1. 5765 mol
First of all, let's find the volume of the oxygen, which is equal to the volume of the room:
[tex]V=7.0 m\cdot 8.0 m \cdot 2.50 m=140 m^3[/tex]
Then the other data of the oxygen that we have are
[tex]p=1.00 atm = 1.01\cdot 10^5 Pa[/tex] is the pressure
[tex]T=22.0^{\circ}+273=295 K[/tex] is the absolute temperature
Using the ideal gas equation:
[tex]pV=nRT[/tex]
we can find the number of moles of the gas, n:
[tex]n=\frac{pV}{RT}=\frac{(1.01\cdot 10^5 Pa)(140 m^3)}{(8.314 J/mol K)(295 K)}=5765 mol[/tex]
2. 184.5 kg
The mass of the oxygen contained in the room is given by:
[tex]m=n M_m[/tex]
where
n = 5765 mol is the number of moles
Mm = 32.0 g/mol is the molar mass (the mass contained in 1 mol)
Using the equation, we find
[tex]m=(5765 mol)(32.0 g/mol)=184,480 g=184.5 kg[/tex]
Answer:
[tex]n = 5785 moles[/tex]
[tex]n = 185.1 kg[/tex]
Explanation:
As we know that volume of the room is defined as
[tex]V = 7.00 \times 8.00 \times 2.50 m^3[/tex]
[tex]V = 140 m^3[/tex]
now we know that
pressure of the gas at given conditions is
[tex]P = 1 atm = 1.013 \times 10^5 Pa[/tex]
also the temperature of gas in room temperature is given as
[tex]T = 22^o C = 22 + 273 = 295 K[/tex]
now from ideal gas equation we know
[tex]n = \frac{PV}{RT}[/tex]
[tex]n = \frac{(1.013 \times 10^5)(140)}{(8.31)(295)}[/tex]
[tex]n = 5785 moles[/tex]
Now for mass of oxygen we can use the relation of mass and atomic mass
[tex]n = \frac{m}{M}[/tex]
[tex]5785 = \frac{m}{32}[/tex]
[tex]n = 185.1 kg[/tex]
