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). The formula we will use is:

\[ \Delta H = \Delta U + \Delta n_g \cdot R \cdot T \]

Where:
- ΔH = Enthalpy change (given as 40.66 kJ)
- ΔU = Internal energy change (what we want to find)
- Δn_g = Change in the number of moles of gas (for vaporization of 1 mole of water, Δn_g = 1)
- R = Universal gas constant (8.314 J/mol·K or 0.008314 kJ/mol·K)
- T = Temperature in Kelvin (100°C = 373 K)

### Step-by-step Solution:

1. **Identify the values:**
- ΔH = 40.66 kJ
- Δn_g = 1 (since 1 mole of water is vaporizing)
- R = 8.314 J/mol·K = 0.008314 kJ/mol·K
- T = 100°C = 373 K

2. **Substitute the values into the formula:**
\[ \Delta H = \Delta U + \Delta n_g \cdot R \cdot T \]
\[ 40.66 = \Delta U + 1 \cdot 0.008314 \cdot 373 \]

3. **Calculate Δn_g \cdot R \cdot T:**
\[ \Delta n_g \cdot R \cdot T = 1 \cdot 0.008314 \cdot 373 = 3.1 \text{ kJ} \]

4. **Rearrange the equation to solve for ΔU:**
\[ \Delta U = \Delta H - \Delta n_g \cdot R \cdot T \]
\[ \Delta U = 40.66 - 3.1 \]

5. **Calculate ΔU:**
\[ \Delta U = 37.56 \text{ kJ} \]

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