
The work of 146 KJ in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by ${7^o}C$. The gas is $\left( {R = 8.3Jmo{l^{ - 1}}{K^{ - 1}}} \right)$
A) Monoatomic
B) Diatomic
C) Triatomic
D) A mixture of monoatomic and diatomic
Answer
218.7k+ views
Hint: To solve this problem one must be aware of the concept of adiabatic index. The adiabatic index of a gas is the ratio of the specific heat of constant pressure of the gas to its specific heat at constant volume. In order to find if the gas is monatomic, diatomic or triatomic we must find out its adiabatic index $(\gamma )$.
Complete step by step answer:
It is given in the problem that the amount of gas is 1 kilo mole, i.e. $n = 1 \times {10^3}mol$
Also, the work done on the gas is 146 KJ, i.e. $W = 146 \times {10^3}J$
The rise in temperate is given to be ${7^o}C$, i.e. $\Delta T = {7^o}C$
We know that work done for a closed system in an adiabatic process is given by,
$W = \dfrac{{nR\Delta T}}{{\gamma - 1}}$
Hence, from the above equation we can derive an expression to find the value of adiabatic index and it is given below,
$ \Rightarrow \gamma = 1 + \dfrac{{nR\Delta T}}{W}$ ………. (1)
Putting all the given values in equation (1), we get,
$\gamma = 1 + \dfrac{{{{10}^3} \times 8.3 \times 7}}{{146 \times {{10}^3}}}$
$ \Rightarrow \gamma = 1.39$
$\therefore \gamma \approx 1.4$
Here we have found that the value of adiabatic index is approximately equal to 1.4. We know that a gas with an adiabatic index of value 1.4 is said to be a diatomic gas.
Hence, option B is the correct answer option.
Note: Following key points regarding the adiabatic index must be kept in mind.
Adiabatic index has no unit as it is a ratio of two similar quantities.
1. If adiabatic index of a gas has a value of 1.66, i.e $\gamma = 1.66$ then the gas is said to be monatomic.
2. If the adiabatic index of a gas has a value of 1.4, i.e $\gamma = 1.4$ then the gas is said to be diatomic.
3. If the adiabatic index of a gas has a value of 1.33, i.e $\gamma = 1.33$ then the gas is said to be triatomic.
Complete step by step answer:
It is given in the problem that the amount of gas is 1 kilo mole, i.e. $n = 1 \times {10^3}mol$
Also, the work done on the gas is 146 KJ, i.e. $W = 146 \times {10^3}J$
The rise in temperate is given to be ${7^o}C$, i.e. $\Delta T = {7^o}C$
We know that work done for a closed system in an adiabatic process is given by,
$W = \dfrac{{nR\Delta T}}{{\gamma - 1}}$
Hence, from the above equation we can derive an expression to find the value of adiabatic index and it is given below,
$ \Rightarrow \gamma = 1 + \dfrac{{nR\Delta T}}{W}$ ………. (1)
Putting all the given values in equation (1), we get,
$\gamma = 1 + \dfrac{{{{10}^3} \times 8.3 \times 7}}{{146 \times {{10}^3}}}$
$ \Rightarrow \gamma = 1.39$
$\therefore \gamma \approx 1.4$
Here we have found that the value of adiabatic index is approximately equal to 1.4. We know that a gas with an adiabatic index of value 1.4 is said to be a diatomic gas.
Hence, option B is the correct answer option.
Note: Following key points regarding the adiabatic index must be kept in mind.
Adiabatic index has no unit as it is a ratio of two similar quantities.
1. If adiabatic index of a gas has a value of 1.66, i.e $\gamma = 1.66$ then the gas is said to be monatomic.
2. If the adiabatic index of a gas has a value of 1.4, i.e $\gamma = 1.4$ then the gas is said to be diatomic.
3. If the adiabatic index of a gas has a value of 1.33, i.e $\gamma = 1.33$ then the gas is said to be triatomic.
Recently Updated Pages
Arithmetic, Geometric & Harmonic Progressions Explained

Cartesian Form of Vector Explained: Formula, Examples & Uses

Apparent Frequency Explained: Formula, Uses & Examples

Calorimetry: Definition, Principles & Calculations

Centrifugal Force Explained: Definition, Formula & Examples

Charge in a Magnetic Field: Definition, Formula & Examples

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Understanding Atomic Structure for Beginners

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

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

How to Convert a Galvanometer into an Ammeter or Voltmeter

