Question

For an ideal gas of molar mass M, the slope of the plot between the rms velocity ($v_{rms}$ along y-axis) and the square root of absolute temperature ($\sqrt{T}$ along x-axis) is ________.

• A. $\sqrt{\frac{M}{3R}}$
• B. $\sqrt{\frac{3R}{M}}$
• C. $\sqrt{\frac{R}{3M}}$
• D. $\frac{R}{3M}$
• E. $\frac{3R}{M}$

Hint:

$v_{rms}$ is directly proportional to $\sqrt{T}$ and inversely proportional to $\sqrt{M}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The root-mean-square (rms) velocity of gas molecules is a measure of their average speed. It is related to the absolute temperature and the molar mass of the gas. The question asks for the slope of a graph plotting \(v_{rms}\) versus \(\sqrt{T}\), which requires us to express \(v_{rms}\) as a function of \(\sqrt{T}\) and identify the slope.
Step 2: Key Formula or Approach
The formula for the rms velocity of an ideal gas is:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \] where R is the ideal gas constant, T is the absolute temperature in Kelvin, and M is the molar mass of the gas (in kg/mol).
We need to compare this equation to the equation of a straight line passing through the origin, \(y = mx\), where \(y = v_{rms}\) and \(x = \sqrt{T}\).
Step 3: Detailed Explanation
1. Rearrange the formula for \(v_{rms}\).
We can rewrite the formula as:
\[ v_{rms} = \sqrt{\frac{3R}{M}} \cdot \sqrt{T} \] 2. Compare with the equation of a straight line.
The problem describes a plot with:
- y-axis = \(v_{rms}\)
- x-axis = \(\sqrt{T}\)
The equation of a straight line passing through the origin is \(y = (\text{slope}) \cdot x\).
Comparing our rearranged formula with this line equation:
\[ \underbrace{v_{rms}}_{y} = \underbrace{\left(\sqrt{\frac{3R}{M}}\right)}_{\text{slope}} \underbrace{\sqrt{T}}_{x} \] 3. Identify the slope.
From the comparison, the slope of the plot of \(v_{rms}\) versus \(\sqrt{T}\) is the constant term multiplying \(\sqrt{T}\).
\[ \text{Slope} = \sqrt{\frac{3R}{M}} \] Step 4: Final Answer
The slope of the plot is \(\sqrt{\frac{3R}{M}}\).