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 3.70 L ? (The temperature was held constant.)

To solve this problem, we can use the ideal gas law equation: PV = nRT.

In this case, we are told that the pressure (P) remains constant, so we can rewrite the equation as V1/n1 = V2/n2, where V1 and n1 are the initial volume and amount of helium in the cylinder respectively, and V2 and n2 are the final volume and amount of helium respectively.

We are given that the initial volume, V1, is 2.00 L and the final volume, V2, is 3.70 L. We are also given that the amount of helium initially is 2.00 g.

First, we need to find the initial amount of helium, n1. We can calculate this using the molar mass of helium (4.00 g/mol):
n1 = mass/molar mass = 2.00 g / 4.00 g/mol = 0.50 mol

Now, we can use the equation V1/n1 = V2/n2 to find n2:
V1/n1 = V2/n2
n2 = (V2 * n1) / V1
n2 = (3.70 L * 0.50 mol) / 2.00 L
n2 = 0.925 mol

Finally, we can calculate the mass of the additional helium added to the cylinder by subtracting the initial amount of helium from the final amount:
mass of additional helium = (n2 - n1) * molar mass
mass of additional helium = (0.925 mol - 0.50 mol) * 4.00 g/mol
mass of additional helium = 0.425 mol * 4.00 g/mol
mass of additional helium = 1.70 g

Therefore, 1.70 grams of helium were added to the cylinder.