The most probable velocity of a gas at 7200 | vedclass

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

Question

(D) Temperature of helium $= 27^{\circ} C = 300 \ K$. The root-mean-square velocity of helium is given by $\mu_{rms} = \sqrt{\frac{3RT}{M_{He}}} = \sq

Vedclass · Accepted Answer

$SO_2$

Vedclass · Answer

(D) Temperature of helium $= 27^{\circ} C = 300 \ K$. The root-mean-square velocity of helium is given by $\mu_{rms} = \sqrt{\frac{3RT}{M_{He}}} = \sqrt{\frac{3R 	imes 300}{4}}$. The most probable speed of the unknown gas is $\mu_{mp} = \sqrt{\frac{2RT}{M}} = \sqrt{\frac{2R 	imes 7200}{M}}$. Equating $\mu_{rms} = \mu_{mp}$: $\sqrt{\frac{3R 	imes 300}{4}} = \sqrt{\frac{2R 	imes 7200}{M}}$. Squaring both sides: $\frac{900R}{4} = \frac{14400R}{M}$. $225 = \frac{14400}{M}$. $M = \frac{14400}{225} = 64 \ g/mol$. The molar mass of $SO_2$ is $32 + 2 	imes 16 = 64 \ g/mol$. Therefore, the gas is $SO_2$.

The most probable velocity of a gas at $7200 \ K$ is equal to the $RMS$ velocity of $He$ gas at $27^{\circ} C$. The gas is

Similar Questions

If the distance between two gas molecules becomes half,what change occurs in the London dispersion forces?

Two gases $X$ and $Y$ are taken in equal molar proportions. The ratio of their rates of diffusion is $1 : 2$. What will be the mole fraction of $X$ in a mixture containing $2 \ g$ of $X$ and $3 \ g$ of $Y$?

The correct value of the universal gas constant $R$ is:

The three states of matter are the result of what?

$A$ mixture of $He$ and $CH_4$ in a molar ratio of $4:1$ is present in a container at a pressure of $20 \ bar$. Due to a hole in the container,the gas mixture leaks out. What will be the composition of the gas mixture effusing out initially (in $: 1$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module