To calculate the work done by the water when it vaporizes and the increase in internal energy, we can use the first law of thermodynamics, which states:
\[ \Delta U = Q - W \]
Where:
- \(\Delta U\) is the change in internal energy,
- \(Q\) is the heat added to the system,
- \(W\) is the work done by the system.
Given Data:
- Volume of water (1 g) = \(1 , \text{cm}^3 = 1 \times 10^{-6} , \text{m}^3\)
- Volume of steam = \(1671 , \text{cm}^3 = 1671 \times 10^{-6} , \text{m}^3\)
- Heat of vaporization \(L_v = 2.256 \times 10^6 , \text{J/kg}\)
- Mass of water = \(1 , \text{g} = 0.001 , \text{kg}\)
- Pressure \(P = 1 , \text{atm} = 1.013 \times 10^5 , \text{Pa}\)
Step 1: Calculate the Work Done (W)
When water vaporizes, the work done by the water can be calculated using the formula:
\[ W = P \Delta V \]
Where:
- \(\Delta V = V_{\text{steam}} - V_{\text{water}} = (1671 \times 10^{-6}) - (1 \times 10^{-6})\)
- \(P\) is the pressure in pascals.
Calculating \(\Delta V\):
\[ \Delta V = 1671 \times 10^{-6} , \text{m}^3 - 1 \times 10^{-6} , \text{m}^3 = 1670 \times 10^{-6} , \text{m}^3 = 0.001671 , \text{m}^3 \]
Now, substituting this into the work done equation:
\[ W = (1.013 \times 10^5 , \text{Pa}) \times (0.001671 , \text{m}^3) \]
Calculating \(W\):
\[ W = 1.013 \times 10^5 \times 0.001671 = 169.72 , \text{J} \]
Step 2: Calculate the Heat Added (Q)
The heat added to vaporize the water is given by:
\[ Q = m L_v \]
Where:
- \(m = 0.001 , \text{kg}\)
- \(L_v = 2.256 \times 10^6 , \text{J/kg}\)
Calculating \(Q\):
\[ Q = 0.001 , \text{kg} \times 2.256 \times 10^6 , \text{J/kg} = 2260.256 , \text{J} \]
Step 3: Calculate the Increase in Internal Energy (\(\Delta U\))
Using the first law of thermodynamics:
\[ \Delta U = Q - W \]
Substituting the calculated values:
\[ \Delta U = 2260.256 , \text{J} - 169.72 , \text{J} = 2090.536 , \text{J} \]
Summary of Results:
- The work done by the water when it vaporizes: \(W \approx 169.72 , \text{J}\)
- The increase in internal energy: \(\Delta U \approx 2090.54 , \text{J}\)