Root Mean Square Velocity Calculate the root mean square velocity of oxygen molecules at 25 °C.

Question

Root Mean Square Velocity

Calculate the root mean square velocity of oxygen molecules at 25 °C.

SORT You are given the kind of molecule and the temperature and asked to find the root mean square velocity.

GIVEN O[latex]_{2}[/latex], t = 25 °C
FIND u[latex]_{rms}[/latex]

STRATEGIZE The conceptual plan for this problem shows how you can use the molar mass of oxygen and the temperature (in kel vins) with the equation that defines the root mean square velocity to calculate the root mean square velocity.

CONCEPTUAL PLAN

M,T → u[latex]_{rms}[/latex]

. u[latex]_{rms}=\sqrt{\frac{3RT}{M}}[/latex]

SOLVE First, gather the required quantities in the correct units.
Note that molar mass must be in kg/mol and temperature must be in kelvins.

Substitute the quantities into the equation to calculate root mean square velocity. Note that 1 J = 1 kg · m²/s².

Accepted Answer

T = 25 + 273 = 298 K
M = [latex]\frac{32.00 \cancel{g} O_{2}}{1 mol O_{2}} imes\frac{ 1 kg}{1000 \cancel{g}} = \frac{32.00 imes 10^{-3} kg O_{2}}{1 mol O_{2}}[/latex]
u[latex]_{rms} = \sqrt{\frac{3RT}{M}}[/latex]
=[latex]\sqrt{\frac{3(8.314\frac{ J}{\cancel{mol}. \cancel{K}})(298 \cancel{K})}{\frac{32.00 imes 10^{-3} kg \cancel{O_{2}}}{1 \cancel{mol} O_{2}}}}[/latex]
= [latex]\sqrt{2.32 imes 10^{5}\frac{ J}{kg}}[/latex]
= [latex]\sqrt{2.32 imes 10^{5}\frac{\frac{\cancel{kg}. m^{2}}{s^{2}}}{\cancel{kg}}}[/latex] = 482 m/s
CHECK The units of the answer are correct. The magnitude of the answer seems reasonable because oxygen is slightly heavier than nitrogen and should therefore have a slightly lower root mean square velocity at the same temperature. (Recall that earlier we stated the root mean square velocity of nitrogen to be 515 m/s at 25 °C.)

Question 5.14: Root Mean Square Velocity Calculate the root mean square vel...

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