What is the work done against the atmosphere when `25` grams of water vaprorizes at `373 K` against a constant external pressure of `1 atm` ? Assume that steam obeys perfect gas law. Given that the molar enthalpy of vaporization is `9.72 kcal//"mole"`, what is the change of internal energy in the above process?

Options

A. `1294.0 cal, 11247 cal`
B. `921.4 cal , 11074 cal`
C. `1025.6 cal , 12474.3 cal`
D. `1129.3 cal, 10207 cal`

Explanation

To solve the problem, we need to calculate the work done against the atmosphere when 25 grams of water vaporizes at 373 K and then determine the change in internal energy for this process.

### Step 1: Calculate the number of moles of water
To find the number of moles (n) of water, we use the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]

Given:
- Mass of water = 25 grams
- Molar mass of water (H₂O) = 18 grams/mole

Calculating the number of moles:
\[ n = \frac{25 \text{ g}}{18 \text{ g/mole}} \approx 1.39 \text{ moles} \]

### Step 2: Calculate the work done against the atmosphere
The work done (W) against the atmosphere can be calculated using the formula:
\[ W = -P \Delta V \]

For an ideal gas, we can express this as:
\[ W = -nRT \]

Where:
- \( R \) (gas constant) = 2 cal/(K·mol)
- \( T \) = 373 K

Substituting the values:
\[ W = -nRT = -\left(1.39 \text{ moles}\right) \left(2 \text{ cal/(K·mol)}\right) \left(373 \text{ K}\right) \]

Calculating the work done:
\[ W = -1.39 \times 2 \times 373 \approx -1036.11 \text{ cal} \]

Since work done against the atmosphere is considered positive:
\[ W = 1036.11 \text{ cal} \]

### Step 3: Calculate the change in internal energy (ΔU)
To find the change in internal energy, we use the relationship:
\[ \Delta H = \Delta U + \Delta n_g RT \]

Where:
- \( \Delta H \) = molar enthalpy of vaporization = 9.72 kcal/mole = 9720 cal/mole
- \( \Delta n_g \) = change in number of moles of gas = moles of products - moles of reactants

In this case:
- Reactants: 1 mole of liquid water (H₂O) → 0 moles of gas
- Products: 1 mole of gas (H₂O vapor)

Thus, \( \Delta n_g = 1 - 0 = 1 \).

Substituting into the equation:
\[ \Delta H = \Delta U + \Delta n_g RT \]
\[ 9720 \text{ cal} = \Delta U + (1) \times (2 \text{ cal/(K·mol)}) \times (373 \text{ K}) \]

Calculating \( \Delta n_g RT \):
\[ \Delta n_g RT = 2 \times 373 = 746 \text{ cal} \]

Now substituting back:
\[ 9720 = \Delta U + 746 \]
\[ \Delta U = 9720 - 746 = 8974 \text{ cal} \]

### Step 4: Adjust for the number of moles
Since we calculated ΔU for 1 mole, we need to adjust it for the actual number of moles (1.39 moles):
\[ \Delta U_{\text{total}} = 8974 \text{ cal/mole} \times 1.39 \text{ moles} \]
\[ \Delta U_{\text{total}} \approx 12464.46 \text{ cal} \]

### Final Results
- Work done against the atmosphere: **1036.11 cal**
- Change in internal energy: **12464.46 cal**