Problem Statement
State Trouton’s rule for the entropy of vaporisation and verify it for water and benzene.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: Trouton’s rule (1884): For most non-associated liquids, the molar entropy of vaporisation at the normal boiling point is approximately:
Step 2 — Apply the relevant physical law or equation: $$\Delta S_{vap} = \frac{L_{vap}}{T_b} \approx 88\ \text{J/mol·K}$$
Step 3 — Solve algebraically for the unknown: Verification:
Step 4 — Substitute numerical values with units:
- Benzene: $L_{vap}=30.8\ \text{kJ/mol}$, $T_b=353\ \text{K}$: $\Delta S = 30800/353 = 87.3\ \text{J/mol·K}$ ✓
- Water: $L_{vap}=40.7\ \text{kJ/mol}$, $T_b=373\ \text{K}$: $\Delta S = 40700/373 = 109\ \text{J/mol·K}$ — anomalously high due to hydrogen bonding.
Step 5 — Compute and check the result: Trouton’s rule fails for strongly associated liquids (water, alcohols) and for low-boiling-point substances.
Worked Calculation
$$\Delta S_{vap} = \frac{L_{vap}}{T_b} \approx 88\ \text{J/mol·K}$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
$$\boxed{\Delta S_{vap} = \frac{L_{vap}}{T_b} \approx 88\ \text{J/mol·K}}$$
Trouton’s rule (1884): For most non-associated liquids, the molar entropy of vaporisation at the normal boiling point is approximately:
$$\Delta S_{vap} = \frac{L_{vap}}{T_b} \approx 88\ \text{J/mol·K}$$
Verification:
- Benzene: $L_{vap}=30.8\ \text{kJ/mol}$, $T_b=353\ \text{K}$: $\Delta S = 30800/353 = 87.3\ \text{J/mol·K}$ ✓
- Water: $L_{vap}=40.7\ \text{kJ/mol}$, $T_b=373\ \text{K}$: $\Delta S = 40700/373 = 109\ \text{J/mol·K}$ — anomalously high due to hydrogen bonding.
Trouton’s rule fails for strongly associated liquids (water, alcohols) and for low-boiling-point substances.
Answer
$$\boxed{\Delta S_{vap} = \frac{L_{vap}}{T_b} \approx 88\ \text{J/mol·K}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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