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

Explanation:

Answer 2

Answer:

a)1.13×10³

b)1.6×10³

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

$Vrms=\sqrt{\frac{3KT}{m} }$

where

K= Boltzmann's constant

T= temperature

m=mass of the gas

$Vrms=\sqrt{\frac{3\times 1.38\times 10^{-23}\times 206}{6.64\times 10^{-27}}}$

$Vrms=1.13\times 10^{3}m/s$

b) RMS speed of helium when the temperature is doubled

$Vrms=\sqrt{\frac{3\times 1.38\times 10^{-23}\times 2\times 206}{6.64\times 10^{-27}}}$

$Vrms=1.598\times 10^{3}m/s$