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olchik [2.2K]
3 years ago
11

A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and t

he 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.)
Chemistry
1 answer:
aliya0001 [1]3 years ago
5 0

Hey There!

Ideal gas equation: PV = nRT where P is pressure, V is volume, n is moles of gas, R is molar gas constant and T is temperature.

Since P, R and T are constant, V/n = constant

V1 = 2.00 L, V2 = 3.30 L

n1 = (2.00 g)/(4.00 g/mol) = 0.5 mol, n2 = ?

V1/n1 = V2/n2

2.00/0.5 = 3.30/n2

n2 = 0.825 mol

Moles of He added = n2 - n1

= 0.825 - 0.5 = 0.325 mol

Mass of He added = moles x atomic mass

= 0.325 * 4.00 = 1.30 g of He

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