# How to calculate the root mean square speed, vrms, in m·s^-1 at 227 Celsius and 900 torr, of a gas having a molar mass of 16.0 g·mol^-1 ?

Feb 28, 2015

There are two important things to look out for when doing root-mean-square speed calculations

1. You must use the value for $R$ expressed in Joules per mol K $\to$ R = 8.31446 "J"/("mol" * "K");
2. You must use molar mass in kilograms;

Using these units will get you to the required units for gas velocity, $m \cdot {s}^{- 1}$.

So, root-mean-square speed allows you to have some measure on the speed of particles in a gas. Mathematically, the equation looks like this

${v}_{\text{rms}} = \sqrt{\frac{3 R T}{M} _ m}$, where

$R$ - the universal gas constant;
$T$ - the temperature of the gas in Kelvin;
${M}_{m}$ - the molar mass of the gas;

I'll convert the molar mass from g per mole to kg per mole first, and then plug all the value into the equation for root-mean-square speed

$16.0 \text{g"/"mol" * "1 kg"/"1000 g" = 0.016"kg"/"mol}$

Therefore,

v_("rms") = sqrt((3 * 8.31446"J"/("mol" * "K") * (273.15 + 227)"K")/(0.016"kg"/"mol"))

${v}_{\text{rms") = 883.01 * sqrt("J"/"kg}}$

Use the fact that "Joule" = ("kg" * "m"^2)/("s"^2) to get

v_("rms") = 883.01 * sqrt(("kg" * "m"^2)/("kg" * "s"^2)) = 883.01 "m"/"s" = "111.7 m" * "s"^(-1)

Rounded to three sig figs, the answer will be

v_("rms") = "883 m" * "s"^(-1)