Q1)Calculate the root mean square velocity and kinetic energy of CO,CO2and SO3 a

Question

Q1)Calculate the root mean square velocity and kinetic energy of CO,CO2and SO3 at 298K. •Which has the greatest velocity? •the greatest kinetic energy? •the greatest effusion rate?
Q2)Calculate the rate effusion rates for Ar and Kr. Q1)Calculate the root mean square velocity and kinetic energy of CO,CO2and SO3 at 298K. •Which has the greatest velocity? •the greatest kinetic energy? •the greatest effusion rate?
Q2)Calculate the rate effusion rates for Ar and Kr. •Which has the greatest velocity? •the greatest kinetic energy? •the greatest effusion rate?
Q2)Calculate the rate effusion rates for Ar and Kr.

Explanation / Answer

1. Root mean square velocity , Crms = (3RT/M)    where, M = molar mass

Kinetic energy , Ek = 3RT/2            So, kinetic energy is independent of the mass of the molecule, it is a function of absolute temperature only. Every gas molecule has same kinetic energy at a particular temepature.

Ek = 3RT/2 = 3*8.314 J.mol-1.K-1*298 K/2 = 3.716 kJ/mol

Crms:

For CO = (3RT/M) = (3*8.314 kg.m2.s-2 .mol-1.K-1*298 K/ 0.028 kg.mol-1 )

= 515.2 m/s

For CO2 = (3RT/M) = (3*8.314 kg.m2.s-2 .mol-1.K-1*298 K/ 0.044 kg.mol-1 )

= 411 m/s

For SO3 = (3RT/M) = (3*8.314 kg.m2.s-2 .mol-1.K-1*298 K/ 0.080 kg.mol-1 )

= 304.8 m/s

Graham's law of effusions says that the rate of effusion of a gas is inversely proportional to the square root of its molar mass ,i.e.,   Rate of effusion      (1/M)

CO has the lowest molar mass among the three, so it has the highest effusion rate.

Q2) Ratio of effusion rates for Ar and Kr = (MKr/MAr) = (83.8/40) = 1.45

So Ar moves 45% faster than Kr.

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