**v = (3RT) / MYou might, if you wish, review even more about the over equation here.The basic concept is that, if you take into consideration each gas molecule"s velocity (which has actually components of both speed and direction), the average velocity of all gas molecules in a sample is zero. That stems from the reality that the gas molecules are relocating in all directions in a random method and each random speed in one direction is cancelled out by a molecule randomly relocating in the precise opposite direction, through the precise very same rate (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 relocating in one method have a positive sign for their direction, those moving in the opposite direction have actually an unfavorable direction). You then average all of these squared values and also take the square root of that average.It"s referred to as a "root intend square" and also technically, it is a speed, not a velocity. However before, in chemistry, we neglect the distinction in between rate and also velocity and use velocity.Some crucial information:R = 8.31447 J mol¯1 K¯1M = the molar mass of the substance, expressed in kilograms**Look at exactly how the systems cancel in v = (3RT) / M

**Example #1:**Calculate the rms rate of an oxygen gas molecule, O2, at 31.0 °C

**Solution:**

**v = (3RT) / Mv = <(3) (8.31447) (304.0)> / 0.0319988v = 486.8 m/sHere is the over set-up done via units:v = <(3) (8.31447 kg m2 s-2 K-1 mol-1) (304.0 K) / 0.0319988 kg/molRemember that kg m2 s-2 is referred to as a Joule and also that the unit on R is commonly created J/K mol. The more extfinished unit of J need to be provided 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 exact same temperature and also pressure?

**Equipment route one:**1) Let us determine the velocity of Kr atoms at 273 K:

**v = (3RT) / Mv = <(3) (8.31447) (273)> / 0.083798v = 285 m/s2) Let us identify the velocity of N2 atoms at 273 K:v = <(3) (8.31447) (273)> / 0.028014v = 493 m/s3) Let us expresss the ratio both ways:N2 : Kr proportion = 493 / 285 = 1.73 to 1Kr : N2 proportion = 285 / 493 = 0.578 to 1Equipment route two:**Please note that the 3RT term is common to both velocity calculations above. 3R is a constant and also TKr = TN2. That implies 3RT have the right to be cancelled out if we are being asked for a proportion. Let"s perform the Kr : N2 ratio:

**vKr / vN2 = (3RTKr / 0.083798) / (3RTN2 / 0.028014)vKr / vN2 = 0.028014 / 0.083798vKr / vN2 = 0.578Example #3:**At the very same temperature, which molecule travels much faster, O2 or N2? How much faster?

**Solution:**1) O2 velocity

You are watching: Calculate the rms speed of an oxygen gas molecule, o2, at 23.0 ∘c .

273 K:v = <(3) (8.31447) (273)> / 0.028014v = 493 m/s3) N2 moves much faster. How a lot faster?493 / 461.3 = 1.0687N2 moves about 1.07 times as fast as O2, at the exact same temperature.Example #4: A pure gas sample at 25 °C has actually an average molecular speed of 682 m/s.

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What is the identity of the gas?

**A) CH4 B) H2 C) HCl D) NO E) CO2Solution:**

**v = (3RT) / M682 = <(3) (8.31447) (298)> / M465124 = 7433.13618 / MM = 0.01598 kg/molCH4 weighs 0.0160426 kg/mol, a distinction of just 0.4%. I think we"re safe is picking CH4 as the correct answer.Example #5:**Analyze the gas velocity equation to determine the systems on molar mass.

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