What is w when 3.37 kg of H2O(l), initially at 25.0oC, is converted into water vapour at 111oC against a constant external pressure of 1.00 atm? Assume that the vapour behaves ideally and that the density of liquid water is 1.00 g/mL. (Remember to include a "+" or "-" sign as appropriate.)

Question

DrBob222 · Answer

dE = q + w
q1 = mass H2O x specific heat H2O x (Tfinal-Tinitial) to get from 25 C to 100C.

q2 = mass H2O x heat vaporization to get from liquid water at 100 to steam at 100

q3 = mass steam x specific heat steam x (Tfinal-Tinitial) to get from steam at 100 C to steam at 111 C.

Add q1+q2+q3 to obtain total q and that is delta H. The sign is + because you ar adding heat for q1 and q2 and q3.

For work:
Use PV = nRT. You can calculate n, you know P and R and T. Solve for V. I assume you are expected to ignore the volume of the liquid but if you aren't you should calculate the volume of liquid water at 100 C. Then p*delta V will be the work. Since this is expanding into the atmosphere it will be -. Then substitute into dE = dH+w. Remember dH will be + and work will be -; dE can come out with either sign.