This is the equation you need to use:

v = $\sqrt{\mathrm{(3RT)\; /\; M}}$

You may, if you wish, read more about the above equation here.

The basic idea is that, if you consider each gas molecule's velocity (which has components of both speed and direction), the average velocity of all gas molecules in a sample is zero. That stems from the fact that the gas molecules are moving in all directions in a random way and each random speed in one direction is cancelled out by a molecule randomly moving in the exact opposite direction, with the exact same speed (when the gas sample is considered in a random way).

So, what you do is square each velocity, which gets rid of the negative sign (molecules moving in one way have a positive sign for their direction, those moving in the opposite direction have a negative direction). You then average all of these squared values and take the square root of that average.

It's called a 'root mean square' and technically, it is a speed, not a velocity. However, in chemistry, we ignore the distinction between speed and velocity and use velocity.

Some necessary information:

R = 8.31447 J mol¯^{1}K¯^{1}M = the molar mass of the substance, expressed in

kilograms

Look at how the units cancel in v = $\sqrt{\mathrm{(3RT)\; /\; M}}$

**Example #1:** Calculate the rms speed of an oxygen gas molecule, O_{2}, at 31.0 °C

**Solution:**

v = $\sqrt{\mathrm{(3RT)\; /\; M}}$v = $\sqrt{\mathrm{[(3)\; (8.31447)\; (304.0)]\; /\; 0.0319988}}$

v = 486.8 m/s

Here is the above set-up done with units:

v = $\sqrt{\mathrm{[(3)\; (8.31447\; kg\; m2s\xaf2K\xaf1mol\xaf1)\; (304.0\; K)\; /\; 0.0319988\; kg/mol}}$

Remember that kg m

^{2}s¯^{2}is called a Joule and that the unit on R is usually written J/K mol. The more extended unit of J must be used in this particular type of problem.

**Example #2:** What is the ratio of the average velocity of krypton gas atoms to that of nitrogen molecules at the same temperature and pressure?

**Solution path one:**

1) Let us determine the velocity of Kr atoms at 273 K:

v = $\sqrt{\mathrm{(3RT)\; /\; M}}$v = $\sqrt{\mathrm{[(3)\; (8.31447)\; (273)]\; /\; 0.083798}}$

v = 285 m/s

2) Let us determine the velocity of N_{2} atoms at 273 K:

v = $\sqrt{\mathrm{[(3)\; (8.31447)\; (273)]\; /\; 0.028014}}$v = 493 m/s

3) Let us expresss the ratio both ways:

N_{2}: Kr ratio = 493 / 285 = 1.73 to 1Kr : N

_{2}ratio = 285 / 493 = 0.578 to 1

**Solution path two:**

Please note that the 3RT term is common to both velocity calculations above. 3R is a constant and T_{Kr} = T_{N2}. That means 3RT can be cancelled out if we are being asked for a ratio. Let's do the Kr : N_{2} ratio:

v_{Kr}/ v_{N2}= $\sqrt{\mathrm{(3RTKr/\; 0.083798)\; /\; (3RTN2/\; 0.028014)}}$v

_{Kr}/ v_{N2}= $\sqrt{\mathrm{0.028014\; /\; 0.083798}}$v

_{Kr}/ v_{N2}= 0.578

**Example #3:** At the same temperature, which molecule travels faster, O_{2} or N_{2}? How much faster?

**Solution:**

1) O_{2} velocity @ 273 K:

v = $\sqrt{\mathrm{3RT\; /\; M}}$v = $\sqrt{\mathrm{[(3)\; (8.31447)\; (273)]\; /\; 0.0319988}}$

v = 461.3 m/s

2) N_{2} velocity @ 273 K:

v = $\sqrt{\mathrm{[(3)\; (8.31447)\; (273)]\; /\; 0.028014}}$v = 493 m/s

3) N_{2} moves faster. How much faster?

493 / 461.3 = 1.0687N

_{2}moves about 1.07 times as fast as O_{2}, at the same temperature.

**Example #4:** A pure gas sample at 25 °C has an average molecular speed of 682 m/s. What is the identity of the gas?

A) CH_{4}B) H_{2}C) HCl D) NO E) CO_{2}

**Solution:**

v = $\sqrt{\mathrm{(3RT)\; /\; M}}$682 = $\sqrt{\mathrm{[(3)\; (8.31447)\; (298)]\; /\; M}}$

465124 = 7433.13618 / M

M = 0.01598 kg/mol

CH

_{4}weighs 0.0160426 kg/mol, a difference of only 0.4%. I think we're safe is picking CH_{4}as the correct answer.

**Example #5:** Analyze the gas velocity equation to determine the units on molar mass.

**Solution:**