Answer:
This problem can be solved by using the expression
[tex]v_{rms}=\sqrt{\frac{3RT}{M_{m}}}[/tex]
where R is the gas constant, T is absolute temperature and M is the molar mass in kg/mol. By calculating for both Helium and oxygen we have
[tex]v_{rms-He}=\sqrt{\frac{2RT}{M_{He}}}=\sqrt{\frac{2(8.314)(330)}{4*10^{-3}}}=1171.2\frac{m}{s}\\v_{rms-O_{2}}=\sqrt{\frac{2RT}{M_{0_{2}}}}=\sqrt{\frac{2(8.314)(330)}{16*10^{-3}}}=585.6\frac{m}{s}\\\frac{v_{rms-He}}{v_{rms-O_{2}}}=1.999\approx 2[/tex]
Hence, the vrms of the monoatomic helium is two times the vrms of the oxygen
I hope this is useful for you
best regards
Answer:
The ratio is 2:1
Explanation:
Given their atomic masses, Mh = 4.0g/mol and Mo = 16g/mol
Vrms = √(3RT/M)
R and T are constant so Vrms ∝√(1/M)
So (Vrms)Helium/(Vrms)oxygen = √(1/Mh)/√(1/Mo) = √(Mo/Mh)
(Vrms)Helium/(Vrms)oxygen = √(Mo/Mh)
= √(16.0/4.0) = √(4) = 2
