Ideal gas law problem in outer space

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SUMMARY

The discussion centers on calculating the average speed and pressure of hydrogen atoms in outer space, given a density of one atom per cm³ and a temperature of 3.4K. The average speed of hydrogen atoms is calculated to be approximately 300 m/s using the equation m/2 = 3kT/2. The pressure calculation initially yielded an incorrect value of 5 x 10^-24 atm, which was later clarified to be approximately 4.8546 x 10^-22 atm after correcting the volume conversion from cubic centimeters to cubic meters.

PREREQUISITES
  • Understanding of the Ideal Gas Law
  • Familiarity with Boltzmann's constant (K = 1.381 x 10^-23 J/K)
  • Basic knowledge of atomic mass units (M_H = 1.0079 au)
  • Proficiency in unit conversions between cubic centimeters and cubic meters
NEXT STEPS
  • Study the Ideal Gas Law and its applications in astrophysics
  • Learn about the kinetic theory of gases and its mathematical formulations
  • Explore unit conversion techniques, especially for volume measurements
  • Investigate the properties of hydrogen and its behavior in low-density environments
USEFUL FOR

Students and educators in physics, particularly those focusing on thermodynamics and astrophysics, as well as anyone interested in the behavior of gases in outer space.

Freye
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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?
 
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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?
 
Freye said:
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.
 
Freye said:
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.
 

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