Answer:
[tex]m=840.88g[/tex]
Explanation:
Hello,
In this case, considering the ideal gas equation as:
[tex]PV=nRT[/tex]
We can first compute the moles of air at the given conditions of 195 kPa (1.92 atm), 10 °C (283K) and 350 L:
[tex]n=\frac{PV}{RT}=\frac{1.92atm*350L}{0.082\frac{atm*L}{mol*K}*283K}\\ \\n=29.03molAir[/tex]
Next, since the molar mass of air is 28.97 g/mol, the mass is computed to be:
[tex]m=29.03mol*\frac{28.97g}{1mol} \\\\m=840.88g[/tex]
Best regards.
