
The ratio \[{{C}_{p}}/{{C}_{v}}\]for ${{H}_{2}}$ is:
A. 1.40
B. 1.67
C. 1.33
D. None of these
Answer
164.4k+ views
Hint: \[{{C}_{p}}\] is the molar heat capacity of the molecule at constant pressure and ${{C}_{v}}$ is the molar heat capacity at constant volume. Identify the relationship between the heat capacity at constant volume and pressure. Find a diatomic gas's heat capacity at constant pressure. Using this value as a reference point for the relationship between the heat capacity at constant pressure and volume. Find the ratio of the two to get the required answer.
Formula Used: \[{{C}_{p}}={{C}_{v}}+R\], where \[{{C}_{p}}\]is the molar heat capacity of the molecule at constant pressure and ${{C}_{v}}$ is the molar heat capacity at constant volume and R is the universal gas constant
Complete Step by Step Answer:
It is possible to define the molar heat capacity of a material or substance as the amount of energy needed to raise the temperature of one mole of the substance by one degree, or one unit. \[{{C}_{p}}\]is the molar heat capacity of the molecule at constant pressure and ${{C}_{v}}$ is the molar heat capacity at constant volume. Since ${{H}_{2}}$ is a diatomic molecule we’ll find the value of \[{{C}_{p}}/{{C}_{v}}\] according to that. We have a diatomic gas,
The ${{C}_{v}}$ for an ideal gas is given by ${{C}_{v}}=\dfrac{5}{2}R$, where R is the universal gas constant.
We will make use of the formula given below:
\[{{C}_{p}}={{C}_{v}}+R\]
Therefore, ${{C}_{p}}=\dfrac{5}{2}R+R$
And hence we get ${{C}_{p}}=\dfrac{7}{2}R$
Now, in the question we have been asked to find the ratio of \[{{C}_{p}}/{{C}_{v}}\]. So we will divide the values of \[{{C}_{p}}\]and ${{C}_{v}}$
\[\dfrac{{{C}_{p}}}{{{C}_{v}}}=\dfrac{\dfrac{7}{2}R}{\dfrac{5}{2}R}=\dfrac{7}{5}=1.40\]
Hence, the correct option is A. 1.40.
Note: The specific heat capacity will vary for various gases. The specific heat capacity at constant volume of a monatomic gas is $\dfrac{3}{2}R$. The specific heat capacity at constant volume of a polyatomic gas is 3R.
Formula Used: \[{{C}_{p}}={{C}_{v}}+R\], where \[{{C}_{p}}\]is the molar heat capacity of the molecule at constant pressure and ${{C}_{v}}$ is the molar heat capacity at constant volume and R is the universal gas constant
Complete Step by Step Answer:
It is possible to define the molar heat capacity of a material or substance as the amount of energy needed to raise the temperature of one mole of the substance by one degree, or one unit. \[{{C}_{p}}\]is the molar heat capacity of the molecule at constant pressure and ${{C}_{v}}$ is the molar heat capacity at constant volume. Since ${{H}_{2}}$ is a diatomic molecule we’ll find the value of \[{{C}_{p}}/{{C}_{v}}\] according to that. We have a diatomic gas,
The ${{C}_{v}}$ for an ideal gas is given by ${{C}_{v}}=\dfrac{5}{2}R$, where R is the universal gas constant.
We will make use of the formula given below:
\[{{C}_{p}}={{C}_{v}}+R\]
Therefore, ${{C}_{p}}=\dfrac{5}{2}R+R$
And hence we get ${{C}_{p}}=\dfrac{7}{2}R$
Now, in the question we have been asked to find the ratio of \[{{C}_{p}}/{{C}_{v}}\]. So we will divide the values of \[{{C}_{p}}\]and ${{C}_{v}}$
\[\dfrac{{{C}_{p}}}{{{C}_{v}}}=\dfrac{\dfrac{7}{2}R}{\dfrac{5}{2}R}=\dfrac{7}{5}=1.40\]
Hence, the correct option is A. 1.40.
Note: The specific heat capacity will vary for various gases. The specific heat capacity at constant volume of a monatomic gas is $\dfrac{3}{2}R$. The specific heat capacity at constant volume of a polyatomic gas is 3R.
Recently Updated Pages
Hess Law of Constant Heat Summation: Definition, Formula & Applications

Disproportionation Reaction: Definition, Example & JEE Guide

Environmental Chemistry Chapter for JEE Main Chemistry

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

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

Difference Between Natural and Whole Numbers: JEE Main 2024

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

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Types of Solutions

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

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE
