The molar mass of the gas : 18 x 10⁻³ kg/mol
Further explanation
Given
An unknown gas has one third the root mean square speed of H2 at 300 K
Required
the molar mass of the gas
Solution
Average velocities of gases can be expressed as root-mean-square (V rms)
[tex]\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}[/tex]
T = temperature, Mm = molar mass of the gas particles , kg/mol
R = gas constant 8,314 J / mol K
v rms An unknown gas = 1/3 v rms H₂
v rms H₂ :
[tex]\tt v_{rms}=\sqrt{\dfrac{3\times 8.314\times 300}{2.10^{-3}} }\\\\v_{rms}=1934.22[/tex]
V rms of unknown gas =
[tex]\tt \dfrac{1}{3}\times 1934.22=644.74[/tex]
[tex]\tt 644.74^2=\dfrac{3\times 8.314\times 300}{M_{gas}}\\\\M_{gas}=18\times 10^{-3}~kg/mol[/tex]
