Question

What is the volume of oxygen occupied by 2 moles at 1.3 atm pressure and 300 K? Use PV = nRT.

Answer 1

we can use the ideal gas law equation to find the volume occupied by O₂
PV = nRT
where P - pressure - 1.3 atm x 101 325 Pa/atm = 131 723 Pa
V - volume
n - number of moles - 2 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 300 K
Substituting these values in the equation
131 723 Pa x V = 2 mol x 8.314 Jmol⁻¹K⁻¹ x 300 K
V = 37.9 L
volume occupied is 37.9 L

Answer 2

The volume of oxygen gas that occupied by 2 moles at 1.3 atm pressure and 300 k is calculated using the ideal gas equation that is Pv =nRT

P(pressure) = 1.3 atm
R(gas constant) =0.082 l.atm/mol.k
n (moles)= 2 moles
T(temperature) =300k
V(volume)=?

by making the v the subject of the formula V =nRT/P

=(2 moles x 0.082 l.atm/mol.k x300 K)/ 1.3 atm= 37.85 Liters