Question

What is the RMS speed of Helium atoms when the temperature of the Helium gas is 206.0 K? (Possibly useful constants: the atomic mass of Helium is 4.00 AMU, the Atomic Mass Unit is: 1 AMU = 1.66x10-27 kg, Boltzmann's constant is: kg = 1.38x1023 J/K.) Submit AnswerTries 0/20 What would be the RMS speed, if the temperature of the Helium gas was doubled?

Answer 1

Answer: a)1.13×10³ b)1.6×10³

Step-by-step explanation:

Answer 2

a)1.13×10³

b)1.6×10³

Step-by-step explanation:

Given:

Boltzmann's constant (K)=1.38×10^-23 J/K

atmoic mass of helium = 4 AMU or 4×1.66×10^-27kg

a)The formula for RMS speed (Vrms) is given as

`Vrms=\sqrt{(3KT)/(m) }`

where

K= Boltzmann's constant

T= temperature

m=mass of the gas

`Vrms=\sqrt{(3* 1.38* 10^(-23)* 206)/(6.64* 10^(-27))}`

`Vrms=1.13* 10^(3)m/s`

b) RMS speed of helium when the temperature is doubled

`Vrms=\sqrt{(3* 1.38* 10^(-23)* 2* 206)/(6.64* 10^(-27))}`

`Vrms=1.598* 10^(3)m/s`