Question

A 5.0-liter gas tank holds 2.7 moles of monatomic helium (He) and 0.90 mole of diatomic oxygen (O2), at a temperature of 330 K. The ATOMIC masses of helium and oxygen are 4.0 g/mol and 16.0 g/mol, respectively. What is the ratio of the root-mean-square (thermal) speed of helium to that of oxygen?

Answer 1

The ratio is 2:1

Step-by-step 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

Answer 2

This problem can be solved by using the expression

`v_(rms)=\sqrt{(3RT)/(M_(m))}`

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

`v_(rms-He)=\sqrt{(2RT)/(M_(He))}=\sqrt{(2(8.314)(330))/(4*10^(-3))}=1171.2(m)/(s)\\v_{rms-O_(2)}=\sqrt{\frac{2RT}{M_{0_(2)}}}=\sqrt{(2(8.314)(330))/(16*10^(-3))}=585.6(m)/(s)\\\frac{v_(rms-He)}{v_{rms-O_(2)}}=1.999\approx 2`

Hence, the vrms of the monoatomic helium is two times the vrms of the oxygen

I hope this is useful for you

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