Question

If the rms velocity of a gas is v, then

• A. $v^{2}T = \text{constant}$
• B. $v^{2}/T = \text{constant}$
• C. $vT^{2} = \text{constant}$
• D. v is independent of T

Hint:

If the rms velocity of a gas is v, then

Answer 1

The Correct Option is B

The question requires us to find the relationship between the root mean square (RMS) velocity of a gas and temperature. Let's break down the problem using the physics behind the ideal gas law and kinetic theory.

The formula for the RMS velocity (\(v_{rms}\)) of a gas is given by:

1. \(v_{rms} = \sqrt{\frac{3kT}{m}}\),

where \(k\) is the Boltzmann constant, \(T\) is the temperature, and \(m\) is the mass of a single gas molecule.

Square both sides to eliminate the square root:

1. \(v_{rms}^2 = \frac{3kT}{m}\).

To establish a constant relationship, multiply both sides by \(\frac{1}{T}\):

1. \(\frac{v_{rms}^2}{T} = \frac{3k}{m}\).

Notice that \(\frac{3k}{m}\) is a constant because both \(k\) and \(m\) are constants for a given gas.

Thus, we can say:

1. \(\frac{v^2}{T} = \text{constant}\).

This derivation matches the given answer options, confirming that "\(v^{2}/T = \text{constant}\)" is the correct relationship.

Therefore, the correct answer is:

$v^{2}/T = \text{constant}$ 
 

All other options do not conform to this established relation under ideal gas behavior as explained above.