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??
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.
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.
1.70
Why did the helium go to therapy? Because it wanted to become more stable!
To solve this problem, you're on the right track with the equation n1/v1 =n2/v2. First, let's find n1, the initial number of moles of helium. You correctly calculated it as 0.499675 moles.
Now, we need to find n2, the final number of moles of helium. Since the pressure is constant and the temperature is constant, according to the ideal gas law, n1/v1 =n2/v2. Plugging in the values, we have 0.499675 moles / 2.00L = n2 / 3.90L.
Solving for n2, we get n2 = (0.499675 moles / 2.00L) * 3.90L. The moles cancel out, and you're left with n2 = 0.97470625 moles.
Finally, to find the mass of the helium added, we use the formula mass = moles * molar mass. The molar mass of helium is approximately 4.00260 g/mol. So, the mass of the helium added is 0.97470625 moles * 4.00260 g/mol = 3.9000314 grams.
Therefore, approximately 3.90 grams of helium were added to the cylinder.
To solve this problem, you're on the right track using the ideal gas law equation:
PV = nRT
However, since the problem states that the pressure is constant, you can simplify the equation to:
n1/V1 = n2/V2
Let's assign the variables to make it clearer:
n1 = initial moles of helium (2.00g)
V1 = initial volume (2.00L)
n2 = final moles of helium (unknown)
V2 = final volume (3.90L)
Now, we can plug in the given values into the equation:
2.00g / 2.00L = n2 / 3.90L
To solve for n2 (final moles of helium), you can cross-multiply:
(2.00g)(3.90L) = (2.00L)(n2)
Simplifying:
7.80g∙L = 2.00L∙n2
Now, divide both sides by 2.00L to solve for n2:
n2 = (7.80g∙L) / (2.00L)
n2 = 3.90g
Therefore, you added 3.90 grams of helium to the cylinder.