One gram of water `(1 cm^(3))` becomes 1671 `cm^(3)` of steam at a pressure of 1 atm. The latent heat of vaporization at this pressure is 2256 J/g. Calculate the external work and the increase in internal energy.

Answer

To solve the problem, we need to calculate the external work done and the increase in internal energy when 1 gram of water is converted into steam.

### Step-by-Step Solution:

1. **Calculate the Heat Transfer (Q)**:
The heat transfer during the phase change from water to steam is given by the formula:
\[ Q = M \times L \]
where:
- \( M \) is the mass of the water (1 gram = 0.001 kg),
- \( L \) is the latent heat of vaporization (2256 J/g).

Converting the mass to grams:
\[ Q = 1 \, \text{g} \times 2256 \, \text{J/g} = 2256 \, \text{J} \]

2. **Calculate the External Work (W)**:
The external work done during the expansion can be calculated using the formula:
\[ W = P \Delta V \]
where:
- \( P \) is the pressure (1 atm),
- \( \Delta V \) is the change in volume.

First, we need to convert the pressure from atm to Pascals:
\[ 1 \, \text{atm} = 101325 \, \text{Pa} \]

Next, calculate the change in volume:
- Initial volume of water = 1 cm³ = \( 1 \times 10^{-6} \, \text{m}^3 \)
- Final volume of steam = 1671 cm³ = \( 1671 \times 10^{-6} \, \text{m}^3 \)

Therefore, the change in volume \( \Delta V \) is:
\[ \Delta V = V_{final} - V_{initial} = (1671 - 1) \times 10^{-6} \, \text{m}^3 = 1670 \times 10^{-6} \, \text{m}^3 \]

Now, substituting the values into the work formula:
\[ W = 101325 \, \text{Pa} \times 1670 \times 10^{-6} \, \text{m}^3 = 169.0 \, \text{J} \]

3. **Calculate the Increase in Internal Energy (\( \Delta U \))**:
Using the first law of thermodynamics:
\[ Q = \Delta U + W \]
Rearranging gives:
\[ \Delta U = Q - W \]

Substituting the values we calculated:
\[ \Delta U = 2256 \, \text{J} - 169 \, \text{J} = 2087 \, \text{J} \]

### Final Results:
- External Work Done (\( W \)): **169 J**
- Increase in Internal Energy (\( \Delta U \)): **2087 J**