Question

What is the average velocity of atoms in 1.00 mol of neon (a monatomic gas) at 288 K? Use the equation: 1/2 mv^2 = 3/2 nRT

For m, use 0.01000 kg. Remember that R = 8.31 J/(mol x K)

Answer 1

Answer: The average velocity of the atoms 847.33 m/s.

Step-by-step explanation:

Moles of the neon = 1.00

Temperature of the gas : 288 K

Mass of the gas = 0.01000

R = 8.31 J/mol K

`(1)/(2)mv^2=(3)/(2)* nRT`

`(1)/(2)0.01000 kg* v^2=(3)/(2)* 1.00* 8.31 J/mol.K* 288 K`

`v^2=(2* 3* 1.00* 8.31 J/mol.K* 288 K)/(1* 2* 0.01000 kg)=717,948`

`v=847.33 m/s`

The average velocity of the atoms 847.33 m/s.