If V_{rms} for O_2 molecules is V at temperat | vedclass

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

Question

(B) The root mean square velocity is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M_w}}$,where $R$ is the universal gas constant,$T$ is the temper

Vedclass · Accepted Answer

$2V$

Vedclass · Answer

(B) The root mean square velocity is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M_w}}$,where $R$ is the universal gas constant,$T$ is the temperature,and $M_w$ is the molar mass.From this,we can see that $V_{rms} \propto \sqrt{\frac{T}{M_w}}$.For $O_2$ molecules,the molar mass is $M_w = 32 \ g/mol$. Given $V_{rms} = V$ at temperature $T$.For oxygen atoms $(O)$,the molar mass is $M_w' = 16 \ g/mol$. The new temperature is $T' = 2T$.Now,calculating the new $V_{rms}'$:$V_{rms}' = V 	imes \sqrt{\frac{T'}{T} 	imes \frac{M_w}{M_w'}}$$V_{rms}' = V 	imes \sqrt{\frac{2T}{T} 	imes \frac{32}{16}}$$V_{rms}' = V 	imes \sqrt{2 	imes 2} = V 	imes \sqrt{4} = 2V$.

If $V_{rms}$ for $O_2$ molecules is $V$ at temperature $T \ K$. Then $V_{rms}$ for oxygen atoms at temperature $2T$ will be

Similar Questions

The rms speed of an oxygen molecule in a vessel at a particular temperature is $\left(1+\frac{5}{x}\right)^{\frac{1}{2}} v$,where $v$ is the average speed of the molecule. The value of $x$ will be: (Take $\pi=\frac{22}{7}$)

The $rms$ speed of oxygen at $27^{\circ}C$ will be .... $ms^{-1}$.

The r.m.s. velocity of gas molecules kept at temperature $27^{\circ} C$ in a vessel is $61 \ m/s$. The molecular weight of the gas is nearly:
$[R = 8.31 \ J \ mol^{-1} \ K^{-1}]$

If the $rms$ speeds of helium and oxygen are equal,then the ratio of the temperatures of helium and oxygen is

The temperature at which the r.m.s. velocity of a gas triples its r.m.s. velocity at $0^{\circ} C$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module