# Ideal gas law problem in outer space

## Homework Statement

In outer space, the density of matter is about one atom per cm^3, mainly hydrogen atoms, and the temperature is about 3.4K. Calculate the average speed of the hydrogen atoms, and the pressure (in atmospheres)

M_H = 1.0079au = 1.67 X 10^-27 Kg
1 Hydrogen atom per cm^3 = 100 hydrogen atoms per m^3
K (Boltzmann's constant) = 1.381 X 10^-23
T = 3.4K

## Homework Equations

m$$\bar{}v$$^2 /2 = 3kT/2
PV = Nm$$\bar{}v$$^2 /3

## The Attempt at a Solution

part 1:
m$$\bar{}v$$^2 /2 = 3kT/2
$$\bar{}v$$ = $$\sqrt{}3(1.381 X 10^-23)(3.4)/(1.67 X 10^-27)$$
$$\bar{}v$$ $$\approx$$ 300m/s

part 2:
Consider V = 1m^3

PV = Nm$$\bar{}v$$^2 /3
P =Nm$$\bar{}v$$^2 /3V
=(100)(1.67 X 10^-25)(300)^2/3(1)
P $$\approx$$ 5X 10^-19 N/m^2 $$\approx$$ 5 X10^-24 atm

As usual, I get the first part of the question right, but the answer to the second part is 5 X 10^-22 not ^-24. Anyone know what I'm doing wrong?

Related Introductory Physics Homework Help News on Phys.org
How many cc in a cubic metre?

Hint 1 cc = sugar cube, 1m^3 = refridgerator

Isn't there 100 cubic centimetres in a cubic metre? I'll be really embarrassed if that's where my mistake is. If there is 100 cc in 1m^3, then my "1 Hydrogen atom per cm^3 = 100 hydrogen atoms per m^3" is correct, is it not?

Isn't there 100 cubic centimetres in a cubic metre? I'll be really embarrassed if that's where my mistake is. If there is 100 cc in 1m^3, then my "1 Hydrogen atom per cm^3 = 100 hydrogen atoms per m^3" is correct, is it not?
1 cubic centimeter = 1.0 × 10-6 cubic meters.
When you use a million, you should get: 5.01E-17 Pa, when converted, you end up with 4.8546E-22 atm.

Isn't there 100 cubic centimetres in a cubic metre? I'll be really embarrassed if that's where my mistake is.
A cubic metre is a 100cm on one side, 100cm on the width and 100cm on the height.

As I said, picture it
1 cc is a sugar cube, 100 sugar cubes would be the size of your hand , 1000 sugar cubes would be 10x10x10cm the size of a saucepan.

Oh gawd, now I am embarrassed, lol. Thanks guys, that clears that up.