A student asked me today (thanks, Austin) why there are two different expressions given on the AP exam for calculating the root mean square speed of a gas, and how this can be explained in terms of the use of k, R, m and M. Here’s the answer that I gave.

You will find two expressions on the data packet that comes with the AP chemistry exam that deal with the calculation of u_{rms} (root mean square speed).

u_{rms} = SQRT (3kT/m)

**AND**

u_{rms} = SQRT (3RT/M)

In the 2nd expression, R is the universal gas constant. The universal gas constant = 8.31 J mol^{-1} K^{-1} (i.e., one of the values that you see on the data packet).

In the 1st expression, k is the Boltzmann constant. The Boltzmann constant = the universal gas constant / 6.022 x 10^{23} = 1.38 x 10^{-23} J K^{-1} (i.e., the value that you see on the data packet).

So, since in the first expression, we have replaced a term in the *numerator* of the 2nd expression, R, by dividing *it* by Avogadro’s number, in order to keep the expression consistent, i.e., in order to STILL calculate u_{rms}, we have to change a term in the *denominator* of the 2nd expression in the same way. Since the only term in the denominator in the 2nd expression is M, dividing *it* by 6.022 x 10^{23} and substituting it in the 1st expression, does the job.

So, M = mass of 1 mole of the gas in kg, and in the other case, m = mass of a SINGLE particle of the gas, also in kg.

As an example, if we are calculating the u_{rms} for say, O_{2} at 30^{o}C, then we can either use;

u_{rms} = SQRT [(3)(8.314)(303)/(0.032)] = **486 ms ^{-1}**

(this calculation is based on a mole of molecules)

**OR**

u_{rms} = SQRT [(3)(1.381 x 10^{-23})(303)/(0.032/6.022 x 10^{23})] = **486 ms ^{-1}**

(this calculation is based on a single molecule)

I don’t EVER recall seeing a calculation on an AP exam that could NOT be dealt with by the R based expression, rather than the k based expression, so I wonder why it’s there. I hope that this helps.