Question 909: Thermodynamics & KTG

Question

If $$\mathrm{m}$$' represents the mass of each molecules of a gas and $$\mathrm{T}$$' its absolute temperature then the root mean square speed of the gas molecule is proportional to

Accepted Answer

The root mean square (rms) speed of gas molecules is a measure of the speed of particles in a gas and is given by the following formula:
$$ v_{	ext{rms}} = \sqrt{\frac{3k_BT}{m}} $$
where:
$$v_{	ext{rms}}$$ is the root mean square speed
$$k_B$$ is the Boltzmann constant
$$T$$ is the absolute temperature
$$m$$ is the mass of a gas molecule
In order to find how the rms speed is proportional to the mass of each molecule $$m$$' and the absolute temperature $$T$$', we can analyze the proportionality directly from the equation given.
Rewriting the proportionality aspect of the equation:
$$ v_{	ext{rms}} \propto \sqrt{\frac{T}{m}} $$
This implies that:
$$ v_{	ext{rms}} \propto T^{\frac{1}{2}} m^{-\frac{1}{2}} $$
Thus, the root mean square speed of the gas molecule is proportional to
Option A:
$$\mathrm{m^{-\frac{1}{2}}T^{\frac{1}{2}}}$$

Answer

$$\mathrm{m^{-\frac{1}{2}}T^{\frac{1}{2}}}$$

Answer

mT

Answer

$$\mathrm{m^{\frac{1}{2}}T^{-\frac{1}{2}}}$$

Answer

$$\mathrm{m^{\frac{1}{2}}T^{\frac{1}{2}}}$$

Thermodynamics & KTG

If $\mathrm{m}' represents the mass of each molecules of a gas and \mathrm{T}$' its absolute temperature then the root mean square speed of the gas molecule is proportional to

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