Question

A 0.50 mº gas tank holds 3.0 moles of ideal monatomic Helium gas at a temperature of 250 K. What is the mms speed of the molecules? (The Boltzmann constant is 1.38 x 10-23 J/K, NA = 6.022 x 1023 molecules/mol.)

Answer 1

v = 1247.92 m/s

Step-by-step explanation:

The formula for kinetic energy is given as follows:

`K.E = (1)/(2)mv^2`

Another formula that is used for Kinetic Energy is given as:

`K.E = (3)/(2)KT`

Comparing both formulae for K.E:

`(1)/(2)mv^2 = (3)/(2)KT\\\\mv^2 = 3KT\\v = \sqrt{ (3KT)/(m)}`

where,

v = rms speed of helium molecule = ?

K = Boltzmann Constant = 1.38 x 10⁻²³ J/k

T = Absolute Temperature = 250 K

m = mass of helium molecule = 6.646 x 10⁻²⁷ kg

Therefore,

`v = \sqrt{((3)(1.38\ x\ 10^(-23)\ J/k)(250\ k))/(6.646\ x\ 10^(-27)\ kg)} \\\\`

v = 1247.92 m/s