Answer:
The amount nitrogen gas in the given balloon = 23.07 g
Explanation:
Given: Total volume of balloon: V = 18 L, Pressure = 1.4 atm, Temperature = 25°C = 25 + 273 = 298K (∵ 0°C = 273 K)
Volume of Nitrogen gas in balloon: V₁ = V × 80% = 18 L × (80 / 100) = 14.4 L
Molar mass of nitrogen gas (N₂): M = 28 g/mol
According to the ideal gas law:
PV= nRT
Here, Gas constant: R = 0.08206 L·atm/(mol·K)
n = total number of moles of gas
P = Pressure in atm
T = Temperature in K
V = Volume in L
Therefore, the number of moles of oxygen gas (n₁) is given by:
[tex]\Rightarrow n_{1} = \frac{PV_{1}}{RT}[/tex]
[tex]n_{1} = \frac{1.4 atm\times 14.4 L}{0.08206 L.atm/(mol.K)\times298K}[/tex]
[tex]n_{1} = 0.824 mole[/tex]
As number of moles: [tex]n = \frac{mass}{Molar\, mass}[/tex]
[tex]\Rightarrow Mass\, of \, nitrogen \, gas = n_{1} \times M = 0.824 mol \times 28 g/mol = 23.07 g[/tex]
Therefore, the amount nitrogen gas in the given balloon = 23.07 g