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iren [92.7K]
4 years ago
5

If He gas has an average kinetic energy of 4310 J/mol under certain conditions, what is the root mean square speed of O2 gas mol

ecules under the same conditions? (given that He and O_2 gas are at the same temperature what can you conclude about 0_2's average kinetic energy?) variable equations used to find the solution are greatly appreciated!!!
Chemistry
1 answer:
Free_Kalibri [48]4 years ago
6 0

Answer:

The root mean square speed of O2 gas molecules is

<u>519.01 m/s</u>

<u></u>

Explanation:

The root mean square velocity  :

v_{rms}=\sqrt{\frac{3RT}{M}}

K.E_{avg}=\frac{3}{2}RT

K.E =\frac{1}{2}mv_{rms}^{2}

Molar mass , M

For He = 4 g/mol

For O2 = 2 x 16 = 32 g/mol

O2 = 32/1000 = 0.032 Kg/mol

First calculate the temperature at which the K.E of He is 4310J/mol

K.E of He =

K.E_{avg}=\frac{3}{2}RT

T=\frac{2(K.E)}{3(R)}

K.E of He = 4310 J/mol

T=\frac{2(4310J/mol)}{3(8.314J/Kmol)}

T=345.60K

<u>Now , Use Vrms to calculate the velocity of O2</u>

v_{rms}=\sqrt{\frac{3(8.314J/Kmol)(345.60K)}{0.032Kg/mol}}

v_{rms}=\sqrt{\frac{8619.9552}{0.032}}

v_{rms}=\sqrt{26935.001}

v_{rms}=519.01m/s

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