Pressure Ratio of He to N2 at Equal Density

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Discussion Overview

The discussion revolves around calculating the pressure ratio of helium (He) to nitrogen (N2) when both gases have the same density at equal temperatures. The context includes a homework problem involving gas laws and density calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant outlines the relationship between the densities of He and N2, suggesting that if both gases occupy the same volume, their masses can be compared directly.
  • The same participant proposes using the ideal gas law (PV = nRT) to relate the pressure ratios to the number of moles of each gas.
  • Another participant expresses confusion about the original question, questioning the use of neon and mentioning inconsistencies in the problem statement.
  • A later reply indicates that the initial errors have been corrected, clarifying the focus on He and N2.
  • One participant concludes that the ratio of the number of moles (7:1) leads to a corresponding pressure ratio, suggesting that the pressures can be directly compared based on the mole ratio.

Areas of Agreement / Disagreement

Participants generally agree on the approach of using the ideal gas law to find the pressure ratio based on the mole ratio, although there was initial confusion regarding the problem statement and some inconsistencies in the calculations.

Contextual Notes

There were references to potential typos and inconsistencies in the problem statement, as well as the need for clarification on the gases involved. The discussion reflects a progression from confusion to a clearer understanding of the problem.

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Homework Statement



Calculate the pressure ratio of He to N2 at which helium would have the same density as nitrogen if their temperatures were the same.

Homework Equations



I used D = m/v

The Attempt at a Solution



DHe = mHe/v
DN2 = mN2/v
Both gases occupy the same volume, so just v for both.

Since DHe = DN2,
mHr/v = mN2/v and mHe = mN2

For some x and y,
x mol He(4.003 g/mol He) = mHe
y mol N2(28.01 g/mol N2) = mN2

4.003x g = 28.01y g
x = 7y

To me it looks like there are 7 times as many moles of He as N2 but I doubt that would directly apply to their pressure ratios. I think I'd have to use PV = nRT but I'm not sure how I'd put it in.
I was actually helping some chemistry students earlier today with this and am hoping I can have the answer ready for them tomorrow morning. :smile:
 
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Honestly, I have no idea what the question asks. Oxygen? And why do you use neon in your calculations?

Could be what you did is OK, but with all these typos/inconsistencies it is not.
 
I fixed the Ne and oxygen; I have an older edition of the book than that which the students are using and I hadn't quite changed everything to match the problem in their book. Should be alright now.
 
OK.

Now, knowing ratio of numbers of moles try to calculate ratio of pressures using PV=nRT. Don't be surprised if everything cancels out :wink:
 
I think I got it now (don't know why I didn't look at it like this before)

I found that nHe/nN2 = 7/1, and using P = nRT/V,

PHe/PN2 = (nHeRT/V)/(nN2RT/V)
PHe/PN2 = nHe/nN2 = 7/1
 

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