Answer :
Answer:
[tex]m=3279.85g[/tex]
Explanation:
Hello,
This example could be analyzed by assuming the ideal gas behavior, thus, recalling the formula:
[tex]PV=nRT[/tex]
The explicit unknown are the moles, because we could get mass by including the molecular mass as follows:
[tex]m=\frac{PVM}{RT}[/tex]
Whereas:
[tex]P=0.85*31.8torr*\frac{1atm}{760torr} =0.03557 atm[/tex]
[tex]V=6m*10m*2.2m=132m^3*\frac{1000L}{1m^3} =1.32x10^5L[/tex]
[tex]M=18\frac{gH_2O}{mol H_2O}[/tex]
[tex]R=0.082\frac{atm*L}{mol*K}[/tex]
[tex]T=30+273.15=303.15K[/tex]
Now, solving for the mass of water we get:[tex]m=\frac{0.03557atm*1.32x10^5L*18g/mol}{0.082\frac{atm*L}{mol*K}*303.15K } \\\\m=3279.85g[/tex]
Best regards.