
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
163.2k+ views
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 \[\frac{3}{{\rm{2}}}{\rm{RT}}\].
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 \[\frac{3}{{\rm{2}}}{\rm{RT}}\].
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Types of Solutions

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

Instantaneous Velocity - Formula based Examples for JEE
