
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
233.1k+ 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
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

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

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

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

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

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

