[Cyberconvict]

A dirigible holds 2.0x10⁵ m³ of hydrogen gas at 23C and 1.0 atm, what mass of hydrogen is present?

I'm not sure where I'm going wrong, but here's my process.

I'm going to use the Ideal-Gas Equation and plug-in the variables.
(1.0atm)(2.0x10⁵)=n(0.0821)(296K)

That comes out to n=8229.9mol.

I then divide that by 2 and I would think that would give me the number of grams of H₂. Using those numbers I get 4114g, but the back of the book says 1.7x10⁴ kg. Where's my mistake?

EDIT: Nevermind, very stupid mistake that I just kept on missing. Obviously I should have multipled by the atomic weight instead of dividing... Time to get some sleep...

[xiankai]

even if u multiply by 2, u will not get the answer.

your mistake lies in the fact that the units of the gas constant, (L atm K^-1 mol^-1) is not constant with your unit for volume (m³)

[lemonoman]

Good catch Xiankai ! 

(1.0 atm)(2.0 x 10⁵ m³)= n(8.2057459 x 10^-5 m³ atm K^-1 mol^-1)(296 K)
so
n  = 8234223 mol
= 16599206 g H₂
~ 1.7 × 10⁴ kg (as it says in the back of the book)

I assumed he used the correct constant, and just didn't put his exponent in the right place.  The answer he got is probably off by an exponent of 3...(the orders of magnitude of difference between L and m³)...but didn't notice that his answer was in g, not kg.

But anyways, he believes the problem is fixed, and may not be back to this post (though with 13 posts, he will be back to the forums ) ... so let this be a lesson to the rest of you...WATCH YOUR UNITS!

[Cyberconvict]

Opps, conversion factor between m³ and L is 10³ isn't it? I saw that on some site and failed to put it into the problem when I posted it.