The internal energy change for the vapourisation of one mole water at 1 atm and 100°C will be (Enthalpy of vapourization of water is 40.66 kJ

Options

A. Zero
B. 37.56 kJ
C. -37.56 kJ
D. 40.6 kJ

Answer

To find the internal energy change (ΔU) for the vaporization of one mole of water at 1 atm and 100°C, we can use the relationship between enthalpy change (ΔH) and internal energy change (ΔU).

### Step-by-Step Solution:

1. **Identify Given Values:**
- Enthalpy of vaporization (ΔH) = 40.66 kJ
- Temperature (T) = 100°C = 373 K (convert Celsius to Kelvin by adding 273)
- Ideal gas constant (R) = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K) (conversion from J to kJ)

2. **Determine Change in Moles of Gas (ΔNG):**
- For the vaporization of water (H2O(l) → H2O(g)), we have:
- Moles of gaseous products = 1 (H2O(g))
- Moles of gaseous reactants = 0 (H2O(l))
- Therefore, ΔNG = 1 - 0 = 1.

3. **Use the Formula Relating ΔH and ΔU:**
- The relationship is given by:
\[ ΔH = ΔU + ΔNG \cdot R \cdot T \]
- Rearranging gives:
\[ ΔU = ΔH - ΔNG \cdot R \cdot T \]

4. **Substitute the Values into the Equation:**
- Substitute ΔH, ΔNG, R, and T into the equation:
\[ ΔU = 40.66 \, \text{kJ} - (1) \cdot (0.008314 \, \text{kJ/(mol·K)}) \cdot (373 \, \text{K}) \]

5. **Calculate the Second Term:**
- Calculate \(ΔNG \cdot R \cdot T\):
\[ ΔNG \cdot R \cdot T = 1 \cdot 0.008314 \cdot 373 = 3.096 \, \text{kJ} \]

6. **Final Calculation of ΔU:**
- Now substitute back into the equation for ΔU:
\[ ΔU = 40.66 \, \text{kJ} - 3.096 \, \text{kJ} = 37.564 \, \text{kJ} \]
- Rounding gives:
\[ ΔU \approx 37.56 \, \text{kJ} \]

### Conclusion:
The internal energy change (ΔU) for the vaporization of one mole of water at 1 atm and 100°C is approximately **37.56 kJ**.