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.

Homework Equations

I used D = m/v

The Attempt at a Solution

D_He = m_He/v
D_N2 = m_N2/v
Both gases occupy the same volume, so just v for both.

Since D_He = D_N2,
m_Hr/v = m_N2/v and m_He = m_N2

For some x and y,
x mol He(4.003 g/mol He) = m_He
y mol N₂(28.01 g/mol N₂) = m_N2

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.

 

[Borek]

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.

 

[Bohrok]

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.

 

[Borek]

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

 

[Bohrok]

I think I got it now (don't know why I didn't look at it like this before)

I found that n_He/n_N2 = 7/1, and using P = nRT/V,

P_He/P_N2 = (n_HeRT/V)/(n_N2RT/V)
P_He/P_N2 = n_He/n_N2 = 7/1