A cylinder with a movable piston contains 2.00 g
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 2.00 L
to 4.10 L
? (The temperature was held constant.)
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (constant)
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature (constant)
Since the pressure and temperature are constant, we can simplify the equation to:
V1 * n1 = V2 * n2
Given:
V1 = 2.00 L
V2 = 4.10 L
n1 = 2.00 g / molar mass of helium
n2 = x (unknown number of moles of helium added)
Molar mass of helium = 4.00 g/mol
We can set-up the equation:
2.00 * (2.00 / 4.00) = 4.10 * x
Solving for x:
x = (2.00 * 2.00) / 4.10
x = 0.9756 moles
Now, we can convert moles of helium to grams:
0.9756 moles * 4.00 g/mol = 3.9024 g
Therefore, 3.9024 grams of helium were added to the cylinder.