Calculate the work done on or by the system when 2.45 mol of liquid h2o vaporizes.

We can use the ideal gas law (PV = nRT) to calculate the volume. We already have P (1.0atm), n (2.5 mol), R (.0821 atm*L/mol*K), and T (25C, which we will need to convert to K) Converting temperature: C + 273 = K so, 25 + 273 = 298K Keep in mind i'm rounding off the decimals, but significant figure rules would round it off for us in this case anyway. Finding volume: PV = nRT lets plug in the numbers we have... (1.0atm)V=(2.5 mol)(.0821atm*L/mol*K)(298K) Now, if we do some algebra to get volume by itself, you'll see that V = 2.5*.0821*298/1.0 = 61.1645L That means that this system would expand 61.1645L at a pressure of 1.0atm. Based on the units, we would have to multiply the change in volume by the pressure to start off: 61.1645L * 1.0atm = 61.1645L*atm Now, we can convert that into Joules using the conversion rate provided (1L*atm = 101.3J) The math: (61.1645L*atm)(101.3J/1L*atm) = 6,195.9J. If we round that off to the appropriate number of significant figures, it should be 6,200J or 6.2kJ. So it looks like B is your answer. Its off slightly from the mult. choice answers because of rounding error.

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