Question

Equal volumes of monoatomic and diatomic gases of the same initial temperature and pressure are mixed. The ratio of the specific heats of the mixture $ \left ( \dfrac{C_p}{C_V} \right )$ will be
A. 1.53
B. 1.52
C. 1.5
D. 1

Answer 1

Hint: The quantity of heat required to enhance the temperature of the system by \[{\rm{1^\circ C}}\] is called heat capacity. If the mass of the system is one gram then it is called specific heat capacity.

Complete Step by Step Solution:
Specific heat capacity of a gaseous system if determined at constant pressure is denoted by \[{{\rm{C}}_{\rm{p}}}\].
Specific heat capacity of a gaseous system if determined at constant pressure is denoted by \[{{\rm{C}}_{\rm{v}}}\].
We know that for a gaseous system, \[{{\rm{C}}_{\rm{p}}}{\rm{ - }}{{\rm{C}}_{\rm{v}}}{\rm{ = R}}\].

The degree of freedom is the number of independent variable coordinates needed to deduce a molecule in space.
can be expressed in terms of the degree of freedom by the following expression:-
\[{{\rm{C}}_{\rm{v}}}{\rm{ = }}\dfrac{{{\rm{Rf}}}}{{\rm{2}}}\]
where
f=degree of freedom
R=universal gas constant.
\[{{\rm{C}}_{\rm{p}}}{\rm{ - }}{{\rm{C}}_{\rm{v}}}{\rm{ = R}}\]
\[ \Rightarrow {C_p} = R + {C_v}\]
\[ \Rightarrow {{\rm{C}}_{\rm{p}}}{\rm{ = R + }}\dfrac{{{\rm{Rf}}}}{{\rm{2}}}\]
So, \[{{\rm{C}}_{\rm{p}}}{\rm{ = R}}\left( {{\rm{1 + }}\dfrac{{\rm{f}}}{{\rm{2}}}} \right)\]

For monoatomic gases and diatomic gases, the degree of freedom is 3 and 5 respectively.
\[{{\rm{C}}_{\rm{p}}}\] for monoatomic gases=\[{\rm{R}}\left( {{\rm{1 + }}\dfrac{{\rm{3}}}{{\rm{2}}}} \right){\rm{ = }}\dfrac{{\rm{5}}}{{\rm{2}}}{\rm{R}}\]
 \[{{\rm{C}}_{\rm{p}}}\] for diatomic gases=\[{\rm{R}}\left( {{\rm{1 + }}\dfrac{{\rm{5}}}{{\rm{2}}}} \right){\rm{ = }}\dfrac{{\rm{7}}}{{\rm{2}}}{\rm{R}}\]

So, \[{{\rm{C}}_{\rm{p}}}\] for the mixture of these gases is given by the expression:-
\[{{\rm{C}}_{\rm{p}}}{\rm{ = }}\dfrac{{{\rm{12}}}}{{\rm{2}}}{\rm{R}}\]
So, \[{{\rm{C}}_{\rm{p}}}{\rm{ = 6R}}\]
\[{{\rm{C}}_{\rm{v}}}\] for monoatomic gases=\[\dfrac{{\rm{3}}}{{\rm{2}}}{\rm{R}}\]

\[{{\rm{C}}_{\rm{v}}}\] for diatomic gases=\[\dfrac{5}{{\rm{2}}}{\rm{R}}\]
So, \[{{\rm{C}}_{\rm{v}}}\] for the mixture of these gases is given by the expression:-
\[{{\rm{C}}_{\rm{v}}}{\rm{ = }}\dfrac{{\rm{8}}}{{\rm{2}}}{\rm{R = 4R}}\]
So, \[{{\rm{C}}_{\rm{v}}}{\rm{ = 4R}}\]
Hence, \[\dfrac{{{C_p}}}{{{C_v}}} = \dfrac{{6R}}{{4R}}\]
\[ \Rightarrow \dfrac{{{C_p}}}{{{C_v}}} = \dfrac{6}{4}\]
\[ \Rightarrow \dfrac{{{C_p}}}{{{C_v}}} = 1.5\]
So, option C is correct.

Note: It must be noted that according to the theory of equipartition of energy, each degree of freedom will hold a contribution of \[\frac{{\rm{1}}}{{\rm{2}}}{\rm{RT}}\] to the molar internal energy. Monoatomic gases have only translational motion which is of three kinds. So, the degree of freedom for these types of gases is 3. Thus, U for a monatomic gas is \[\frac{3}{{\rm{2}}}{\rm{RT}}\].