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.
A cylinder with a movable piston contains 2.00g of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00L to 3.90L? (The temperature was held constant.)
mass of helium added=___g???
for this question...I tried to solve for moles of He. I went 2g He*1mol/4.00260gHe. I got: 0.499675moles of He.
Where do I apply this info to the equation & which equation would be suggested for this problem? Do I have to include room temp of 22.2*C (if this is room temp??)= 295.35K?
so...m=MRT/PV?? Should I not somewhere that the V1 changed from 2L -> V2=3.9L & use 3.90L, or subract & get a displacement of 1.9L??
is it something like n1/v1 =n2/v2??
5 answers
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 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   n1RT
---- = ------
P2V2   n2RT
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.
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   n1RT
---- = ------
P2V2   n2RT
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
3 answers