Two containers A and B contain an ideal gas a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $p_B$?

Similar Questions

At a given temperature,the $rms$ velocity of a gas molecule of mass $m$ is proportional to:

The r.m.s. speed of the molecules of a gas at $100^{\circ} C$ is $v$. The temperature at which the r.m.s. speed will be $\sqrt{3} v$ is: (in $^{\circ} C$)

For a given gas at $1\,atm$ pressure,the $rms$ speed of the molecules is $200\,m/s$ at $127\,^oC$. At $2\,atm$ pressure and at $227\,^oC$,the $rms$ speed of the molecules will be:

When pressure remains constant,at what temperature will the $r.m.s.$ speed of gas molecules increase by $10 \%$ of the $r.m.s.$ speed at $STP$?

Five molecules of a gas have speeds $1, 2, 4, 8$ and $16 \, m/s$ at some instant. The root mean square velocity of the gas molecules is ..... $m/s$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module