# What temperature does Cl2 need to escape Earth?

## Homework Statement

At what temperature will Cl2 molecules have sufficient velocity to escape the Earth’s gravitational pull? The escape velocity from the Earth is 25,000 mph. (Note: Cl2 is diatomic, so you’ll need to double the atomic weight shown on the periodic table.)
Cl=35.5g/mol v=11.2km/s

v=√3kT/m

## The Attempt at a Solution

Cl=35.5g/mol Cl2≈71g/mol = 0.071kg/mol

.071kg/mol/6.02x10^23=1.18x10^-25

11.2km/s=√3(1.38x10^-23)T/(1.18x10^-25)

This is where I am stuck. I don't know where to go from here. I'm not looking for an answer, just an explanation of the steps. My professor doesn't explain anything. Thank you!

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tms
You should always solve problems symbolically before plugging in numbers. That is the only way to learn the physics.

Look at a single molecule, not a whole mole. Relate the energy to temperature. There is a distribution function that does that.

gneill
Mentor

## Homework Statement

At what temperature will Cl2 molecules have sufficient velocity to escape the Earth’s gravitational pull? The escape velocity from the Earth is 25,000 mph. (Note: Cl2 is diatomic, so you’ll need to double the atomic weight shown on the periodic table.)
Cl=35.5g/mol v=11.2km/s

v=√3kT/m

## The Attempt at a Solution

Cl=35.5g/mol Cl2≈71g/mol = 0.071kg/mol

.071kg/mol/6.02x10^23=1.18x10^-25

11.2km/s=√3(1.38x10^-23)T/(1.18x10^-25)

This is where I am stuck. I don't know where to go from here. I'm not looking for an answer, just an explanation of the steps. My professor doesn't explain anything. Thank you!
Won't the particle speeds follow a statistical distribution (Maxwell-Boltzmann) for any given temperature? Given that, some fraction of the molecules must always have a speed equal to or greater than escape speed.

Perhaps there are some unstated assumptions? They may want you to find the temperature where the average speed is sufficient. Or maybe the RMS speed?

ehild
Homework Helper

## The Attempt at a Solution

Cl=35.5g/mol Cl2≈71g/mol = 0.071kg/mol

.071kg/mol/6.02x10^23=1.18x10^-25

11.2km/s=√3(1.38x10^-23)T/(1.18x10^-25)
Do not forget that the square root refers to the whole expression on the right-hand side. But you do not need the square root. Write up the expression for v^2. Transform km/s to m/s and isolate T.

ehild