Respuesta :
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
[tex]n = \frac{Mass}{Molar \ mass} = \frac{2.00 \ g}{4.003 \ g/mol} = 0.49963 \ moles[/tex]
The pressure of the gas in the cylinder is given by;
[tex]P = \frac{nRT}{V} = \frac{0.49963 \times 8.3145 \times 294.15}{0.002} = 610,969.31676 \ Pa[/tex]
Therefore, when the volume is increased to 2.7 by adding more helium, we have
[tex]n = \frac{PV}{RT} = \frac{610,969.31676 \times 0.0027 }{8.3145 \times 294.15} = 0.67449 \ moles[/tex]
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 = â€0.69998111 ≈ 0.7 grams.
