Calculating Pressure of a Dilute Gas in Space

AI Thread Summary
The discussion focuses on calculating the pressure of a dilute hydrogen gas in outer space at a temperature of 3.0K. The initial approach involved using the ideal gas law, PV = nRT, but the user struggled with determining the correct number of moles due to the low density of hydrogen. After clarification that one atom corresponds to 1/6.022x10^23 moles, the user was able to correctly compute the pressure as 1.4x10^-23 Pa. The conversation highlights the importance of accurately converting units and understanding mole calculations in gas law applications.
Bashkir
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Homework Statement



The ambient temperature in outer space is about 3.0K. Most of the matter in space is hydrogen gas at a density of about one atom per 3.0m^3. What is the pressure of this dilute gas if it is in thermal equilibrium at the ambient temperature.

Homework Equations



PV = nRT

The Attempt at a Solution



I started with saying that PV = nRT and substituting volume for mass/density since I was given density. I cannot seem to get the correct answer playing with the numbers for mass and density of the hydrogen. The correct answer is 1.4x10-23Pa.
 
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Bashkir said:

Homework Statement



The ambient temperature in outer space is about 3.0K. Most of the matter in space is hydrogen gas at a density of about one atom per 3.0m^3. What is the pressure of this dilute gas if it is in thermal equilibrium at the ambient temperature.

Homework Equations



PV = nRT

The Attempt at a Solution



I started with saying that PV = nRT and substituting volume for mass/density since I was given density. I cannot seem to get the correct answer playing with the numbers for mass and density of the hydrogen. The correct answer is 1.4x10-23Pa.
Show us what you did (in detail), please.
 
Well, like I said, I made the equation

P = nRT/V

Then, I replaced volume with mass/density

P = nRTρ/m

After that, the units on density are kg/m3, so I converted the mass of hydrogen into kilograms.

I wasn't given how many moles I had, so I just assumed (probably a bad idea) 1.00mol for simplicity.

After putting this into my calculator it did not give me the desired answer, but I don't know of any other way to look at the problem unless I am given a number a number of moles.
 
Bashkir said:
Well, like I said, I made the equation

P = nRT/V

Then, I replaced volume with mass/density

P = nRTρ/m

After that, the units on density are kg/m3, so I converted the mass of hydrogen into kilograms.

I wasn't given how many moles I had, so I just assumed (probably a bad idea) 1.00mol for simplicity.

After putting this into my calculator it did not give me the desired answer, but I don't know of any other way to look at the problem unless I am given a number a number of moles.
You are given the number of moles in the 3m^3. How many moles is one atom?
 
1/6.022x10^23.

That makes a lot more sense. I am now getting the correct answer. Thank you very much!
 
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