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

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

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