If 15.0 g CO2 (mm = 44.01 g/mol) has a volume of 0.30L at 27oC, what is the pressure in atm? Group of answer choices 27.97 atm 1230.9 atm 2.52 atm 325.0 atm

P = 27.97 atm

Explanation:

PV = nRT = > P = nRT/V

P = unknown

n = moles CO₂ = (15g/44.01g/mol) = 0.3408 mole

R = 0.08206 L·atm/mol·K

T = 27°C = (27 + 273) K = 300K

V = 0.30L

P = (0.3408 mole) (0.08206 L·atm/mol·K) (300K) / (0.30L) = 27.97 atm