When water is boiled at `2` atm pressure the latent heat of vaporization is `2.2xx10^(6) J//kg` and the boiling point is `120^(@)C` At `2` atm pressure `1` kg of water has a volume of `10^(-3)m^(-3)` and `1` kg of steam has volume of `0.824m^(3)` . The increase in internal energy of `1` kg of water when it is converted into steam at `2` atm pressure and `120^(@)C` is [`1` atm pressure `=1.013xx10^(5)N//m^(2)`]

Options

A. `2.033 J`
B. `2.033 xx 10^(6)J`
C. `0.167 xx 10^(6)J`
D. `2.267 xx 10^(6)J`

Explanation

To find the increase in internal energy of 1 kg of water when it is converted into steam at 2 atm pressure and 120°C, we will follow these steps:

### Step 1: Calculate the heat absorbed (Q) during vaporization
The heat absorbed during the phase change from water to steam can be calculated using the formula:
\[ Q = m \times L \]
where:
- \( m \) = mass of water = 1 kg
- \( L \) = latent heat of vaporization = \( 2.2 \times 10^6 \, \text{J/kg} \)

Substituting the values:
\[ Q = 1 \, \text{kg} \times 2.2 \times 10^6 \, \text{J/kg} = 2.2 \times 10^6 \, \text{J} \]

### Step 2: Calculate the work done (W) during vaporization
The work done by the gas during the phase change can be calculated using the formula:
\[ W = P \Delta V \]
where:
- \( P \) = pressure = \( 2 \, \text{atm} = 2 \times 1.013 \times 10^5 \, \text{N/m}^2 = 2.026 \times 10^5 \, \text{N/m}^2 \)
- \( \Delta V \) = change in volume = \( V_{\text{steam}} - V_{\text{water}} \)

Given:
- Volume of 1 kg of water \( V_{\text{water}} = 10^{-3} \, \text{m}^3 \)
- Volume of 1 kg of steam \( V_{\text{steam}} = 0.824 \, \text{m}^3 \)

Calculating \( \Delta V \):
\[ \Delta V = 0.824 \, \text{m}^3 - 10^{-3} \, \text{m}^3 = 0.823 \, \text{m}^3 \]

Now substituting the values into the work done formula:
\[ W = 2.026 \times 10^5 \, \text{N/m}^2 \times 0.823 \, \text{m}^3 \]
\[ W \approx 1.669 \times 10^5 \, \text{J} \]

### Step 3: Calculate the change in internal energy (ΔU)
Using the first law of thermodynamics:
\[ \Delta U = Q - W \]

Substituting the values we calculated:
\[ \Delta U = 2.2 \times 10^6 \, \text{J} - 1.669 \times 10^5 \, \text{J} \]
\[ \Delta U \approx 2.033 \times 10^6 \, \text{J} \]

### Final Answer
The increase in internal energy of 1 kg of water when it is converted into steam at 2 atm pressure and 120°C is approximately:
\[ \Delta U \approx 2.033 \times 10^6 \, \text{J} \]

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