If there is no loss of energy. Two monatomic ideal gases at temperature ${T_1}$ and ${T_2}$ are mixed. If the masses of molecules of two gases are ${m_1}$ and ${m_2}$ number of their molecules are ${n_1}$ and ${n_2}$.then the temperature of mixture will be:
(A) $\dfrac{{{T_1} + {T_2}}}{{{n_1} + {n_2}}}$
(B) \[\dfrac{T_1}{n_1}\] + \[\dfrac{T_1}{n_2}\]
(C) $\dfrac{{{n_2}{T_1} + {n_1}{T_2}}}{{{n_1} + {n_2}}}$
(D) $\dfrac{{{n_1}{T_1} + {n_2}{T_2}}}{{{n_1} + {n_2}}}$
Answer
275.1k+ views
Hint An ideal gas is a theoretical gas composed of many randomly moving point particles that are not subject to interparticle interactions. The ideal gas concept is useful because it obeys the ideal gas law, a simplified equation of state, and is amenable to analysis under statistical mechanics. The average molecular kinetic energy is proportional to the ideal gas law's absolute temperature.
Complete Step by Step Solution
1. The average kinetic energy per molecule of a perfect gas = $\dfrac{3}{2}RT$
2. Average kinetic energy of the molecules of the first gas = $\dfrac{3}{2}{n_1}R{T_1}$
Average kinetic energy of the molecules of the second gas = $\dfrac{3}{2}{n_2}R{T_2}$
Therefore, the total kinetic energy of the two molecules of gas before they are mixed is ${K_i} = \dfrac{3}{2}{n_1}R{T_1} + \dfrac{3}{2}{n_2}R{T_2} = \dfrac{3}{2}\left( {{n_1}{T_1} + {n_2}{T_2}} \right)R.......(1)$
If T is the temperature of the mixture, then the kinetic energies of the molecules after mixing is ${K_f} = \dfrac{3}{2}\left( {{n_1} + {n_2}} \right)RT.............(2)$
Since there is no loss of energy
∴ Sum of change of internal energies must be zero
Equating (1) and (2) we get,
$
{K_i} = {K_f} \\
\dfrac{3}{2}\left( {{n_1}{T_1} + {n_2}{T_2}} \right)R = \dfrac{3}{2}\left( {{n_1} + {n_2}} \right)RT \\
T = \dfrac{{\left( {{n_1}{T_1} + {n_2}{T_2}} \right)}}{{\left( {{n_1} + {n_2}} \right)}} \\
$
Hence, the option (d) is the correct answer.
Note In the kinetic model of gases, the pressure is equal to the force exerted by the atoms hitting and rebounding from a unit area of the gas container surface. The product of pressure and volume per mole is proportional to the average (translational) molecular kinetic energy.
Complete Step by Step Solution
1. The average kinetic energy per molecule of a perfect gas = $\dfrac{3}{2}RT$
2. Average kinetic energy of the molecules of the first gas = $\dfrac{3}{2}{n_1}R{T_1}$
Average kinetic energy of the molecules of the second gas = $\dfrac{3}{2}{n_2}R{T_2}$
Therefore, the total kinetic energy of the two molecules of gas before they are mixed is ${K_i} = \dfrac{3}{2}{n_1}R{T_1} + \dfrac{3}{2}{n_2}R{T_2} = \dfrac{3}{2}\left( {{n_1}{T_1} + {n_2}{T_2}} \right)R.......(1)$
If T is the temperature of the mixture, then the kinetic energies of the molecules after mixing is ${K_f} = \dfrac{3}{2}\left( {{n_1} + {n_2}} \right)RT.............(2)$
Since there is no loss of energy
∴ Sum of change of internal energies must be zero
Equating (1) and (2) we get,
$
{K_i} = {K_f} \\
\dfrac{3}{2}\left( {{n_1}{T_1} + {n_2}{T_2}} \right)R = \dfrac{3}{2}\left( {{n_1} + {n_2}} \right)RT \\
T = \dfrac{{\left( {{n_1}{T_1} + {n_2}{T_2}} \right)}}{{\left( {{n_1} + {n_2}} \right)}} \\
$
Hence, the option (d) is the correct answer.
Note In the kinetic model of gases, the pressure is equal to the force exerted by the atoms hitting and rebounding from a unit area of the gas container surface. The product of pressure and volume per mole is proportional to the average (translational) molecular kinetic energy.
Recently Updated Pages
Mass vs Weight: Key Differences Explained for Students

Circuit Switching vs Packet Switching: Key Differences Explained

Uniform Acceleration Explained: Formula, Examples & Graphs

Wheatstone Bridge – Principle, Formula, Diagram & Applications

Classification of Drugs in Chemistry: Types, Examples & Exam Guide

Area vs Volume: Key Differences Explained for Students

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

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Derivation of Equation of Trajectory Explained for Students

Electron Gain Enthalpy and Electron Affinity Explained

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2026-27

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

