Question

(0) A cook puts 1.30 g of water in a 2.00 L pressure cooker that is then warmed from the kitchen temperature of 20°C to 118°C. What is the pressure (in atm) inside the container? 113.7 Assume that all of the water has been converted to vapor, atm g) that the cook can put inside the pressure cooker before warming it to (D) What if? If the gauge pressure inside the container cannot exceed 3.00 atm, what is the maximum amount of water ( 118°C? 0.03 x9 Need Help? Read

Answer 1

a) The pressure has two contribution: Pinside = Pwater + Pair Air pressure Pair = P. Constant volume process, then: (1 atm) (118 + 273) = 1.334 atm Pair = 20 + 273 Remember, the temperature must be in kelvin unit. Water pressure, the water has converted into steam, then: NRT: Pwater ve mol Remember that: n = m, where M = 18 mRT (1.30)(0.082)(118 + 273) -= 1.158 atm 18 * 2 Pwater = MV. The total pressure is: P = 1.334 + 1.158 = 2.492 atm b) The maximum gauge pressure is: Pmax gauge = Pair + Pwater – Po = 3.00 atm Then: Pwater = 3.00 + Po – Pair = 3.00 + 1.00 – 1.334 = 2.666 atm MRT m(0.082)(118 + 273) -= 2.666 18 * 2 Solving for m we get: 2.993 g Pwater – MV,