To solve this problem, we will use the first law of thermodynamics and some concepts from thermodynamics related to phase changes.
Given Data:
- Volume of water: \( V_{\text{water}} = 1 , \text{cm}^3 = 1 \times 10^{-6} , \text{m}^3 \)
- Volume of steam: \( V_{\text{steam}} = 1671 , \text{cm}^3 = 1.671 \times 10^{-3} , \text{m}^3 \)
- Heat of vaporization: \( L_v = 2256 , \text{J/g} = 2.256 \times 10^6 , \text{J/kg} \)
- Mass of water: \( m = 1 , \text{g} = 0.001 , \text{kg} \)
1. Compute the work done by the water when it vaporizes.
The work done by the gas during the vaporization process at constant pressure can be calculated using the formula:
\[ W = P \Delta V \]
where:
- \( P \) is the pressure,
- \( \Delta V \) is the change in volume.
First, we find \( \Delta V \):
\[ \Delta V = V_{\text{steam}} - V_{\text{water}} = 1.671 \times 10^{-3} , \text{m}^3 - 1 \times 10^{-6} , \text{m}^3 = 1.669 \times 10^{-3} , \text{m}^3 \]
Now, substituting the values into the work formula:
\[ W = (1.013 \times 10^5 , \text{Pa}) \times (1.669 \times 10^{-3} , \text{m}^3) \]
Calculating this gives:
\[ W = 1.013 \times 10^5 \times 1.669 \times 10^{-3} = 169.55 , \text{J} \]
So, the work done by the water when it vaporizes is approximately:
\[ \boxed{169.55 , \text{J}} \]
2. Compute the increase in internal energy.
According to the first law of thermodynamics, the change in internal energy \( \Delta U \) is given by:
\[ \Delta U = Q - W \]
where \( Q \) is the heat added to the system. The heat added for vaporization can be calculated as:
\[ Q = m L_v = 0.001 , \text{kg} \times 2.256 \times 10^6 , \text{J/kg} = 2256 , \text{J} \]
Now we can find the increase in internal energy \( \Delta U \):
\[ \Delta U = Q - W = 2256 , \text{J} - 169.55 , \text{J} = 2086.45 , \text{J} \]
So, the increase in internal energy is approximately:
\[ \boxed{2086.45 , \text{J}} \]