Difficulty: Easy
Correct Answer: both ΔH_mix = 0 and ΔV_mix = 0
Explanation:
Introduction / Context:When liquid mixtures behave ideally, design calculations simplify dramatically because activities equal mole fractions. Recognizing the thermodynamic hallmarks of ideality helps in VLE modeling and quick feasibility studies.
Given Data / Assumptions:
Concept / Approach:Ideality implies no enthalpy or volume effects on mixing. Hence, ΔH_mix = 0 (no heat released/absorbed) and ΔV_mix = 0 (no contraction/expansion). This reflects indistinguishable unlike/like interactions and activity coefficients near unity (γ_i ≈ 1).
Step-by-Step Solution:State the ideal-solution criteria.Map them to thermodynamic statements: ΔH_mix = 0 and ΔV_mix = 0.Choose the option containing both equalities.
Verification / Alternative check:Empirical data for near-ideal pairs (like benzene–toluene) show negligible heat and volume changes upon mixing, validating these criteria.
Why Other Options Are Wrong:Options with nonzero ΔH or ΔV describe nonideal mixtures exhibiting positive/negative deviations from Raoult’s law.
Common Pitfalls:Believing only one of the two (ΔH or ΔV) must be zero; both are required for strict ideality.
Final Answer:both ΔH_mix = 0 and ΔV_mix = 0
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