Question 150: Thermodynamics & KTG

Question

A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300 K in a container. The ratio of the rms speeds $$\left( {{{{v_{rms}}\left( {helium} \right)} \over {{v_{rms}}\left( {\arg on} \right)}}} \right)$$ is

Accepted Answer

The root mean square (rms) speed of a gas is given by the formula :
$$v_{rms} = \sqrt{\frac{3kT}{m}}$$
where :
$$k$$ is the Boltzmann constant
$$T$$ is the temperature in Kelvin
$$m$$ is the mass of one molecule of the gas
Given that the atomic mass of helium is 4 amu and that of argon is 40 amu, we can find the rms speed ratio by comparing the rms speeds of helium and argon, as temperature $$T$$ and Boltzmann constant $$k$$ are the same for both gases :
$$\frac{v_{rms} \left( 	ext{helium} ight)}{v_{rms} \left( 	ext{argon} ight)} = \sqrt{\frac{m_{	ext{argon}}}{m_{	ext{helium}}}}$$
Substituting the masses :
$$m_{	ext{helium}} = 4 	ext{ amu}$$
$$m_{	ext{argon}} = 40 	ext{ amu}$$
we get :
$$\frac{v_{rms} \left( 	ext{helium} ight)}{v_{rms} \left( 	ext{argon} ight)} = \sqrt{\frac{40}{4}} = \sqrt{10} \approx 3.16$$
Thus, the ratio of the rms speeds of helium to argon is approximately 3.16.
The correct answer is :
Option D : 3.16

Answer

0.32

Answer

0.45

Answer

2.24

Answer

3.16

Thermodynamics & KTG

A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300 K in a container. The ratio of the rms speeds $\left( {{{{v_{rms}}\left( {helium} \right)} \over {{v_{rms}}\left( {\arg on} \right)}}} \right)$ is

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