`1 cm^(3)` of water at its boiling point absorbs `540 cal` of heat to become steam with a volume of `1671 cm^(3)`. If the atmospheic pressure is `1.013 xx 10^(5) N//m^(2)` and the mechanical equivalent of heat `= 4.19 J//cal`, the energy spent in this process in overcoming intermolecular forces is

Options

A. `540 cal`
B. `40 cal`
C. `500 cal`
D. Zero

Explanation

To solve the problem, we need to find the energy spent in overcoming intermolecular forces during the phase change from water to steam. We can use the first law of thermodynamics, which states that the change in internal energy (ΔU) is equal to the heat added to the system (ΔQ) minus the work done by the system (ΔW).

### Step-by-Step Solution:

1. **Identify Given Values:**
- Heat absorbed (ΔQ) = 540 cal
- Volume of water (V_initial) = 1 cm³
- Volume of steam (V_final) = 1671 cm³
- Atmospheric pressure (P) = 1.013 × 10^5 N/m²
- Mechanical equivalent of heat = 4.19 J/cal

2. **Convert Heat from Calories to Joules:**
\[ \Delta Q = 540 \, \text{cal} \times 4.19 \, \text{J/cal} = 2262.6 \, \text{J} \]

3. **Calculate Change in Volume (ΔV):**
\[ \Delta V = V_{\text{final}} - V_{\text{initial}} = 1671 \, \text{cm}^3 - 1 \, \text{cm}^3 = 1670 \, \text{cm}^3 \]
Convert cm³ to m³:
\[ \Delta V = 1670 \, \text{cm}^3 \times 10^{-6} \, \text{m}^3/\text{cm}^3 = 0.00167 \, \text{m}^3 \]

4. **Calculate Work Done (ΔW):**
\[ \Delta W = P \times \Delta V = 1.013 \times 10^5 \, \text{N/m}^2 \times 0.00167 \, \text{m}^3 = 169.1711 \, \text{J} \]

5. **Calculate Change in Internal Energy (ΔU):**
Using the first law of thermodynamics:
\[ \Delta U = \Delta Q - \Delta W \]
\[ \Delta U = 2262.6 \, \text{J} - 169.1711 \, \text{J} = 2093.4289 \, \text{J} \]

6. **Convert ΔU back to Calories:**
\[ \Delta U = \frac{2093.4289 \, \text{J}}{4.19 \, \text{J/cal}} \approx 499.1 \, \text{cal} \]

### Final Answer:
The energy spent in this process in overcoming intermolecular forces is approximately **499.1 cal**.