An antique copper water heater (closed cylinder) contains 3 mol of water in form of steam at pressure of 20 bar.(a) If the pressure in the heater drops to 12 bar, and the temperature remains constant, how much water must have leaked out of the heater? I tried using the equation V1P1V2P2, but i‘m not getting the right answer can anyone tell me what im doing wrong?
V1P1 V2P2 if (and only if) the moles of gas (H2O vapor) and temperature remain constant. The moles of H2O (steam) did not remain constant. (The Volume remains constant.) use PV nRT at first n 3 mol 20 bar x V 3 x R x T divide by V 20 bar 3 x (RT/V); note: RT/V a constant 12 bar n x (RT/V); RT/V the same constant n 3 x (12/20) 1.8 mole 1.2 mole of steam leaked out
1 bar 0.987 atm 20bar(0.987 atm/bar) 19.74 atm 12 bar(0.987 atm/bar) 11.84 atm Assume the steam is at 100 C or 373 K Initial volume of steam in heater nRT / P (3 moles)(0.0821 l atm / mole K)(373 K) / 19.74 atm 4.65 liters Final volume of steam in heater nRT / P (3 moles)(.0821 l atm/mole K)(373 K) / 11.84 atm 7.76 liters Difference in volumes 7.76 - 4.65 3.11 liters Answer: 3.11 liters of water must have leaked out.
The volume is remaining the same. What is changing is the number of moles,n. As n goes down P goes down so P is directly proportional to n. This means that P1/n1P2/n2 solve for n2 20 bar/3mol 12 bar/n2 20n236 n21.8 mol H20 (This can be converted to g by multiplying by the molar mass of H2O - 18.01 g/mol) mass of water is 32 g.
