacities of Gases Que for Short Ans
HC Verma Solutions of Ch-27 Vol-2 Concept of Physics for Class-12
Question 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 1 :-
No, a gas doesn’t have just two specific heat capacities, as the heat capacities depend on the process followed. There are infinite processes; therefore, there can be infinite number of specific heat capacities.
Question 2 :-
Can we define specific heat capacity at constant temperature?
Answer 2 :-
Question 3 :-
Can we define specific heat capacity for an adiabatic process?
Answer 3 :-
Specific heat capacity, s =△Q/m△T, where △Q/m is the heat supplied per unit mass of the substance and ΔT is the change in temperature produced. In an adiabatic process, no heat exchange is allowed; so, ΔQ = 0 and hence, s = 0. Therefore, in an adiabatic process, specific heat capacity is zero.
Question 4 :-
Does a solid also have two kinds of molar heat capacities Cp and Cv? If yes, is Cp > Cv? Or is Cp − Cv = R?
Yes, a solid also has two kinds of molar heat capacities, Cp and Cv. In a solid, expansion coefficient is quite small; therefore dependence of heat capacity on the process is negligible. So, Cp > Cv with just a small difference, which is not equal to R.
Question 5 :-
In a real gas, the internal energy depends on temperature and also on volume. The energy increases when the gas expands isothermally. Examining 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.
In a real gas, as the internal energy depends on temperature and volume, the derived equation for an ideal gas
(dQ)P = (dQ)v + nRdT will change to
(dQ)P = (dQ)v + nRdT+ k ,where k is the change in internal energy (positive) due to change in volume when pressure is kept constant. So, in the case of a real gas, for n=1 mole (say),
CP -Cv =R + k/dt
⇒ CP – Cv > R,
where Cp and Cv are the specific heat capacities at constant pressure and volume, respectively.
Question 6 :-
Can a process on an ideal gas be both adiabatic and isothermal?
According to the first law of thermodynamics, change in internal energy, ΔU is equal to the difference between heat supplied to the gas, Δ Q and the work done on the gas, ΔW,
such that ΔQ = ΔU +ΔW . In an adiabatic process, ΔQ =0 and in an isothermal process, change in temperature, Δ T =0. Therefore,
ΔQ = ΔU + ΔW
⇒ ΔQ = nCvΔT +ΔW
⇒ 0 =nCv(0) + Δ W
⇒ Δ W = 0 ,
if the process is adiabatic as well as isothermal, no work will be done. So, a process on an ideal gas cannot be both adiabatic and isothermal.
Question 7 :-
Show that the slope of the p−V diagram is greater for an adiabatic process compared to an isothermal process.
In an isothermal process,
PV = k …(i)
On differentiating it w.r.t V, we get
k = constant
In an adiabatic process,
PVγ = K …(ii)
On differentiating it w.r.t V, we get
are the slope of the curve and the ratio of heat capacities at constant pressure and volume, respectively; P is pressure and V is volume of the system.
By comparing the two slopes and keeping in mind that γ >1 , we can see that the slope of the P-V diagram is greater for an adiabatic process than an isothermal process.
Question 8 :-
Is a slow process always isothermal? Is a quick process always adiabatic?
For an isothermal process, PV =K , where P is P is pressure, V is volume of the system and Kis constant. In an isothermal process, a small change in V produces only a small change in p, so as to keep the product constant. On the other hand, in an adiabatic process, 
is the ratio of heat capacities at constant pressure and volume, respectively, and k is a constant. In this process, a small increase in volume produces a large decrease in pressure. Therefore, an isothermal process is considered to be a slow process and an adiabatic process a quick process.
Question 9 :-
Can two states of an ideal gas be connected by an isothermal process as well as an adiabatic process?
For two states to be connected by an isothermal process,
P1V1 = P2V2 … (i)
For the same two states to be connected by an adiabatic process,
P1V1γ = P2V2γ …(ii)
If both the equations hold simultaneously then, on dividing eqaution (ii) by (i) we get
V1γ-1 = V2γ-1
Let the gas be monatomic. Then,
γ =53
⇒V12/3=V22/3
⇒ V1 = V2
If this condition is met, then the two states can be connected by an isothermal as well as an adiabatic process.
Question 10 :-
The ratio Cp / Cv for a gas is 1.29. What is the degree of freedom of the molecules of this gas?
For the molecules of a gas, γ = CP/CV=1+2/f
where f is the degree of freedom.
Given : γ =1.29
⇒ 1 + 2/f = 1.29 = 9/7
⇒ 2/f = 9/7 – 1
⇒ 2/f = 2/7
⇒ f =7
Therefore, the molecules of this gas have 7 degrees of freedom.
But in reality, no gas can have more than 6 degrees of freedom.
— : End of Specific Heat Capacities of Gases HC Verma vol -2 Que for Short Ans :–
Return to — HC Verma Solutions Vol-2 Concept of Physics
Thanks