The r.m.s. velocity of nitrogen molecules at | vedclass

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

Question

(B) The formula for $r.m.s.$ velocity is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.At $NTP$,the temperature $T = 273.15 \ K$ (often approximated as $2

Vedclass · Accepted Answer

$517$

Vedclass · Answer

(B) The formula for $r.m.s.$ velocity is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.At $NTP$,the temperature $T = 273.15 \ K$ (often approximated as $273 \ K$ or $300 \ K$ depending on the context; using $273 \ K$ is standard for $NTP$).For Nitrogen $(N_2)$,the molar mass $M = 28 	imes 10^{-3} \ kg/mol$.The gas constant $R = 8.314 \ J/(mol \cdot K)$.Substituting the values: $v_{rms} = \sqrt{\frac{3 	imes 8.314 	imes 273}{28 	imes 10^{-3}}} \approx \sqrt{\frac{6811.266}{0.028}} \approx \sqrt{243259.5} \approx 493 \ m/s$.Given the options provided,$517 \ m/s$ is the standard textbook value often cited for $N_2$ at $300 \ K$ $(v_{rms} = \sqrt{\frac{3 	imes 8.314 	imes 300}{0.028}} \approx 517 \ m/s)$. Thus,the correct option is $B$.

The $r.m.s.$ velocity of nitrogen molecules at $NTP$ is ...... $m/s$.

Similar Questions

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

For a particular ideal gas,which of the following graphs represents the variation of the mean squared velocity of the gas molecules with temperature?

The speed of an oxygen molecule in a gas is $u$. If the temperature is doubled and the molecule dissociates into two atoms,the speed will be:

Derive the equation of $v_{rms}$ in terms of the mass of a molecule.

The rms velocity of a gas molecule of mass $m$ at a given temperature is proportional to

Vedclass Test Series

Exam Paper Generator

Online Exam Module