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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Option (D) is the correct answer.

Explanation:

When we use an object or substance in excess then it is likely that the substance will run out with time. But at the same time, if we use that substance according to its requirement and try to conserve it as much as we can then we can preserve the substance.

Hence, it is good to avoid wastage of a substance and balance its use according to the need.

Therefore, we can conclude that in the given situation yes, there is a problem with the solution as the need to conserve copper must be balanced with use.