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Question: The specific heats, \({C_p}{\text{ and }}{{\text{C}}_v}\) of a gas of diatomic molecules, A, are giv...

The specific heats, Cp and Cv{C_p}{\text{ and }}{{\text{C}}_v} of a gas of diatomic molecules, A, are given (in units of Jmol−1K−1Jmo{l^{ - 1}}{K^{ - 1}}) by 29 and 22, respectively. Another gas of diatomic molecules, B, has the corresponding values 30 and 21. If they are treated as ideal gases, then?
(A)\left( A \right) A has one vibrational mode and B has two
(B)\left( B \right) Both A and B have a vibrational mode each
(C)\left( C \right) A is rigid but B has a vibrational mode
(D)\left( D \right) A has a vibrational mode but B has none

Explanation

Solution

Hint – In this question simply use the given specific heat capacity at constant pressure and at constant volume of the gases that is Cp and Cv{C_p}{\text{ and }}{{\text{C}}_v} respectively to find the degree of freedom of both the gases using the direct relationship that is Cv=fR2{C_v} = \dfrac{{fR}}{2} where R is the difference of the specific heats. This will help approach the solution of this problem.

Complete step-by-step answer:
As we know a diatomic molecule has 5 degrees of freedom (d.o.f) often denoted by f.
In these 5 degrees there are 3 translations and 2 rotations degrees are present.
Now if any diatomic molecule has a degree of freedom greater than 5, then the remaining degrees are called vibrational degrees.
Now diatomic molecules of A has specific heat constants are given as,
Cp=29,Cv=22{C_p} = 29,{C_v} = 22
Now as we know that
Cv=f1R2{C_v} = \dfrac{{{f_1}R}}{2}, where f1{f_1}degree of freedom of A, and R is nothing but the difference of the specific heats.
Now from here calculate degree of freedom we have,
⇒f1=2CvR\Rightarrow {f_1} = \dfrac{{2{C_v}}}{R}
Now substitute the values we have,
⇒f1=2(22)29−22=447=6.285≃6\Rightarrow {f_1} = \dfrac{{2\left( {22} \right)}}{{29 - 22}} = \dfrac{{44}}{7} = 6.285 \simeq 6
So the degree of freedom of (A) molecules is 6, so it has 3 translations, 2 rotation and 1 vibration.
Now diatomic molecules of B has specific heat constants are given as,
Cp=30,Cv=21{C_p} = 30,{C_v} = 21
Now as we know that
Cv=f2R2{C_v} = \dfrac{{{f_2}R}}{2}, where f2{f_2}degree of freedom of (B), and R is nothing but the difference of the specific heats.
Now from here calculate degree of freedom we have,
⇒f2=2CvR\Rightarrow {f_2} = \dfrac{{2{C_v}}}{R}
Now substitute the values we have,
⇒f2=2(21)30−21=429=4.666≃5\Rightarrow {f_2} = \dfrac{{2\left( {21} \right)}}{{30 - 21}} = \dfrac{{42}}{9} = 4.666 \simeq 5
So the degree of freedom of B molecules is 5, so it has 3 translations, 2 rotations only.
So A has a vibrational mode and B has none.
So this is the required answer.
Hence option (D) is the correct answer.

Note – The understanding of physical interpretation of the degree of freedom is very important in dealing with problems of this kind. Degree of freedom can be considered as a parameter that helps understanding the formal description of the state of the system. It can also be defined as the number of the total variables that can be used to define the motion of the gas.