Question

Four balloons, each with a mass of 10.0 g, are inflated to a volume of 20.0 L, each with a different gas: helium, neon, carbon monoxide, or nitrogen monoxide. If the temperature is 25.0°C and the atmospheric pressure is1.00 atm, what is the density of each filled balloon? (Note: The density of the filled balloon refers to both the contents of the balloon and the balloon itself.)

Answer 1

On temperature 25°C (298,15K) and pressure of 1 atm each gas has same amount of substance:
n(gas) = p·V ÷ R·T = 1 atm · 20L ÷ 0,082 L·atm/K·mol · 298,15 K
n(gas) = 0,82 mol.
1) m(He) = 0,82 mol · 4 g/mol = 3,28 g.
d(He) = 10 g + 3,28 g ÷ 20 L = 0,664 g/L.
2) m(Ne) = 0,82 mol · 20,17 g/mol = 16,53 g.
d(Ne) = 26,53 g ÷ 20 L = 1,27 g/L.
3) m(CO) = 0,82 mol ·28 g/mol = 22,96 g.
d(CO) = 32,96 g ÷ 20L = 1,648 g/L.
4) m(NO) = 0,82 mol ·30 g/mol = 24,6 g.
d(NO) = 34,6 g ÷ 20 L = 1,73 g/L.

Answer 2

Answer:

$T=25^o C+273 K + 25= 298 K(0^oC=273 K)$

Pressure ,P = 1 atm

Volume, V = 20.0 L

R = 0.0821 atm L/mol K

Balloon 1

$PV=n_{He}RT=\frac{\text{mass of He}}{\text{molar mass of He}}\times RT$

$\text{mass of He}=\frac{PV\times \text{molar mass of He}}{RT}=\frac{1atm \times 20.0 L\times 4 g/mol}{0.0821 atm L/mol K\times 298 K}=3.26 grams$

Density of balloon 1 = $\rho _1=\frac{\text{mass of balloon+mass of He}}{volume}=\frac{10.0 g+3.26 g}{20 L}=0.663 g/L$

Balloon 2

$PV=n_{Ne}RT=\frac{\text{mass of Ne}}{\text{molar mass of Ne}}\times RT$

$\text{mass of Ne}=\frac{PV\times \text{molar mass of Ne}}{RT}=\frac{1atm \times 20.0 L\times 20.18 g/mol}{0.0821 atm L/mol K\times 298 K}=16.49 grams$

Density of balloon 2= $\rho _2=\frac{\text{mass of balloon+mass of Ne}}{volume}=\frac{10.0 g+16.49 g}{20 L}=1.32 g/L$

Balloon 3

$PV=n_{CO}RT=\frac{\text{mass of CO}}{\text{molar mass of CO}}\times RT$

$\text{mass of CO}=\frac{PV\times \text{molar mass of CO}}{RT}=\frac{1atm \times 20.0 L\times 28 g/mol}{0.0821 atm L/mol K\times 298 K}=22.88 grams$

Density of balloon 3 = $\rho _3=\frac{\text{mass of balloon+mass of CO}}{volume}=\frac{10.0 g+22.88 g}{20 L}=1.64 g/L$

Balloon 4

$PV=n_{NO}RT=\frac{\text{mass of NO}}{\text{molar mass of NO}}\times RT$

$\text{mass of NO}=\frac{PV\times \text{molar mass of NO}}{RT}=\frac{1atm \times 20.0 L\times 30 g/mol}{0.0821 atm L/mol K\times 298 K}=24.52 grams$

Density of balloon 4= $\rho _4=\frac{\text{mass of balloon+mass of NO}}{volume}=\frac{10.0 g+24.52 g}{20 L}=1.72 g/L$