CHEM

I tired 7.79g of He added & it told me this:

This number looks like the final mass of He multiplied by 2. Instead, find the difference between the final and initial masses.

I don't know what to make of the note you received.
I would do this.
PV = nRT
Solve for P.
V = 2.00 L, n is 0.5 mol, you know R, and you can use any T since it isn't changing. I used 298. That gives the pressure for the original material of 2.00 g.

Then use this P for a new PV = nRT.
V = 3.90 L, solve for n, you know R and use the same T as above. Then n = g/atomic mass or g = n x 4 = ??

This will give total grams of the original + added amount so subtract 2.00 from that to get the added amount.

Check my thinking. It looks like about 1.90 grams added but check my arithmetic.

I worked the info above out &, yes, your thinking was indeed correct. Thank you heeps.

I thought about this after I went to bed and there is an easier way to do it.
P1V1=n1RT for one set of conditions.
P2V2=n2RT for the second set of conditions. Now divide the first equation by the second equation.
P1V1&nbsp&nbsp&nbspn1RT
---- = ------
P2V2&nbsp&nbsp&nbspn2RT

P1=P2 so they cancel
V1=2.00 L
V2 = 3.90 L
n1 = 0.5 mol
n2 = solve for this
R and T are equal and they cancel.
We are left with
2.00/3.90 = 0.5/n2
n2 = 0.975 mols
0.975 mols x 4 g/mol = 3.90 g
3.90-2.00 = 1.90 g added.

1.70