
Molecular velocities of two gases at the same temperature are \[{U_1}\] and \[{U_2}\]and their molecular masses \[{m_1}\] and \[{m_2}\] respectively. Which of the following expressions is correct?
A. \[\dfrac{{{m_1}}}{{{U_1}^2}} = \dfrac{{{m_2}}}{{{U_2}^2}}\]
B. \[{m_1}{U_1} = {m_2}{U_2}\]
C. \[\dfrac{{{m_1}}}{{{U_1}}} = \dfrac{{{m_2}}}{{{U_2}}}\]
D. \[{m_1}{U_1}^2 = {m_2}{U_2}^2\]
Answer
160.8k+ views
Hint: The distance covered by a molecule of gas in one unit of time is called molecular velocity. The molecular velocities of different molecules are different because of differences in molecular masses of them.
Complete Step by Step Solution:
The formula used for calculation of molecular velocity is \[{u_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \] , where, R stands for gas constant, T stands for temperature, M stands for molar mass of the gas.
Here, two gases are given.
For gas 1,
Molecular mass = \[{m_1}\]
Molecular velocity = \[{U_1}\]
For gas 2,
Molecular mass = \[{m_2}\]
Molecular velocity = \[{U_2}\]
For 1st gas,
\[{u_1} = \sqrt {\dfrac{{3RT}}{{{m_1}}}} \]
\[T = \dfrac{{{u_1}^2{m_1}}}{{3RT}}\] …… (1)
For 2nd gas,
\[{u_2} = \sqrt {\dfrac{{3RT}}{{{m_2}}}} \]
\[T = \dfrac{{{u_2}^2{m_2}}}{{3RT}}\] …… (2)
From (1) and (2),
\[\dfrac{{{u_1}^2{m_1}}}{{3RT}} = \dfrac{{{u_2}^2{m_2}}}{{3RT}}\]
\[{u_1}^2{m_1} = {u_2}^2{m_2}\]
Therefore, option D is right.
Additional Information: Let’s understand how molecular mass is responsible for rms velocity. The rms velocity can be find out by the following formula, \[{v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \] .As R is gas constant and if given that all gases are at same temperature, molar masses of each gas decide its rms speed. So, \[{v_{rms}} = \sqrt {\dfrac{1}{M}} \], that means, rms velocity is indirectly proportional to molar mass of gases. So, the gas that possesses the highest molar mass has the lowest rms velocity.
Note: It is to be noted that average velocity is different from rms velocity. Average velocity defines the arithmetic mean calculation of velocities of different gaseous molecules at a particular temperature. The average velocity can be found by the formula \[{v_{av}} = \sqrt {\dfrac{{8RT}}{{\pi M}}} \] . The rms velocity is always greater than average velocity.
Complete Step by Step Solution:
The formula used for calculation of molecular velocity is \[{u_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \] , where, R stands for gas constant, T stands for temperature, M stands for molar mass of the gas.
Here, two gases are given.
For gas 1,
Molecular mass = \[{m_1}\]
Molecular velocity = \[{U_1}\]
For gas 2,
Molecular mass = \[{m_2}\]
Molecular velocity = \[{U_2}\]
For 1st gas,
\[{u_1} = \sqrt {\dfrac{{3RT}}{{{m_1}}}} \]
\[T = \dfrac{{{u_1}^2{m_1}}}{{3RT}}\] …… (1)
For 2nd gas,
\[{u_2} = \sqrt {\dfrac{{3RT}}{{{m_2}}}} \]
\[T = \dfrac{{{u_2}^2{m_2}}}{{3RT}}\] …… (2)
From (1) and (2),
\[\dfrac{{{u_1}^2{m_1}}}{{3RT}} = \dfrac{{{u_2}^2{m_2}}}{{3RT}}\]
\[{u_1}^2{m_1} = {u_2}^2{m_2}\]
Therefore, option D is right.
Additional Information: Let’s understand how molecular mass is responsible for rms velocity. The rms velocity can be find out by the following formula, \[{v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \] .As R is gas constant and if given that all gases are at same temperature, molar masses of each gas decide its rms speed. So, \[{v_{rms}} = \sqrt {\dfrac{1}{M}} \], that means, rms velocity is indirectly proportional to molar mass of gases. So, the gas that possesses the highest molar mass has the lowest rms velocity.
Note: It is to be noted that average velocity is different from rms velocity. Average velocity defines the arithmetic mean calculation of velocities of different gaseous molecules at a particular temperature. The average velocity can be found by the formula \[{v_{av}} = \sqrt {\dfrac{{8RT}}{{\pi M}}} \] . The rms velocity is always greater than average velocity.
Recently Updated Pages
Two pi and half sigma bonds are present in A N2 + B class 11 chemistry JEE_Main

Which of the following is most stable A Sn2+ B Ge2+ class 11 chemistry JEE_Main

The enolic form of acetone contains a 10sigma bonds class 11 chemistry JEE_Main

The specific heat of metal is 067 Jg Its equivalent class 11 chemistry JEE_Main

The increasing order of a specific charge to mass ratio class 11 chemistry JEE_Main

Which one of the following is used for making shoe class 11 chemistry JEE_Main

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

JEE Main 2025: Derivation of Equation of Trajectory in Physics

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

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

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

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

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

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction
