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**Hint:**We will use the formula of Specific heat at constant pressure and Specific heat at constant volume to find the ratio.

First, we will use the specific heat at constant pressure equation i.e. \[Cp = Cv + R\] and then the equation of specific heat at constant volume i.e. \[Cv = \dfrac{3}{2}R\] .

Then we will use the above equations and find the ratio of \[\dfrac{{{C_P}}}{{{C_V}}}\] . After that we will be able to choose the correct option.

**Complete step by step solution**

We know that according to the first law of thermodynamics, for constant volume process with a monatomic ideal gas the molar specific heat can be derived that is molar specific heat at constant pressure:

\[Cp = Cv + R\] , where Cp is the specific heat at constant pressure and Cv is the specific heat at constant volume and R is the is the molar gas constant i.e.R=8.3144598(48) \[Jmo{l^{ - 1}}\] .

And we also know the formula of Cv i.e. \[Cv = \dfrac{3}{2}R\] .

Putting the equation of Cv in the equation of Cp we get:

\[Cp = \dfrac{{3R}}{2} + R = \dfrac{{3R + 2R}}{2} = \dfrac{{5R}}{2}\] .

Now we find the ratio of \[\dfrac{{{C_P}}}{{{C_V}}}\] . After putting the equations in ratio, we get

\[

\dfrac{{{C_P}}}{{{C_V}}} = \dfrac{{5R}}{2} \times \dfrac{2}{{3R}} \\

\dfrac{{{C_P}}}{{{C_V}}} = \dfrac{5}{3} \\

\]

So the value of \[\dfrac{{{C_P}}}{{{C_V}}} = \dfrac{5}{3}\] i.e. it is maximum value and the value of the ratio cannot be greater than \[\dfrac{5}{3}\].

**So option A is correct .**

**Note:**Remember that specific heat of a solid or liquid is the amount of heat that raises the temperature of a unit mass of the solid through \[1^\circ C\] .

Also note that specific heat possesses infinite values as the behaviour of gas when heat is supplied, the pressure and volume change in temperature and the amount of heat required to raise the temperature for 1gm of gas through \[1^\circ C\] depends on the way gas is heated. So, because of this we get infinite values.

Always remember the formulas of specific heat at constant volume and constant temperature as for different conditions the formulas are different.

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