Ideal gas law problem in outer space

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Homework Help Overview

The discussion revolves around applying the ideal gas law in the context of outer space, specifically focusing on the behavior of hydrogen atoms at a low temperature of 3.4K and a low density of one atom per cm³. Participants are tasked with calculating the average speed of hydrogen atoms and the resulting pressure in atmospheres.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to calculate the average speed and pressure using the ideal gas law and provided constants, but questions arise regarding the conversion between cubic centimeters and cubic meters, which may affect the calculations.

Discussion Status

Participants are actively questioning the assumptions made about volume conversions, particularly the relationship between cubic centimeters and cubic meters. Some have provided hints and clarifications regarding the correct number of cubic centimeters in a cubic meter, which may influence the original poster's calculations.

Contextual Notes

There is a noted uncertainty regarding the conversion factor between cubic centimeters and cubic meters, which is critical for the calculations being discussed. The original poster expresses concern about potential errors in their understanding of these units.

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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