Question
Part 1: 1 cm^3 of water is boiled at atmospheric pressure to become 1427.1 cm^3 of steam, also at atmospheric pressure. Calculate the work done by the gas during this process. The latent heat of vaporization of water is 2.26 × 10^6 J/kg. Answer in units of J.
Part 2: Find the amount of heat added to the water to accomplish this process. Answer in units of J.
Part 3: Find the change in internal energy. Answer in units of J.
Part 2: Find the amount of heat added to the water to accomplish this process. Answer in units of J.
Part 3: Find the change in internal energy. Answer in units of J.
Answers
GPT-4o mini
To solve these problems, we will first need to use some principles of thermodynamics.
### Part 1: Calculating Work Done by the Gas
The work done \( W \) by the gas during an expansion at constant pressure can be calculated using the formula:
\[
W = P \Delta V
\]
where \( P \) is the pressure and \( \Delta V \) is the change in volume.
1. **Convert units**:
- The pressure \( P \) at atmospheric pressure is \( 101325 \, \text{Pa} \) (Pascal).
- The change in volume \( \Delta V \) is:
\[
\Delta V = V_{\text{final}} - V_{\text{initial}} = 1427.1 \, \text{cm}^3 - 1 \, \text{cm}^3 = 1426.1 \, \text{cm}^3 = 1426.1 \times 10^{-6} \, \text{m}^3
\]
2. **Calculate the work done**:
Plugging in the values we get:
\[
W = 101325 \, \text{Pa} \times 1426.1 \times 10^{-6} \, \text{m}^3
\]
Calculating this gives:
\[
W = 101325 \times 1426.1 \times 10^{-6} \approx 144.3 \, \text{J}
\]
### Part 2: Amount of Heat Added to the Water
The amount of heat \( Q \) added to the water during this phase change can be calculated using:
\[
Q = m L
\]
where \( m \) is the mass of water and \( L \) is the latent heat of vaporization.
1. **Mass of 1 cm³ of water**:
- The density of water is approximately \( 1000 \, \text{kg/m}^3 \).
Thus, the mass \( m \):
\[
m = 1 \, \text{cm}^3 = 1 \times 10^{-6} \, \text{m}^3 \rightarrow m = 1000 \times 10^{-6} \, \text{kg} = 0.001 \, \text{kg}
\]
2. **Latent heat of vaporization**:
- \( L = 2.26 \times 10^6 \, \text{J/kg} \)
3. **Calculate the heat added**:
\[
Q = 0.001 \, \text{kg} \times 2.26 \times 10^6 \, \text{J/kg} \approx 2260 \, \text{J}
\]
### Part 3: Change in Internal Energy
To find the change in internal energy \( \Delta U \), we can use the first law of thermodynamics:
\[
\Delta U = Q - W
\]
Substituting the calculated values of \( Q \) and \( W \):
1. \( Q = 2260 \, \text{J} \)
2. \( W \approx 144.3 \, \text{J} \)
Now substituting into the equation gives:
\[
\Delta U = 2260 \, \text{J} - 144.3 \, \text{J} \approx 2115.7 \, \text{J}
\]
### Summary of Answers:
- **Part 1 (Work Done)**: \( W \approx 144.3 \, \text{J} \)
- **Part 2 (Heat Added)**: \( Q \approx 2260 \, \text{J} \)
- **Part 3 (Change in Internal Energy)**: \( \Delta U \approx 2115.7 \, \text{J} \)
### Part 1: Calculating Work Done by the Gas
The work done \( W \) by the gas during an expansion at constant pressure can be calculated using the formula:
\[
W = P \Delta V
\]
where \( P \) is the pressure and \( \Delta V \) is the change in volume.
1. **Convert units**:
- The pressure \( P \) at atmospheric pressure is \( 101325 \, \text{Pa} \) (Pascal).
- The change in volume \( \Delta V \) is:
\[
\Delta V = V_{\text{final}} - V_{\text{initial}} = 1427.1 \, \text{cm}^3 - 1 \, \text{cm}^3 = 1426.1 \, \text{cm}^3 = 1426.1 \times 10^{-6} \, \text{m}^3
\]
2. **Calculate the work done**:
Plugging in the values we get:
\[
W = 101325 \, \text{Pa} \times 1426.1 \times 10^{-6} \, \text{m}^3
\]
Calculating this gives:
\[
W = 101325 \times 1426.1 \times 10^{-6} \approx 144.3 \, \text{J}
\]
### Part 2: Amount of Heat Added to the Water
The amount of heat \( Q \) added to the water during this phase change can be calculated using:
\[
Q = m L
\]
where \( m \) is the mass of water and \( L \) is the latent heat of vaporization.
1. **Mass of 1 cm³ of water**:
- The density of water is approximately \( 1000 \, \text{kg/m}^3 \).
Thus, the mass \( m \):
\[
m = 1 \, \text{cm}^3 = 1 \times 10^{-6} \, \text{m}^3 \rightarrow m = 1000 \times 10^{-6} \, \text{kg} = 0.001 \, \text{kg}
\]
2. **Latent heat of vaporization**:
- \( L = 2.26 \times 10^6 \, \text{J/kg} \)
3. **Calculate the heat added**:
\[
Q = 0.001 \, \text{kg} \times 2.26 \times 10^6 \, \text{J/kg} \approx 2260 \, \text{J}
\]
### Part 3: Change in Internal Energy
To find the change in internal energy \( \Delta U \), we can use the first law of thermodynamics:
\[
\Delta U = Q - W
\]
Substituting the calculated values of \( Q \) and \( W \):
1. \( Q = 2260 \, \text{J} \)
2. \( W \approx 144.3 \, \text{J} \)
Now substituting into the equation gives:
\[
\Delta U = 2260 \, \text{J} - 144.3 \, \text{J} \approx 2115.7 \, \text{J}
\]
### Summary of Answers:
- **Part 1 (Work Done)**: \( W \approx 144.3 \, \text{J} \)
- **Part 2 (Heat Added)**: \( Q \approx 2260 \, \text{J} \)
- **Part 3 (Change in Internal Energy)**: \( \Delta U \approx 2115.7 \, \text{J} \)