Answer : The mass of oxygen gas is, 24.94 grams
Explanation :
Using ideal gas equation :
[tex]PV=nRT[/tex]
or,
[tex]PV=\frac{w}{M}RT[/tex]
where,
P = pressure of gas = 1.83 atm
V = volume of gas = 10.5 L
T = temperature of gas = [tex]27.3^oC=273+27.3=300.3K[/tex]
R = gas constant = 0.0821 L.atm/mole.K
n = number of moles of oxygen gas
w = mass of oxygen gas = ?
M = molar mass of oxygen gas = 32 g/mole
Now put all the given values in the above law, we get
[tex]1.83atm\times 10.5L=\frac{w}{32g/mole}\times 0.0821L.atm/mole.K\times 300.3K[/tex]
[tex]w=24.94g[/tex]
Therefore, the mass of oxygen gas is, 24.94 grams
Answer : The mass of oxygen gas contained in the tank is 24.9 grams.
Explanation :
Using ideal gas equation:
[tex]PV=nRT\\\\PV=\frac{w}{M}RT[/tex]
where,
P = pressure of oxygen gas = 1.83 atm
T = temperature of oxygen gas = [tex]27.3^oC=273+27.3=300.3K[/tex]
V = volume of oxygen gas = 10.5 L
M = molar mass of oxygen gas = 32 g/mole
w = mass of oxygen gas = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the above equation, we get:
[tex]PV=\frac{w}{M}RT[/tex]
[tex](1.83atm)\times (10.5L)=\frac{w}{32g/mole}\times (0.0821L.atm/mol.K)\times (300.3K)[/tex]
[tex]w=24.9g[/tex]
Therefore, the mass of oxygen gas contained in the tank is 24.9 grams.
