What is w when 3.93 kg of H2O(l), initially at 25.0 ºC, is converted into water vapour at 157 ºC 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.
6 answers
w is work
3930 g H2O = 3930/18 = approx 2 mol but this an estimate. You need to do a better job when you calculate this an all calculations that follow.
Ideal gas law is obeyed (from the problem) so V = nRT/P or
V (in liters) = 2*0.08206*(273.15+157)/1 atm = ? about 71 L
If you assume the volume of the liquid is negligible (I don't think I would do that), then work done is -P(Vvapor -Vliquid) = -1*(71-0) = about 71 L*atm.
Multiply by 101.325 to convert to J
If you want to include volume of the liquid and a number and not 0, then
3930 grams H2O is volume = mass/density = 3930/1.00 = 3930 cc or 3.93 L = about 4 L. The problem doesn't say to ignore the volume of the liquid and I would not. Post your work if you get stuck.
Ideal gas law is obeyed (from the problem) so V = nRT/P or
V (in liters) = 2*0.08206*(273.15+157)/1 atm = ? about 71 L
If you assume the volume of the liquid is negligible (I don't think I would do that), then work done is -P(Vvapor -Vliquid) = -1*(71-0) = about 71 L*atm.
Multiply by 101.325 to convert to J
If you want to include volume of the liquid and a number and not 0, then
3930 grams H2O is volume = mass/density = 3930/1.00 = 3930 cc or 3.93 L = about 4 L. The problem doesn't say to ignore the volume of the liquid and I would not. Post your work if you get stuck.
for the first part, how did you get 2 mol? 3930/18 = ~218?
Note that I said APPROX 2 mols. 2.18 is correct but I'm trying to show you HOW to do it and leave the calculations to you. I almost always say about or approximate. I expect you to follow up, calculate a more exact answer for each time I've estimated a number.
i meant, how did you get 2.18, 3930/18 is literally 218. where did you get the decimal from? sorry for the misunderstanding!
I get it and thank you for persisting. 3930/18 is 218 and I, of all people, should have recognized that 2.18 is an unrealistic answer. How did that happen? I have an old calculator and some of the pixels don't work anymore. A few dark dots (spots if you will) show up now and then. Those look like decimals to me. You guessed it. One of those sporadic spots showed up and I read the answer as 2.18 and not 218. My bad. Sorry about that. Also, I failed to note that your answer explicitly showed 218 and I mentally added a decimal to your answer to make it read 2.18 but there was no decimal there. By the way, with that being 218 and not 2.18, the answer will be 100 times greater so the volume of the liquid may very well be negligible when compared to V2. Thanks for catching that.