At STP,write the order of root mean square sp | vedclass

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Question

(B) The root mean square speed $(u_{rms})$ of a gas is given by the formula: $u_{rms} = \sqrt{\frac{3RT}{M}}$.At a constant temperature $(T)$,$u_{rms}

Vedclass · Accepted Answer

$H_2 > N_2 > O_2 > HBr$

Vedclass · Answer

(B) The root mean square speed $(u_{rms})$ of a gas is given by the formula: $u_{rms} = \sqrt{\frac{3RT}{M}}$.At a constant temperature $(T)$,$u_{rms} \propto \frac{1}{\sqrt{M}}$,where $M$ is the molar mass of the gas.The molar masses are: $H_2 = 2 \ g/mol$,$N_2 = 28 \ g/mol$,$O_2 = 32 \ g/mol$,and $HBr = 81 \ g/mol$.Since $u_{rms}$ is inversely proportional to the square root of the molar mass,the gas with the lowest molar mass will have the highest $u_{rms}$.The order of molar masses is: $H_2 Therefore,the order of $u_{rms}$ is: $H_2 > N_2 > O_2 > HBr$.

List-$I$	List-$II$
$(a)$ $U_{rms} / U_{av}$	$(i)$ $1.22$
$(b)$ $U_{av} / U_{mp}$	$(ii)$ $1.13$
$(c)$ $U_{rms} / U_{mp}$	$(iii)$ $1.08$

At $STP$,write the order of root mean square speed $(u_{rms})$ for the molecules $H_2, N_2, O_2$,and $HBr$.

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Match the following (where $U_{rms}$ = root mean square speed,$U_{av}$ = average speed,$U_{mp}$ = most probable speed)
List-$I$ List-$II$
$(a)$ $U_{rms} / U_{av}$ $(i)$ $1.22$
$(b)$ $U_{av} / U_{mp}$ $(ii)$ $1.13$
$(c)$ $U_{rms} / U_{mp}$ $(iii)$ $1.08$

At a given temperature,which gas will have a higher speed between $H_2$ and $O_2$? Why?

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