A mixture of 2 moles of helium gas (atomic ma | vedclass

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

Question

(D) The root mean square (rms) speed of a gas molecule is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constan

Vedclass · Accepted Answer

$3.16$

Vedclass · Answer

(D) The root mean square (rms) speed of a gas molecule is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.Since the temperature $T$ is the same for both gases in the mixture,we have $V_{rms} \propto \frac{1}{\sqrt{M}}$.Therefore,the ratio of the rms speeds is given by $\frac{V_{rms}(	ext{helium})}{V_{rms}(	ext{argon})} = \sqrt{\frac{M(	ext{argon})}{M(	ext{helium})}}$.Substituting the given values,$M(	ext{helium}) = 4 \ g/mol$ and $M(	ext{argon}) = 40 \ g/mol$,we get:$\frac{V_{rms}(	ext{helium})}{V_{rms}(	ext{argon})} = \sqrt{\frac{40}{4}} = \sqrt{10} \approx 3.16$.

$A$ 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 $\left( \frac{V_{rms}(\text{helium})}{V_{rms}(\text{argon})} \right)$ is

Similar Questions

$A$ ray of light strikes a plane mirror $M$ at an angle of $45^o$ as shown in the figure. After reflection,the ray passes through a prism of refractive index $1.5$ whose apex angle is $4^o$. The total angle through which the ray is deviated is....$^o$

The contrapositive of the statement "If two numbers are not equal,then their squares are not equal" is

The first step of taxonomy is .......... .

$A$ straight line is equally inclined to all the three coordinate axes. Then, the angle made by the line with the $y$-axis is

$\int_0^\pi \frac{1}{1+\sin x} \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module