Answer:
The mass of air stored in the vessel is 235.34 kilograms.
Explanation:
Let supossed that air inside pressure vessel is an ideal gas, The density of the air ([tex]\rho[/tex]), measured in kilograms per cubic meter, is defined by following equation:
[tex]\rho = \frac{P\cdot M}{R_{u}\cdot T}[/tex] (1)
Where:
[tex]P[/tex] - Pressure, measured in kilopascals.
[tex]M[/tex] - Molar mass, measured in kilomoles per kilogram.
[tex]R_{u}[/tex] - Ideal gas constant, measured in kilopascal-cubic meters per kilomole-Kelvin.
[tex]T[/tex] - Temperature, measured in Kelvin.
If we know that [tex]P = 2026.5\,kPa[/tex], [tex]M = 28.965\,\frac{kg}{kmol}[/tex], [tex]R_{u} = 8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K}[/tex] and [tex]T = 300\,K[/tex], then the density of air is:
[tex]\rho = \frac{(2026.5\,kPa)\cdot \left(28.965\,\frac{kg}{kmol} \right)}{\left(8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} \right)\cdot (300\,K)}[/tex]
[tex]\rho = 23.534\,\frac{kg}{m^{3}}[/tex]
The mass of air stored in the vessel is derived from definition of density. That is:
[tex]m = \rho \cdot V[/tex] (2)
Where [tex]m[/tex] is the mass, measured in kilograms.
If we know that [tex]\rho = 23.534\,\frac{kg}{m^{3}}[/tex] and [tex]V = 10\,m^{3}[/tex], then the mass of air stored in the vessel is:
[tex]m = \left(23.534\,\frac{kg}{m^{3}} \right)\cdot (10\,m^{3})[/tex]
[tex]m = 235.34\,kg[/tex]
The mass of air stored in the vessel is 235.34 kilograms.
