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**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 ${C_v} = \dfrac{5}{2}R$.

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