One kg of water at `373K` is converted into steam at the same temperature. The volume `1 cm^(3)` of water becomes `1671 cm^(3)` on boiling. Calculate the change in internal energy of the system , if heat of vaporisation is `540 cal g^(-1)`. Given standard atmospheric pressure `=1.013xx10^(5) Nm^(-2)`.

Explanation

To calculate the change in internal energy of the system when 1 kg of water at 373 K is converted into steam at the same temperature, we can follow these steps:

### Step 1: Identify the given data
- Mass of water, \( m = 1 \, \text{kg} = 1000 \, \text{g} \)
- Initial volume of water, \( V_1 = 1 \, \text{cm}^3 \)
- Final volume of steam, \( V_2 = 1671 \, \text{cm}^3 \)
- Heat of vaporization, \( L = 540 \, \text{cal/g} \)
- Atmospheric pressure, \( P = 1.013 \times 10^5 \, \text{N/m}^2 \)

### Step 2: Calculate the heat absorbed (dq)
The heat absorbed during the phase change from water to steam can be calculated using the formula:
\[ dq = m \cdot L \]
Substituting the values:
\[ dq = 1000 \, \text{g} \times 540 \, \text{cal/g} = 540000 \, \text{cal} = 540 \, \text{kcal} \]

### Step 3: Calculate the work done (dw)
The work done during the phase change can be calculated using the formula:
\[ dw = P \cdot \Delta V \]
Where \( \Delta V = V_2 - V_1 \).
First, we need to convert the volumes from cm³ to m³:
\[ V_1 = 1 \, \text{cm}^3 = 1 \times 10^{-6} \, \text{m}^3 \]
\[ V_2 = 1671 \, \text{cm}^3 = 1671 \times 10^{-6} \, \text{m}^3 \]
Now, calculate \( \Delta V \):
\[ \Delta V = V_2 - V_1 = (1671 - 1) \times 10^{-6} \, \text{m}^3 = 1670 \times 10^{-6} \, \text{m}^3 \]
Now, substituting the values into the work done formula:
\[ dw = P \cdot \Delta V = (1.013 \times 10^5 \, \text{N/m}^2) \cdot (1670 \times 10^{-6} \, \text{m}^3) \]
Calculating this gives:
\[ dw = 1.013 \times 10^5 \cdot 1670 \times 10^{-6} = 169.171 \, \text{J} \]

### Step 4: Convert work done to calories
To convert joules to calories, we use the conversion factor \( 1 \, \text{cal} = 4.184 \, \text{J} \):
\[ dw = \frac{169.171 \, \text{J}}{4.184 \, \text{J/cal}} \approx 40.5 \, \text{cal} \approx 0.0405 \, \text{kcal} \]

### Step 5: Calculate the change in internal energy (du)
Using the first law of thermodynamics:
\[ dq = du + dw \]
Rearranging gives us:
\[ du = dq - dw \]
Substituting the values:
\[ du = 540 \, \text{kcal} - 0.0405 \, \text{kcal} \approx 539.9595 \, \text{kcal} \]

### Final Answer
The change in internal energy of the system is approximately:
\[ du \approx 539.96 \, \text{kcal} \]