A. What is the RMS speed of Helium atoms when the temperature of the Helium gas is 343.0 K? (Possibly useful constants: the atomic mass of Helium is 4.00 AMU, the Atomic Mass Unit is: 1 AMU = 1.66*10^-27 kg, Boltzmann's constant is: kB = 1.38*10^-23 J/K.)

b. What would be the RMS speed, if the temperature of the Helium gas was doubled?

(a) 1462.38 m/s

(b) 2068.13 m/s

Explanation:

(a)

The Kinetic energy of the atom can be given as:

K. E = (3/2) KT

where,

K = Boltzman's Constant = 1.38 x 10⁻²³ J/k

K. E = Kinetic Energy of atoms = 343 K

T = absolute temperature of atoms

The K. E is also given as:

K. E = (1/2) mv²

Comparing both equations:

(1/2) mv² = (3/2) KT

v² = 3KT/m

v = √[3KT/m]

where,

m = mass of Helium = (4 A. M. U) (1.66 X 10⁻²⁷ kg / A. M. U) = 6.64 x 10⁻²⁷ kg

v = RMS Speed of Helium Atoms = ?

Therefore,

v = √[ (3) (1.38 x 10⁻²³ J/K) (343 K) / (6.64 x 10⁻²⁷ kg) ]

v = 1462.38 m/s

(b)

For double temperature:

T = 2 x 343 K = 686 K

all other data remains same:

v = √[ (3) (1.38 x 10⁻²³ J/K) (686 K) / (6.64 x 10⁻²⁷ kg) ]

v = 2068.13 m/s