The Question is from neet-physics , and it was created by the author VENKATESWARARAO

Question

The root mean square speed of gas molecules is related to their average kinetic energy. Which of the following correctly represents this relationship?

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
 **Primary Concept**: Kinetic Theory of Gases
 **Related Topics**: Root Mean Square Speed, Average Kinetic Energy, Ideal Gas Law
 **Prerequisites**: Understanding of kinetic energy, molar mass, and the ideal gas constant.
## SOLUTION PATHWAY
**Given Information**:
The question asks for the relationship between the root mean square speed ($v_{rms}$) of gas molecules and their average kinetic energy.
**Method & Steps**:
1. The average kinetic energy of a gas molecule is given by: $KE_{avg} = \frac{3}{2}kT$, where k is the Boltzmann constant and T is the temperature.
2. The root mean square speed ($v_{rms}$) is related to the average kinetic energy by: $KE_{avg} = \frac{1}{2}mv_{rms}^2$, where m is the mass of a single molecule.
3. We can relate the mass of a single molecule (m) to the molar mass (M) using Avogadro's number ($N_A$): $m = \frac{M}{N_A}$ 
4. Substituting the expression for $KE_{avg}$ and m into the equation, we get: $\frac{3}{2}kT = \frac{1}{2}\frac{M}{N_A}v_{rms}^2$
5. Rearranging the equation to solve for $v_{rms}$: $v_{rms}^2 = \frac{3kT N_A}{M}$ 
6. Since $R = k N_A$ (where R is the ideal gas constant), we can simplify the equation: $v_{rms}^2 = \frac{3RT}{M}$
7. Taking the square root of both sides: $v_{rms} = \sqrt{\frac{3RT}{M}}$
 Result: The correct relationship is $v_{rms} = \sqrt{\frac{3RT}{M}}$
 Verification: The units are consistent (m/s). The equation shows that $v_{rms}$ increases with temperature and decreases with molar mass, which aligns with our understanding of gas behavior.
 ## OPTION ANALYSIS
A ✓: $v_{rms} = \sqrt{\frac{3RT}{M}}$: This is the correct formula derived from the kinetic theory of gases.
B :x: $v_{rms} = \sqrt{\frac{RT}{M}}$: This formula is missing the factor of 3 from the average kinetic energy equation.
C :x: $v_{rms} = \sqrt{\frac{2RT}{M}}$: This formula uses an incorrect numerical factor in the numerator. 
D :x: $v_{rms} = \sqrt{\frac{3RT}{2M}}$: This formula has an incorrect factor of 2 in the denominator.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Incorrectly using the average speed or most probable speed instead of the root mean square speed.
  2.Mixing up the molar mass (M) with the mass of a single molecule (m).
 **Quick Tips**:
  1.Remember the key equation: $KE_{avg} = \frac{3}{2}kT = \frac{1}{2}mv_{rms}^2$
  2.Use dimensional analysis to check your units.

VENKATESWARARAO · Answer

The correct Option is "$v_{rms} = \sqrt{\frac{3RT}{M}}$"

VENKATESWARARAO · Answer

$v_{rms} = \sqrt{\frac{RT}{M}}$

VENKATESWARARAO · Answer

$v_{rms} = \sqrt{\frac{2RT}{M}}$

VENKATESWARARAO · Answer

$v_{rms} = \sqrt{\frac{3RT}{2M}}$

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