Respuesta :
Answer:
The density of air at 22 °C and 760 torr is 1.195 KG/m³
Explanation:
The solution to the above question is arrived t by considering the given variables and calculating the number of moles in a 1 m³ sample of air by plugging values into the universal gas equation from which the number of moles of the constituent gases can be calculated by Dalton's law of partial pressure, then their masses and lastly the density of air is calculated using the formula, Density = mass/volume
The given variables are
Percentage Nitrogen = 78.1% by volume
Percentage oxygen = 20.9% by volume
Percentage argon = 0.934% by volume
The molar mass of nitrogen = 14.006g/mol
The molar mass of oxygen = 15.999g/mol
The molar mass of argon = 39.948 g/mol hence Considering a unit volume of air of one cubic meter (1m^3) we have
0.781 m³ of nitrogen, 0.209 m³ of oxygen and 0.00934 m³ of argon
The number of moles in 1 m³ of gas at 22 °C and 760 torr is given by
PV = nRT or n = [tex]\frac{PV}{RT}[/tex] = where 760 torr = 101325Pa we have n = [tex]\frac{(101325)(0.001)}{(8.314)(295.15)}[/tex] = 0.00413 mols per liter or 41.29 moles/m³
thus we have number of moles of nitrogen = 42.129 × 78.1% = 32.25 moles and the mass of nitrogen = 32.25×28.02 = 903.6 g
number of moles of oxygen= 42.129 × 20.1% = 8.63 moles and the mass of nitrogen = 8.63×32 = 276.16 g
number of moles of argon= 42.129 × 0.934% = 0.386 moles and the mass of nitrogen = 0.386×40 = 15.43 g
Therefore, mass of one cubic meter of air (1 m³), has a mass of
903.6 g + 276.16 g + 15.43 g = 1195.2 g or 1.195 KG Hence the density of air
is given by Density = [tex]\frac{mass}{volume}[/tex] =[tex]\frac{1.195 KG}{1 m^{3} }[/tex] = 1.195 KG/m³
The density of air at 22 °C and 760 Torr is 1.19 g/L.
Air is 78.1% nitrogen, 20.9% oxygen, and 0.934% argon by moles. We will calculate the average molar mass of the air (M) as a weighted average of the molar masses of its constituents.
[tex]M = 78.1\% \times M(N_2) + 20.9\% \times M(O_2) + 0.934\% \times M(Ar)\\\\M = 78.1\% \times 28.00g/mol + 20.9\% \times 32.00g/mol + 0.934\% \times 39.95 g/mol = 28.93 g/mol[/tex]
Then, we will convert 22 °C to Kelvin using the following expression.
[tex]K = \° C + 273.15 = 22\° C + 273.15 = 295 K[/tex]
Assuming ideal behavior, we can calculate the density (ρ) of the air at 295 K (T) and 760 Torr (P) using the following expression.
[tex]\rho = \frac{P \times M }{R \times T} = \frac{760 Torr \times 28.93g/mol }{(62.4mmHg.L/mol.K) \times 295K} = 1.19 g/L[/tex]
where,
- R: ideal gas constant
The density of air at 22 °C and 760 Torr is 1.19 g/L.
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