Calculate the work done when 1.0 mole water at 373K vaporizes against an atmospheric pressure of 1.0 atmosphere. Assume ideal gas behaviour.

Explanation

To calculate the work done when 1.0 mole of water vaporizes at 373 K against an atmospheric pressure of 1.0 atmosphere, we can follow these steps:

### Step 1: Determine the volume of water before vaporization (V1)
1 mole of water has a mass of 18 grams. The density of water is approximately 1 g/mL, so:
\[ V_1 = \frac{\text{mass}}{\text{density}} = \frac{18 \text{ g}}{1 \text{ g/mL}} = 18 \text{ mL} = 0.018 \text{ L} \]

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