Question
One gram of water (1cm³) becomes 1671 cm³ of steam when boiled at a constant pressure of 1 atm(1.013×10⁵pa). The heat of vaporization at this pressure is Lv=2.256×10⁶J/kg. Calculate:
1: the work done by water when it vaporizes
2: it increases internal energy
1: the work done by water when it vaporizes
2: it increases internal energy
Answers
GPT-4o mini
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:
1. Volume of water (1 g) = \(1 \, \text{cm}^3 = 1 \times 10^{-6} \, \text{m}^3\)
2. Volume of steam = \(1671 \, \text{cm}^3 = 1671 \times 10^{-6} \, \text{m}^3\)
3. Heat of vaporization \(L_v = 2.256 \times 10^6 \, \text{J/kg}\)
4. Mass of water = \(1 \, \text{g} = 0.001 \, \text{kg}\)
5. 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:
1. The work done by the water when it vaporizes: \(W \approx 169.72 \, \text{J}\)
2. The increase in internal energy: \(\Delta U \approx 2090.54 \, \text{J}\)
\[
\Delta U = Q - W
\]
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:
1. Volume of water (1 g) = \(1 \, \text{cm}^3 = 1 \times 10^{-6} \, \text{m}^3\)
2. Volume of steam = \(1671 \, \text{cm}^3 = 1671 \times 10^{-6} \, \text{m}^3\)
3. Heat of vaporization \(L_v = 2.256 \times 10^6 \, \text{J/kg}\)
4. Mass of water = \(1 \, \text{g} = 0.001 \, \text{kg}\)
5. 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:
1. The work done by the water when it vaporizes: \(W \approx 169.72 \, \text{J}\)
2. The increase in internal energy: \(\Delta U \approx 2090.54 \, \text{J}\)