One mole of ideal monoatomic gas ($\gamma $ = 5/3) is mixed with one mole of diatomic gas ($\gamma $ = 7/5). What is $\gamma $ for the mixture? ( $\gamma $ denotes the ratio of specific heat at constant pressure, to that at constant volume.)
(a) 3/2
(b) 23/15
(c) 35/23
(d) 4/3
Answer
256.2k+ views
Hint: For the given mixture, the specific heat for a mixture of two gases is given by\[{{\gamma }_{mix}}=~\dfrac{\left[ {{\mathbf{n}}_{\mathbf{1}}}\text{ }{{\gamma }_{\mathbf{1}}}\text{ }+\text{ }{{\mathbf{n}}_{2}}\text{ }{{\gamma }_{2}} \right]}{\left[ \mathbf{n1}\text{ }+\text{ }\mathbf{n2} \right]}\text{, where n refers to the number of moles of each gas }\]
Now, apply this formula for the given mixture.
Step-by-Step Solution:
Let us first understand the concept of the ratio of specific heats before moving on to the particulars of this question.
In thermal physics and thermodynamics, the heat capacity ratio, also known as the adiabatic index, the ratio of specific heats, or Laplace's coefficient, is the ratio of the heat capacity at constant pressure (CP) to heat capacity at constant volume (CV). It is sometimes also known as the isentropic expansion factor and is denoted by $\gamma $ for an ideal gas or κ, the isentropic exponent for a real gas. The symbol gamma is used by aerospace and chemical engineers.
\[\gamma \text{ }=\text{ }\dfrac{{{C}_{P}}}{{{C}_{V}}}=\dfrac{{{c}_{P}}}{{{c}_{V}}}\]
where C is the heat capacity and c the specific heat capacity (heat capacity per unit mass) of a gas. The suffixes P and V refer to constant pressure and constant volume conditions respectively.
The change in internal energy and enthalpy of mixing ideal gases is zero. According to Gibbs' Theorem, the individual contribution of each species in an ideal gas mixture to the extensive thermodynamic properties of the mixture is the same as that of the pure species at the same temperature and at the partial pressure of the species in the mixture.
Thus, the resulting formula the specific heat for a mixture of two gases is given by\[{{\gamma }_{mix}}=~\dfrac{\left[ {{\mathbf{n}}_{\mathbf{1}}}\text{ }{{\gamma }_{\mathbf{1}}}\text{ }+\text{ }{{\mathbf{n}}_{2}}\text{ }{{\gamma }_{2}} \right]}{\left[ \mathbf{n1}\text{ }+\text{ }\mathbf{n2} \right]}\text{, where n refers to the number of moles of each gas }\]
Now, applying this formula for the given gases
\[\begin{align}
& {{\gamma }_{mix}}=~\dfrac{\left[ \text{1}\text{. 5/3 }+\text{ 1}\text{.7/5} \right]}{1+1}\text{, where n refers to the number of moles of each gas } \\
& {{\gamma }_{mix}}\approx 1.5 \\
\end{align}\]
Therefore, the required answer is a)
Note: Remember that the Ratio of Specific Heat is dimensionless and the value is the same in the SI and the Imperial system of units. Another way of obtaining the specific heats of a gaseous mixture would be calculating the sum of the product of mole fraction times the specific heat of that gas component.
Now, apply this formula for the given mixture.
Step-by-Step Solution:
Let us first understand the concept of the ratio of specific heats before moving on to the particulars of this question.
In thermal physics and thermodynamics, the heat capacity ratio, also known as the adiabatic index, the ratio of specific heats, or Laplace's coefficient, is the ratio of the heat capacity at constant pressure (CP) to heat capacity at constant volume (CV). It is sometimes also known as the isentropic expansion factor and is denoted by $\gamma $ for an ideal gas or κ, the isentropic exponent for a real gas. The symbol gamma is used by aerospace and chemical engineers.
\[\gamma \text{ }=\text{ }\dfrac{{{C}_{P}}}{{{C}_{V}}}=\dfrac{{{c}_{P}}}{{{c}_{V}}}\]
where C is the heat capacity and c the specific heat capacity (heat capacity per unit mass) of a gas. The suffixes P and V refer to constant pressure and constant volume conditions respectively.
The change in internal energy and enthalpy of mixing ideal gases is zero. According to Gibbs' Theorem, the individual contribution of each species in an ideal gas mixture to the extensive thermodynamic properties of the mixture is the same as that of the pure species at the same temperature and at the partial pressure of the species in the mixture.
Thus, the resulting formula the specific heat for a mixture of two gases is given by\[{{\gamma }_{mix}}=~\dfrac{\left[ {{\mathbf{n}}_{\mathbf{1}}}\text{ }{{\gamma }_{\mathbf{1}}}\text{ }+\text{ }{{\mathbf{n}}_{2}}\text{ }{{\gamma }_{2}} \right]}{\left[ \mathbf{n1}\text{ }+\text{ }\mathbf{n2} \right]}\text{, where n refers to the number of moles of each gas }\]
Now, applying this formula for the given gases
\[\begin{align}
& {{\gamma }_{mix}}=~\dfrac{\left[ \text{1}\text{. 5/3 }+\text{ 1}\text{.7/5} \right]}{1+1}\text{, where n refers to the number of moles of each gas } \\
& {{\gamma }_{mix}}\approx 1.5 \\
\end{align}\]
Therefore, the required answer is a)
Note: Remember that the Ratio of Specific Heat is dimensionless and the value is the same in the SI and the Imperial system of units. Another way of obtaining the specific heats of a gaseous mixture would be calculating the sum of the product of mole fraction times the specific heat of that gas component.
Recently Updated Pages
Electricity and Magnetism Explained: Key Concepts & Applications

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

JEE Main 2023 (February 1st Shift 1) Maths Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 2) Maths Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 2) Chemistry Question Paper with Answer Key

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Colleges 2026: Complete List of Participating Institutes

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Hybridisation in Chemistry – Concept, Types & Applications

Other Pages
JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

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

CBSE Notes Class 11 Chemistry Chapter 9 - Hydrocarbons - 2025-26

CBSE Notes Class 11 Chemistry Chapter 5 - Thermodynamics - 2025-26

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

CBSE Notes Class 11 Chemistry Chapter 8 - Organic Chemistry Some Basic Principles And Techniques - 2025-26

