What is the average speed (actually the root-mean-square speed) of a argon atom at 27°c?

The average speed (actually the root-mean-square speed) of the argon atom is 432.8 m/s

The thermodynamic temperature T is proportional to the average kinetic energy of a molecule. The mean translational kinetic energy of an atom or (molecule) of an ideal gas is proportional to the thermodynamic temperature.

That is, the average speed (actually the root-mean-square speed)

1/2m = 3KT/2

Where

m = molar mass

c =  speed

K = Bolzman constant

T = Temperature.

Given that the temperature of a argon atom is 27°C

Mass of an argon m = 6.63 x Kg

Bolzman constant K = 1.38 x J/K

Temperature T = 273 + 27 = 300K

Substitute all the parameters into the equation

1/2m = 3KT/2

m = 3KT

= 3KT/m

= (3 x 1.38 x x 300) / 6.63 x

= 187330.3

c =

c = 432.8 m/s

Therefore, the average speed (actually the root-mean-square speed) of the argon atom is 432.8 m/s

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the answer is d. 74.02