Answer:
1.68 atm
Explanation:
First we must obtain the number of moles of the oxygen = 32.7g/32 g/mol = 1.02 moles of oxygen
V= 15.0 L
T= 302 K
R = 0.082 atmLmol-1K-1
From the ideal gas equation;
PV=nRT
P= nRT/V
P= 1.02 * 0.082 * 302/15.0
P= 1.68 atm
