Example #1: Calculate the rms speed of an oxygen gas molecule, O2, at 31.0 °CSolution:v = (3RT) / Mv = <(3) (8.31447) (304.0)> / 0.0319988v = 486.8 m/sHere is the over set-up done with 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 that the unit on R is typically composed J/K mol. The even more extended unit of J need to be supplied in this certain form 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 pressure?Solution path one:1) Let us identify 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 proportion both ways:N2 : Kr ratio = 493 / 285 = 1.73 to 1Kr : N2 ratio = 285 / 493 = 0.578 to 1Solution route two:Please note that the 3RT term is common to both velocity calculations over. 3R is a continuous and TKr = TN2. That suggests 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 same temperature, which molecule travels quicker, O2 or N2? How much faster?Solution:1) O2 velocity
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273 K:v = <(3) (8.31447) (273)> / 0.028014v = 493 m/s3) N2 moves much faster. How much faster?493 / 461.3 = 1.0687N2 moves around 1.07 times as rapid as O2, at the very same temperature.Example #4: A pure gas sample at 25 °C has an average molecular rate 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 difference 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 units on molar mass.Solution:
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