# Calculate the root mean square velocity of f2,cl2, and br2 at 310 k .

The average kinetic energy remains constant as the particle varies (same for different patracles of different compounds at the same temperature) however the root mean square is dependant on the temperature as well ans the molar mass

Well here are the equations:

root mean square velocity=sqrt(3RT/M)

where R=8.3145kgmeters^2/(seconds^2temperature in kelvinmol)

T=temperature in kelvin

M=molar mass in kilograms (Kg/mol)

average kinetic energy=(3/2)*RT

Where R=8.3145Joules/(temperature in kelvinmoles)

T=temperature in kelvin

The moles can be changed to particles using avagadro’s number

avagodro’s number=6.02214*10^23

on to calculations starting with root mean square velocity:

We first neeg the molar masses so:

2*19.00=38.00 g/mol F2=0.03800 kg/mol F2

2*35.45=70.90 g/mol Cl2=0.07090 kg/mol Cl2

2*79.90=159.80 g/mol Br2=0.15980 kg/mol Br2

We use the above equation to set up the problem:

sqrt(38.3145292/each molar mass in kg)

all your units should cancel except m^2 and s^2 and I will skip strait to the final answers with three signiFI.GΛnt figures (292k is limiting)

1) 438 m/s

2) 320 m/s

3) 213 m/s

The answer makes sense as bigger paticles move slower than smaller particles

The average kinetic energies of the three gasses are the same as they are all at the same temperature however you must change moles to representative particles (molecules) so the setup is:

(3/2)8.3145292/(6.02214*10^23)

Your final answers are all similar, are in Joules per molecules and have three signiFI.GΛnt figures because 292k is limiting

1) 6.05*10^-21 Joules/molecule F2

2) 6.05*10^-21 Joules/molecule Cl2

3) 6.05*10^-21 Joules/molecule Br2