Question

Calculate the work done in joules when 1.50 mol of water vaporizes at 2.00 atm and 100^oC, assuming that the volume of the water is negligible compared with that of the steam formed.

Answer 1

Calculate the amount of work done, in joules, when 2.5 mole of H2O vaporizes at 1.0 atm and 25°C?

Calculate the amount of work done, in joules, when 2.5 mole of H2O vaporizes at 1.0 atm and 25°C. Assume the volume of liquid H2O is negligible compared to that of vapor. [1 L·atm = 101.3 J]
Answer

6,190 kJ
6.19 kJ
61.1 J
5.66 kJ
518 J

Since most of the questions here tend to be questions for basic chemistry classes, I'm assuming your teacher is just giving you a gas law problem with additional conversion units. If you remember going over a different formula for converting gas expansion to energy, I recommend you check out a different answer.


Since you are provided with the amount of energy for every unit of volume (L) and pressure (atm), I'm assuming what we need to do is figure out how much the volume changes. For gases, volumes are additive so all we need to do is figure out how much space the 2.5 moles of water will take up, and that will be our volume change.

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.