Calculate the meansquare speed, of krypton atoms in a sample of gas at a temperature of 22ºC?
Mass of 1 mole of krypton = 0.084kg
Mass of 1 mole of krypton = 0.084kg
1 Answer
Explanation:
Both kinetic energy and the ideal gas law provide expressions for pressure. Those expressions must be numerically equal, so it must be that
#1/2m< v^2 > =3/2kT#
We can solve for the meansquare speed:
#< v^2 > =(3kT)/m#
where:

#k# is Boltzmann's constant#(k_b=1.381xx10^(23)m^2kg//s^2K)# 
#T# is the temperature of the gas (in Kelvin) 
#m# is the mass of one molecule of the gas (in kg)
Then
Given that one mole of the gas has
#(0.084"kg")/"mole"*(1"mole")/(6.022*10^23"molecules")#
#=>m=1.395xx10^25"kg"#
And so we have:
#< v^2 > = (3(1.381xx10^(23)m^2kg//s^2K)(295"K"))/(1.395xx10^25"kg")#
#=>< v^2 > ~~8.8xx10^4"m"//"s"#
Usually, you are asked for the root mean square speed, which is the square root of the above. That would give
You may also see the equation written in terms of the universal gas constant
Here is a question I have answered previously on Socratic pertaining to this expression and similar expressions using the maxwell speed distribution.