Question

By what factor does the average velocity of a gaseous molecule increase when the temperature (in kelvin) is doubled?

• A. 2.8
• B. 4
• C. 1.4
• D. 2

Answer 1

The Correct Option is C

To determine how the average velocity of a gaseous molecule changes when the temperature (in Kelvin) is doubled, we need to use the formula for the average velocity (also known as the root mean square velocity) of gas molecules.

The average velocity \( \overline{v} \) of a gas molecule is given by the equation:

v_{rms} = \sqrt{\frac{3kT}{m}}

where:

• k is the Boltzmann constant,
• T is the absolute temperature in Kelvin,
• m is the mass of a gas molecule.

From this equation, we see that the root mean square velocity is proportional to the square root of temperature, i.e.,

v_{rms} \propto \sqrt{T}

Now, let's consider the scenario where the temperature is doubled:

If the initial temperature is T, the new temperature is 2T. Substituting this into our formula gives:

v_{rms,new} = \sqrt{\frac{3k \cdot 2T}{m}}

This simplifies to:

v_{rms,new} = \sqrt{2} \cdot \sqrt{\frac{3kT}{m}} = \sqrt{2} \cdot v_{rms,initial}

Hence, the average velocity is increased by a factor of \sqrt{2}, which is approximately 1.4.

Therefore, the correct answer is that the average velocity increases by a factor of 1.4.

Thus, the correct option is: 1.4