To calculate the work done when 2.5 moles of H₂O vaporizes at 1.0 atm and 25°C, we will follow these steps:
### Step 1: Identify the formula for work done
The work done (W) during a phase change 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.
### Step 2: Calculate the change in volume (ΔV)
Since we are vaporizing liquid water, we can assume that the volume of the liquid is negligible compared to that of the vapor. Therefore, we need to calculate the volume of the water vapor using the ideal gas law:
\[ PV = nRT \]
Rearranging gives us:
\[ V = \frac{nRT}{P} \]
### Step 3: Substitute the values into the ideal gas law
Given:
- \( n = 2.5 \) moles,
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \),
- \( T = 25°C = 298 \, \text{K} \) (by adding 273),
- \( P = 1.0 \, \text{atm} \).
Substituting these values into the equation for volume:
\[ V = \frac{(2.5 \, \text{mol})(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1})(298 \, \text{K})}{1.0 \, \text{atm}} \]
### Step 4: Calculate the volume
Calculating the volume:
\[ V = \frac{(2.5)(0.0821)(298)}{1} \]
\[ V = \frac{61.16}{1} \]
\[ V = 61.16 \, \text{L} \]
### Step 5: Calculate the work done
Now that we have the volume, we can substitute back into the work done formula:
\[ W = -P \Delta V = - (1.0 \, \text{atm})(61.16 \, \text{L}) \]
### Step 6: Convert work done to joules
Since 1 L atm = 101.3 J, we convert the work done:
\[ W = - (61.16 \, \text{L})(1.0 \, \text{atm}) \times 101.3 \, \text{J/L atm} \]
\[ W = -6195.96 \, \text{J} \]
### Step 7: Convert to kilojoules
To express this in kilojoules:
\[ W = -6195.96 \, \text{J} \div 1000 = -6.19596 \, \text{kJ} \]
Thus, the magnitude of work done is:
\[ W \approx 6.195 \, \text{kJ} \]
### Final Answer
The work done when 2.5 moles of H₂O vaporizes at 1.0 atm and 25°C is approximately **6.195 kJ**.
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