Determine the value of `DeltaH` and `DeltaU` for the reversible isothermal evaporation of 90.0g of water at `100^(@)C`.Assume that water vapour behaves as an ideal gas and heat of evaporation of water is `540 calg^(-)R = 2.0 cal mol^(-1)K^(-1)`. (Answer in calories)

Explanation

To determine the values of ΔH (change in enthalpy) and ΔU (change in internal energy) for the reversible isothermal evaporation of 90.0 g of water at 100°C, follow these steps:

### Step 1: Calculate the number of moles of water
To find the number of moles of water, use the formula:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

Given:
- Mass of water = 90.0 g
- Molar mass of water (H₂O) = 18 g/mol

\[ \text{Number of moles} = \frac{90.0 \, \text{g}}{18 \, \text{g/mol}} = 5 \, \text{moles} \]

### Step 2: Calculate ΔH (change in enthalpy)
The change in enthalpy for the evaporation process can be calculated using the heat of evaporation:

\[ \Delta H = \text{heat of evaporation} \times \text{mass of water} \]

Given:
- Heat of evaporation = 540 cal/g

\[ \Delta H = 540 \, \text{cal/g} \times 90.0 \, \text{g} = 48600 \, \text{cal} \]

### Step 3: Calculate ΔU (change in internal energy)
To find ΔU, we can use the relation between ΔH and ΔU:

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

Where:
- ΔN_g = change in the number of moles of gas (moles of water vapor produced)
- R = gas constant (2 cal/mol·K)
- T = temperature in Kelvin

Since we are evaporating 5 moles of water, ΔN_g = 5. The temperature in Kelvin is:

\[ T = 100°C + 273 = 373 \, K \]

Now, substituting the values into the equation:

\[ \Delta U = \Delta H - \Delta N_g R T \]

Substituting the known values:

\[ \Delta U = 48600 \, \text{cal} - (5 \, \text{moles} \times 2 \, \text{cal/mol·K} \times 373 \, K) \]

Calculating the second term:

\[ \Delta N_g R T = 5 \times 2 \times 373 = 3730 \, \text{cal} \]

Now, substituting back:

\[ \Delta U = 48600 \, \text{cal} - 3730 \, \text{cal} = 44870 \, \text{cal} \]

### Final Results
- ΔH = 48600 cal
- ΔU = 44870 cal