Answer : The number of moles present in the oxygen gas is, 272.625 moles and the pressure of oxygen gas in atmosphere in the cylinder is, 264.56 atm
Solution : Given,
Mass of oxygen gas = 4.362 Kg = 4362 g (1 Kg = 1000 g)
Molar mass of oxygen gas = 16 g/mole
First we have to calculate the moles of oxygen gas.
\(\text{Moles of oxygen gas}=\frac{\text{Mass of oxygen gas}}{\text{Molar mass of oxygen gas}}=\frac{4362g}{16g/mole}=272.625moles\)
Now we have to calculate the pressure of oxygen gas by using ideal gas equation.
\(PV=nRT\\\\P=\frac{nRT}{V}\)
where,
P = pressure of gas
T = temperature of gas = \(22.5^oC=273+22.5=295.5K\)
V = volume of gas = 25 L
n = number of moles of gas = 272.625 moles
R = gas constant = 0.0821 Latm/moleK
Now put all the given values in the ideal gas equation, we get the pressure of gas.
\(P=\frac{(272.625moles)\times (0.0821Latm/moleK)\times (295.5K)}{25L}=264.56atm\)
Therefore, the number of moles present in the oxygen gas is, 272.625 moles and the pressure of oxygen gas in atmosphere in the cylinder is, 264.56 atm
