# Answer to Question #7957 in Physical Chemistry for Ashley

Question #7957

what is th eroot-mean squared velocity of a sulfur dioxide, SO2, molecule at 275C ?

Expert's answer

To solve this, we apply the law stating that the average kinetic energy of a gas molecule is directly proportional to the temperature in Kelvin:

<KE> = 1/2(mv^2) = 3/2(KT), where m is the mass of the gas molecule, v is the velocity, T is the temperature and K is the Boltzmann

constant(1.38x10^-23m^2s^-2K^-1.

Expressing it in terms of v, we get v = √(3KT/m)

Relative Molecular Mass (RMM) of SO

6.02x10^23 molecules of SO

1 molecule of SO

Thus, v = √(3)(1.38x10^-23)(275+273)/1.06x10^-25) = 462.6 m/s (root mean square velocity of an SO

<KE> = 1/2(mv^2) = 3/2(KT), where m is the mass of the gas molecule, v is the velocity, T is the temperature and K is the Boltzmann

constant(1.38x10^-23m^2s^-2K^-1.

Expressing it in terms of v, we get v = √(3KT/m)

Relative Molecular Mass (RMM) of SO

_{2}= 646.02x10^23 molecules of SO

_{2}= 64 g1 molecule of SO

_{2}= 64/6.02x10^23 = 1.06x10^-22g = 1.06x10^-25kgThus, v = √(3)(1.38x10^-23)(275+273)/1.06x10^-25) = 462.6 m/s (root mean square velocity of an SO

_{2}molecule)Need a fast expert's response?

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