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

Heat is involved in chemical change in two ways; chemical changes either result in heat being added to or removed from an object, and chemical changes require a certain amount of heat before they can take place.