Question

A pressure vessel that has a volume of 10m3 is used to store high-pressure air for operating a supersonic wind tunnel. If the air pressure and temperature inside the vessel are 20 atm and 300K, respectively: What is the mass of air stored in the vessel

Answer 1

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 ($\rho$), measured in kilograms per cubic meter, is defined by following equation:

$\rho = \frac{P\cdot M}{R_{u}\cdot T}$ (1)

Where:

$P$ - Pressure, measured in kilopascals.

$M$ - Molar mass, measured in kilomoles per kilogram.

$R_{u}$ - Ideal gas constant, measured in kilopascal-cubic meters per kilomole-Kelvin.

$T$ - Temperature, measured in Kelvin.

If we know that $P = 2026.5\,kPa$, $M = 28.965\,\frac{kg}{kmol}$, $R_{u} = 8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K}$ and $T = 300\,K$, then the density of air is:

$\rho = \frac{(2026.5\,kPa)\cdot \left(28.965\,\frac{kg}{kmol} \right)}{\left(8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} \right)\cdot (300\,K)}$

$\rho = 23.534\,\frac{kg}{m^{3}}$

The mass of air stored in the vessel is derived from definition of density. That is:

$m = \rho \cdot V$ (2)

Where $m$ is the mass, measured in kilograms.

If we know that $\rho = 23.534\,\frac{kg}{m^{3}}$ and $V = 10\,m^{3}$, then the mass of air stored in the vessel is:

$m = \left(23.534\,\frac{kg}{m^{3}} \right)\cdot (10\,m^{3})$

$m = 235.34\,kg$

The mass of air stored in the vessel is 235.34 kilograms.