At 1 atmospheric pressure, 1.000 g of water having a volume of `1.000 cm^3` becomes 1671 `cm^3` of steam when boiled. The heat of vaporization of water at 1 atmosphere is `539 cal//g` . What is the change in internal energy during the process ?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Mass of water (m) = 1.000 g
- Heat of vaporization (L) = 539 cal/g
- Initial volume of water (V_initial) = 1.000 cm³
- Final volume of steam (V_final) = 1671 cm³
- Pressure (P) = 1 atm = \(1.013 \times 10^5\) Pa (for calculations, we'll convert to calories later)

### Step 2: Calculate the heat absorbed (ΔQ)
The heat absorbed during the phase change from water to steam can be calculated using the formula:
\[ \Delta Q = m \cdot L \]
Substituting the values:
\[ \Delta Q = 1.000 \, \text{g} \times 539 \, \text{cal/g} = 539 \, \text{cal} \]

### Step 3: Calculate the change in volume (ΔV)
The change in volume (ΔV) during the process is given by:
\[ \Delta V = V_{\text{final}} - V_{\text{initial}} \]
Substituting the values:
\[ \Delta V = 1671 \, \text{cm}^3 - 1 \, \text{cm}^3 = 1670 \, \text{cm}^3 \]

### Step 4: Convert ΔV to cubic meters
To convert cm³ to m³:
\[ \Delta V = 1670 \, \text{cm}^3 \times 10^{-6} \, \text{m}^3/\text{cm}^3 = 0.001670 \, \text{m}^3 \]

### Step 5: Calculate the work done (W)
The work done during the expansion against constant pressure is given by:
\[ W = P \cdot \Delta V \]
Substituting the values (using P in Pascals):
\[ W = 1.013 \times 10^5 \, \text{Pa} \times 0.001670 \, \text{m}^3 = 169.71 \, \text{J} \]
Now, converting Joules to calories (1 cal = 4.184 J):
\[ W = \frac{169.71 \, \text{J}}{4.184 \, \text{J/cal}} \approx 40.5 \, \text{cal} \]

### Step 6: Calculate the change in internal energy (ΔU)
Using the first law of thermodynamics:
\[ \Delta U = \Delta Q - W \]
Substituting the values:
\[ \Delta U = 539 \, \text{cal} - 40.5 \, \text{cal} \approx 498.5 \, \text{cal} \]

### Final Answer
The change in internal energy during the process is approximately:
\[ \Delta U \approx 499 \, \text{cal} \]