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 vaporisation 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 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:
\[
\boxed{2086.45 \, \text{J}}
\]
**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:
\[
\boxed{2086.45 \, \text{J}}
\]