Answer:
The number of moles of oxygen gas comes out to be 0.0548 mole
Explanation:
Given volume of gas = V = 3.0 L
The mixture contains 30 % oxygen gas by mole.
Pressure of mixture of gas = P = 2.0 atm
Temperature = T = 400 K
Assuming n be the total number of moles of the mixture of gas.
The ideal gas equation is shown below
[tex]\textrm{PV} = \textrm{nRT} \\2.0 \textrm{ atm}\times 3.0\textrm{ L} = n \times 0.0821 \textrm{ L.atm.mol}^{-1}.K^{-1} \times 400 \textrm{ K} \\n = 0.18270 \textrm{ mole}[/tex]
The mixture contains 30% oxygen gas by mole
[tex]\textrm{ Number of moles of oxygen gas} = \displaystyle \frac{30\times 0.18270 \textrm{ mole}}{100} = 0.0548 \textrm{ mole}[/tex]
Number of moles of oxygen gas is 0.0548 mole
