18g of water is taken to prepare the tea. Find out the internal
energy of vaporisation at `100^(@)`C. `(Delta_(vap)H` for water at 373 K
is `40.66kJ mol^(-1))`

Options

A. `37.56kJmol^(-1)`
B. `-37.56kJmol^(-1)`
C. `43.76kJmol^(-1)`
D. `3-43.76kJmol^(-1)`

Explanation

To find the internal energy of vaporization (\( \Delta E_{vap} \)) of water at \( 100^\circ C \), we will use the relationship between enthalpy of vaporization (\( \Delta H_{vap} \)) and internal energy of vaporization (\( \Delta E_{vap} \)) given by the equation:

\[ \Delta H = \Delta E + \Delta N_g RT \]

Where:
- \( \Delta H \) = Enthalpy of vaporization
- \( \Delta E \) = Internal energy of vaporization
- \( \Delta N_g \) = Change in the number of moles of gas
- \( R \) = Universal gas constant (8.314 J/mol·K)
- \( T \) = Temperature in Kelvin

### Step 1: Convert the mass of water to moles
Given that the molar mass of water (H₂O) is approximately 18 g/mol, we can find the number of moles of water in 18 g:

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{18 \, \text{g}}{18 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 2: Identify the values
From the problem statement:
- \( \Delta H_{vap} = 40.66 \, \text{kJ/mol} = 40660 \, \text{J/mol} \) (conversion to Joules)
- \( T = 100^\circ C = 373 \, \text{K} \)
- \( R = 8.314 \, \text{J/mol·K} \)

### Step 3: Calculate \( \Delta N_g \)
In this case, when water vaporizes, it goes from liquid to gas. Therefore:
- \( \Delta N_g = N_p - N_r = 1 - 0 = 1 \)

### Step 4: Substitute values into the equation
Now we can substitute the values into the equation:

\[ \Delta H_{vap} = \Delta E_{vap} + \Delta N_g RT \]

Rearranging gives us:

\[ \Delta E_{vap} = \Delta H_{vap} - \Delta N_g RT \]

Substituting the known values:

\[ \Delta E_{vap} = 40660 \, \text{J/mol} - (1 \cdot 8.314 \, \text{J/mol·K} \cdot 373 \, \text{K}) \]

### Step 5: Calculate \( RT \)
Calculating \( RT \):

\[ RT = 8.314 \, \text{J/mol·K} \cdot 373 \, \text{K} = 3100.622 \, \text{J/mol} \]

### Step 6: Final calculation
Now substituting back into the equation for \( \Delta E_{vap} \):

\[ \Delta E_{vap} = 40660 \, \text{J/mol} - 3100.622 \, \text{J/mol} \]

\[ \Delta E_{vap} = 37559.378 \, \text{J/mol} \]

Converting back to kilojoules:

\[ \Delta E_{vap} = \frac{37559.378 \, \text{J/mol}}{1000} = 37.56 \, \text{kJ/mol} \]

### Final Answer
The internal energy of vaporization at \( 100^\circ C \) is:

\[ \Delta E_{vap} = 37.56 \, \text{kJ/mol} \]

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