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. 4800 kJ
B. `188.494 kJ`
C. `40.8 kJ`
D. `125.03 kJ`

Explanation

To find the value of ΔU (change in internal energy) for the reversible isothermal evaporation of 90 g of water at 100°C, we can use the relationship between ΔU and ΔH (enthalpy change) for phase changes. The equation we will use is:

\[ \Delta U = \Delta H - \Delta n \cdot R \cdot T \]

Where:
- ΔH = change in enthalpy (given as 40.8 kJ/mol)
- Δn = change in the number of moles of gas (for evaporation, it will be the moles of water vapor produced)
- R = ideal gas constant (8.314 J/(K·mol))
- T = temperature in Kelvin

**Step 1: Convert the mass of water to moles.**

The molar mass of water (H₂O) is approximately 18 g/mol. Thus, we can calculate the number of moles of 90 g of water:

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

**Step 2: Determine the change in the number of moles (Δn).**

In the case of evaporation, 1 mole of liquid water produces 1 mole of water vapor. Therefore, for 5 moles of water evaporating, Δn = 5.

\[ \Delta n = 5 \, \text{mol} \]

**Step 3: Convert the temperature to Kelvin.**

The temperature given is 100°C. To convert to Kelvin:

\[ T = 100 + 273.15 = 373.15 \, \text{K} \]

**Step 4: Calculate the change in internal energy (ΔU).**

Now we can substitute the values into the equation:

1. Convert ΔH from kJ to J:
\[ \Delta H = 40.8 \, \text{kJ/mol} = 40.8 \times 1000 \, \text{J/mol} = 40800 \, \text{J/mol} \]

2. Substitute the values into the equation:
\[ \Delta U = \Delta H - \Delta n \cdot R \cdot T \]
\[ \Delta U = 40800 \, \text{J/mol} - (5 \, \text{mol} \cdot 8.314 \, \text{J/(K·mol)} \cdot 373.15 \, \text{K}) \]

3. Calculate the second term:
\[ \Delta n \cdot R \cdot T = 5 \cdot 8.314 \cdot 373.15 = 15567.5 \, \text{J} \]

4. Finally, calculate ΔU:
\[ \Delta U = 40800 \, \text{J} - 15567.5 \, \text{J} = 25232.5 \, \text{J} \]

**Step 5: Convert ΔU back to kJ.**
\[ \Delta U = \frac{25232.5 \, \text{J}}{1000} = 25.23 \, \text{kJ} \]

Thus, the value for ΔU for the reversible isothermal evaporation of 90 g of water at 100°C is approximately **25.23 kJ**.

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