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Question

(B) The root mean square velocity of an ideal gas molecule is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$ or $v_{rms} = \sqrt{\frac{3kT}{m}}$,where $m$

Vedclass · Accepted Answer

$p_A = \left( \frac{m_A}{m_B} \right)^{1/2} p_B$

Vedclass · Answer

(B) The root mean square velocity of an ideal gas molecule is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$ or $v_{rms} = \sqrt{\frac{3kT}{m}}$,where $m$ is the mass of the molecule.The momentum of a molecule is $p = mv$.For container $A$,$p_A = m_A v_A = m_A \sqrt{\frac{3kT}{m_A}} = \sqrt{3kT m_A}$.For container $B$,$p_B = m_B v_B = m_B \sqrt{\frac{3kT}{m_B}} = \sqrt{3kT m_B}$.Taking the ratio of the two momenta:$\frac{p_A}{p_B} = \frac{\sqrt{3kT m_A}}{\sqrt{3kT m_B}} = \sqrt{\frac{m_A}{m_B}} = \left( \frac{m_A}{m_B} \right)^{1/2}$.Therefore,$p_A = \left( \frac{m_A}{m_B} \right)^{1/2} p_B$.

Two containers $A$ and $B$ contain an ideal gas at the same temperature. If the masses of the molecules are $m_A$ and $m_B$ respectively,what is the relationship between their momenta $p_A$ and $p_B$?

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