A cook puts 7.20 g of water in a 2.00-L pressure cooker that is then warmed to 530°C. What is the pressure inside the container? (Assume room temperature is 20°C.)
2 answers
Note: I used PV=nRT to solve (there are 0.400 mol), but the answer I got was incorrect; the hint I received was that all the water has evapourated.
(2.00 L) x (273 / (20 + 273)) / (22.414 L/mol) = 0.083139 mol air in the "empty" container
(7.20 g H2O) / (18.01532 g H2O/mol) = 0.39966 mol H2O added
Supposing all the water vaporized:
PV = nRT
P = nRT / V = (0.083139 mol + 0.39966 mol) x (0.08205746 L atm/K mol) x
(530 + 273 K) / (2.00 L) = 15.9 atm
15.9atm x 101.3kPa = 1611.31
(7.20 g H2O) / (18.01532 g H2O/mol) = 0.39966 mol H2O added
Supposing all the water vaporized:
PV = nRT
P = nRT / V = (0.083139 mol + 0.39966 mol) x (0.08205746 L atm/K mol) x
(530 + 273 K) / (2.00 L) = 15.9 atm
15.9atm x 101.3kPa = 1611.31
1 answer