Question

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 molecules 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!!!

Answer 1

Answer:

The root mean square speed of O2 gas molecules is

519.01 m/s

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

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

$T=345.60K$

Now , Use Vrms to calculate the velocity of O2

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

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

$v_{rms}=\sqrt{26935.001}$

$v_{rms}=519.01m/s$