Answer : The mass of oxygen gas will be, 30.5 grams
Solution :
Using ideal gas equation,
[tex]PV=nRT\\\\PV=\frac{w}{M}\times RT[/tex]
where,
n = number of moles of gas
w = mass of gas = ?
P = pressure of the gas = [tex]883mmHg=\frac{883}{760}atm[/tex]
conversion used : (1 atm = 760 mmHg)
T = temperature of the gas = [tex]24^oC=273+24=297K[/tex]
M = molar mass of oxygen gas = 32 g/mole
R = gas constant = 0.0821 Latm/moleK
V = volume of gas = 20.0 L
Now put all the given values in the above equation, we get the mass of oxygen gas.
[tex](\frac{883}{760}atm)\times (20.0L)=\frac{w}{32g/mole}\times (0.0821Latm/moleK)\times (297K)[/tex]
[tex]w=30.5g[/tex]
Therefore, the mass of oxygen gas will be, 30.5 grams
