Answer
64.8k+ views
Hint: To solve this problem we can use the relationship between the specific heats at constant pressure and volume and the universal gas constant. The answer can be found by finding the values of specific heats at constant pressure and volume and then comparing the ratio of them with known values.
Complete step by step solution:
As explained in the hint, we will use the relation between the molar specific heats at constant pressure and volume and the universal gas constant to find out the required values using the information given in the question.
We know that,
${C_p} - {C_v} = R$----equation 1
Here ${C_p}$ is the molar specific heat capacity at constant pressure and ${C_v}$ is the molar specific heat capacity at constant volume. R is the universal gas constant having value equal to $8.314J.mo{l^{ - 1}}{K^{ - 1}}$
It is given that
$\dfrac{R}{{{C_v}}} = 0.4$
$ \Rightarrow {C_v} = \dfrac{5}{2}R$
Substituting this value in equation 1 we have
${C_p} = R + {C_v}$
$ \Rightarrow {C_p} = R + \dfrac{5}{2}R$
$ \Rightarrow {C_p} = \dfrac{7}{2}R$
The implies that,
$ \Rightarrow \gamma = \dfrac{{{C_p}}}{{{C_v}}} = \dfrac{{\dfrac{7}{2}R}}{{\dfrac{5}{2}R}}$
$ \Rightarrow \gamma = 1.4$
This ratio is known as adiabatic index or the ratio of specific heats or Laplace’s coefficient.
For monoatomic gases the value of $\gamma = 1.67$ and for diatomic gases this value is equal to $1.4$ .
For rigid diatomic gas, there are five degrees of freedom and ${C_v} = \dfrac{5}{2}R$
The calculated value of ${C_v}$ is also the same for the given problem.
Therefore, the gas is made up of molecules which are rigid diatomic.
Thus, option A is the correct option.
Note: Note that for monoatomic gas molecules the values of molar specific heat capacity at constant volume is ${C_v} = \dfrac{3}{2}R$ and value at constant pressure is ${C_p} = \dfrac{5}{2}R$. While for diatomic gas molecules, the values of molar specific heat capacity at constant pressure ${C_p} = \dfrac{7}{2}R$ and at constant volume is ${C_v} = \dfrac{5}{2}R$.
Complete step by step solution:
As explained in the hint, we will use the relation between the molar specific heats at constant pressure and volume and the universal gas constant to find out the required values using the information given in the question.
We know that,
${C_p} - {C_v} = R$----equation 1
Here ${C_p}$ is the molar specific heat capacity at constant pressure and ${C_v}$ is the molar specific heat capacity at constant volume. R is the universal gas constant having value equal to $8.314J.mo{l^{ - 1}}{K^{ - 1}}$
It is given that
$\dfrac{R}{{{C_v}}} = 0.4$
$ \Rightarrow {C_v} = \dfrac{5}{2}R$
Substituting this value in equation 1 we have
${C_p} = R + {C_v}$
$ \Rightarrow {C_p} = R + \dfrac{5}{2}R$
$ \Rightarrow {C_p} = \dfrac{7}{2}R$
The implies that,
$ \Rightarrow \gamma = \dfrac{{{C_p}}}{{{C_v}}} = \dfrac{{\dfrac{7}{2}R}}{{\dfrac{5}{2}R}}$
$ \Rightarrow \gamma = 1.4$
This ratio is known as adiabatic index or the ratio of specific heats or Laplace’s coefficient.
For monoatomic gases the value of $\gamma = 1.67$ and for diatomic gases this value is equal to $1.4$ .
For rigid diatomic gas, there are five degrees of freedom and ${C_v} = \dfrac{5}{2}R$
The calculated value of ${C_v}$ is also the same for the given problem.
Therefore, the gas is made up of molecules which are rigid diatomic.
Thus, option A is the correct option.
Note: Note that for monoatomic gas molecules the values of molar specific heat capacity at constant volume is ${C_v} = \dfrac{3}{2}R$ and value at constant pressure is ${C_p} = \dfrac{5}{2}R$. While for diatomic gas molecules, the values of molar specific heat capacity at constant pressure ${C_p} = \dfrac{7}{2}R$ and at constant volume is ${C_v} = \dfrac{5}{2}R$.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)