The rms speed of a gas at room temperature of | vedclass

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

Question

(A) The formula for $rms$ speed is given by $v_{rms} = \sqrt{\frac{3RT}{M_w}}$,where $R = 8.31 \ J/(mol \cdot K)$,$T = 27 + 273 = 300 \ K$,and $v_{rms

Vedclass · Accepted Answer

$CO_2$

Vedclass · Answer

(A) The formula for $rms$ speed is given by $v_{rms} = \sqrt{\frac{3RT}{M_w}}$,where $R = 8.31 \ J/(mol \cdot K)$,$T = 27 + 273 = 300 \ K$,and $v_{rms} = 412 \ m/s$.Squaring both sides,we get $v_{rms}^2 = \frac{3RT}{M_w}$.Rearranging for molar mass $M_w$: $M_w = \frac{3RT}{v_{rms}^2}$.Substituting the values: $M_w = \frac{3 	imes 8.31 	imes 300}{412^2} \approx \frac{7479}{169744} \approx 0.044 \ kg/mol = 44 \ g/mol$.The molar mass of $CO_2$ is $12 + (2 	imes 16) = 44 \ g/mol$. Therefore,the gas is $CO_2$.

The $rms$ speed of a gas at room temperature of $27^{\circ}C$ is found to be $412 \ m/s$. The gas is:

Similar Questions

The rms speed of an oxygen molecule at a certain absolute temperature is $v$. If the absolute temperature is doubled and the oxygen molecules dissociate into atomic oxygen,then the rms speed would be

Find the $v_{rms}$ of nitrogen gas at $300 \ K$.

To increase the rms speed of gas molecules by $25 \%$,the percentage increase in absolute temperature of the gas is to be

At what temperature (in $K$) is the $rms$ velocity of a hydrogen molecule equal to that of an oxygen molecule at $47^o \ C$?

The $rms$ speed of hydrogen molecules at room temperature $(300 \ K)$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module