The value for `DeltaU` for the reversible isothermal evaporation of 90 g water at `100^(@)C` will be `(DeltaH_("evap")" of water "=40.8" kJ mol"^(-1), R=8.314"J K"^(-1)"mol"^(-1))`

Options

A. (a) `4800 kJ`
B. (b) `188.494 kJ`
C. (c) `40.8 kJ`
D. (d) `125.03 kJ`

Explanation

To find the change in internal energy (ΔU) for the reversible isothermal evaporation of 90 g of water at 100°C, we can follow these steps:

### Step 1: Determine the number of moles of water
First, we need to calculate the number of moles of water in 90 g. The molar mass of water (H₂O) is approximately 18 g/mol.

\[ \text{Number of moles (n)} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{90 \, \text{g}}{18 \, \text{g/mol}} = 5 \, \text{mol} \]

### Step 2: Calculate ΔH for 90 g of water
We are given that the enthalpy of evaporation (ΔH) for water is 40.8 kJ/mol. Therefore, for 5 moles of water, we can calculate ΔH as follows:

\[ \Delta H = 40.8 \, \text{kJ/mol} \times 5 \, \text{mol} = 204 \, \text{kJ} \]

### Step 3: Convert ΔH to Joules
Since we need to work in consistent units, we convert ΔH from kJ to J:

\[ \Delta H = 204 \, \text{kJ} \times 1000 \, \text{J/kJ} = 204000 \, \text{J} \]

### Step 4: Calculate ΔN
Next, we need to find ΔN, which is the change in the number of moles of gas. In this case, we are converting liquid water to vapor. The number of moles of gas after evaporation is 5 moles (from 90 g of water), and before evaporation, it is 0 moles (since we start with liquid). Thus:

\[ \Delta N = 5 - 0 = 5 \, \text{mol} \]

### Step 5: Use the equation ΔH = ΔU + ΔNRT
We can now use the relationship between ΔH, ΔU, and ΔN:

\[ \Delta H = \Delta U + \Delta N R T \]

Where:
- R = 8.314 J/(K·mol)
- T = 373 K (100°C in Kelvin)

### Step 6: Substitute known values into the equation
Now we can substitute the known values into the equation:

\[ 204000 \, \text{J} = \Delta U + (5 \, \text{mol})(8.314 \, \text{J/(K·mol)})(373 \, \text{K}) \]

### Step 7: Calculate ΔNRT
First, calculate ΔNRT:

\[ \Delta N R T = 5 \times 8.314 \times 373 = 15580.61 \, \text{J} \]

### Step 8: Solve for ΔU
Now, substitute this value back into the equation:

\[ 204000 \, \text{J} = \Delta U + 15580.61 \, \text{J} \]

Rearranging gives:

\[ \Delta U = 204000 \, \text{J} - 15580.61 \, \text{J} = 188419.39 \, \text{J} \]

### Step 9: Convert ΔU to kJ
Finally, convert ΔU back to kJ:

\[ \Delta U = \frac{188419.39 \, \text{J}}{1000} = 188.419 \, \text{kJ} \]

### Final Answer
Thus, the change in internal energy (ΔU) for the reversible isothermal evaporation of 90 g of water at 100°C is approximately:

\[ \Delta U \approx 188.419 \, \text{kJ} \]