# Pressure Ratio of He to N2 at Equal Density

• Bohrok

## 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.

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. Last edited:
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 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