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

Answer:

v = 1247.92 m/s

Explanation:

The formula for kinetic energy is given as follows:

$K.E = \frac{1}{2}mv^2$

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

$K.E = \frac{3}{2}KT$

Comparing both formulae for K.E:

$\frac{1}{2}mv^2 = \frac{3}{2}KT\\\\mv^2 = 3KT\\v = \sqrt{ \frac{3KT}{m}}$

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,

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

v = 1247.92 m/s