Question
One gram of water (1cm^3) becomes 1671cm^3 of steam when boiled at a constant pressure of 1atm(1.013×10^5pa). The heat of vaporization at this pressure is Lv= 2256J/g. Compute:
1. The work done by the water when it vaporizes?
2. It increases internal energy?
1. The work done by the water when it vaporizes?
2. It increases internal energy?
Answers
GPT-4o mini
To compute the work done by the water when it vaporizes and the increase in internal energy, we can use the following thermodynamic principles.
### 1. Work Done by the Water When It Vaporizes
The work done during the vaporization can be computed using the formula for work done in a constant pressure process:
\[
W = P \Delta V
\]
where \( P \) is the pressure and \( \Delta V \) is the change in volume.
- **Pressure, \( P \)**: 1 atm = \( 1.013 \times 10^5 \, \text{Pa} \)
- **Initial Volume of Water, \( V_i \)**: \( 1 \, \text{cm}^3 = 1 \times 10^{-6} \, \text{m}^3 \)
- **Final Volume of Steam, \( V_f \)**: \( 1671 \, \text{cm}^3 = 1671 \times 10^{-6} \, \text{m}^3 \)
Now calculate the change in volume, \( \Delta V \):
\[
\Delta V = V_f - V_i = (1671 \times 10^{-6} \, \text{m}^3) - (1 \times 10^{-6} \, \text{m}^3) = 1670 \times 10^{-6} \, \text{m}^3
\]
Now substitute these into the formula for work:
\[
W = P \Delta V = (1.013 \times 10^5 \, \text{Pa}) \times (1670 \times 10^{-6} \, \text{m}^3)
\]
Calculating that gives:
\[
W = 1.013 \times 10^5 \times 1670 \times 10^{-6} = 169.1711 \, \text{J}
\]
So, the work done by the water when it vaporizes is approximately:
\[
W \approx 169.17 \, \text{J}
\]
### 2. Increase in Internal Energy
The total energy change (increase in internal energy) during the phase change can be computed using the heat of vaporization. The heat added for vaporization, \( Q \), can be expressed as:
\[
Q = m L_v
\]
where:
- \( m \) is the mass of the water, \( m = 1 \, \text{g} \) (or \( 0.001 \, \text{kg} \))
- \( L_v \) (heat of vaporization) = \( 2256 \, \text{J/g} \) or \( 2.256 \times 10^6 \, \text{J/kg} \)
So the heat added is:
\[
Q = 1 \, \text{g} \times 2256 \, \text{J/g} = 2256 \, \text{J}
\]
Using the first law of thermodynamics, the increase in internal energy \( \Delta U \) for a closed system can be calculated as:
\[
\Delta U = Q - W
\]
Substituting the values obtained:
\[
\Delta U = 2256 \, \text{J} - 169.17 \, \text{J} \approx 2086.83 \, \text{J}
\]
So, the increase in internal energy is approximately:
\[
\Delta U \approx 2086.83 \, \text{J}
\]
### Summary
1. The work done by the water when it vaporizes is approximately \( 169.17 \, \text{J} \).
2. The increase in internal energy is approximately \( 2086.83 \, \text{J} \).
### 1. Work Done by the Water When It Vaporizes
The work done during the vaporization can be computed using the formula for work done in a constant pressure process:
\[
W = P \Delta V
\]
where \( P \) is the pressure and \( \Delta V \) is the change in volume.
- **Pressure, \( P \)**: 1 atm = \( 1.013 \times 10^5 \, \text{Pa} \)
- **Initial Volume of Water, \( V_i \)**: \( 1 \, \text{cm}^3 = 1 \times 10^{-6} \, \text{m}^3 \)
- **Final Volume of Steam, \( V_f \)**: \( 1671 \, \text{cm}^3 = 1671 \times 10^{-6} \, \text{m}^3 \)
Now calculate the change in volume, \( \Delta V \):
\[
\Delta V = V_f - V_i = (1671 \times 10^{-6} \, \text{m}^3) - (1 \times 10^{-6} \, \text{m}^3) = 1670 \times 10^{-6} \, \text{m}^3
\]
Now substitute these into the formula for work:
\[
W = P \Delta V = (1.013 \times 10^5 \, \text{Pa}) \times (1670 \times 10^{-6} \, \text{m}^3)
\]
Calculating that gives:
\[
W = 1.013 \times 10^5 \times 1670 \times 10^{-6} = 169.1711 \, \text{J}
\]
So, the work done by the water when it vaporizes is approximately:
\[
W \approx 169.17 \, \text{J}
\]
### 2. Increase in Internal Energy
The total energy change (increase in internal energy) during the phase change can be computed using the heat of vaporization. The heat added for vaporization, \( Q \), can be expressed as:
\[
Q = m L_v
\]
where:
- \( m \) is the mass of the water, \( m = 1 \, \text{g} \) (or \( 0.001 \, \text{kg} \))
- \( L_v \) (heat of vaporization) = \( 2256 \, \text{J/g} \) or \( 2.256 \times 10^6 \, \text{J/kg} \)
So the heat added is:
\[
Q = 1 \, \text{g} \times 2256 \, \text{J/g} = 2256 \, \text{J}
\]
Using the first law of thermodynamics, the increase in internal energy \( \Delta U \) for a closed system can be calculated as:
\[
\Delta U = Q - W
\]
Substituting the values obtained:
\[
\Delta U = 2256 \, \text{J} - 169.17 \, \text{J} \approx 2086.83 \, \text{J}
\]
So, the increase in internal energy is approximately:
\[
\Delta U \approx 2086.83 \, \text{J}
\]
### Summary
1. The work done by the water when it vaporizes is approximately \( 169.17 \, \text{J} \).
2. The increase in internal energy is approximately \( 2086.83 \, \text{J} \).