Question

A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium is added to the container and the volume adjusted so that the pressure remains the same. How many grams of helium are added to the cylinder if the volume changes from 2.00 L to 2.70 L

Answer 1

Answer:

The mass of helium added is approximately 0.7 grams

Explanation:

Here we have

Initial mass of helium gas in the cylinder = 2.00 g

Initial pressure of gas = Final pressure of gas

Initial volume of gas = 2.00 L

Final volume of gas = 2.70 L = ‪0.0027 m

Initial temperature of the gas = room temperature = 21 °C

Molar mass of helium gas = 4.003 g/mol

Therefore, the number of moles, n of helium is given as

$n = \frac{Mass}{Molar \ mass} = \frac{2.00 \ g}{4.003 \ g/mol} = 0.49963 \ moles$

The pressure of the gas in the cylinder is given by;

$P = \frac{nRT}{V} = \frac{0.49963 \times 8.3145 \times 294.15}{0.002} = 610,969.31676 \ Pa$

Therefore, when the volume is increased to 2.7 by adding more helium, we have

$n = \frac{PV}{RT} = \frac{610,969.31676 \times 0.0027 }{8.3145 \times 294.15} = 0.67449 \ moles$

That is the number of moles of helium added is given by;

0.67449 - 0.49963 = 0.174864 moles

Mass of helium  added = Number of moles added × Molar mass of helium

0.174864 moles × 4.003 g/mol = ‭0.69998111 ≈ 0.7 grams.