Relative volatility intuition: Given equal normal boiling-point vapor pressures at 100 °C, which compound will have the higher vapor pressure at 150 °C—water or methylcyclohexane?

Introduction / Context:The Clausius–Clapeyron relation qualitatively explains how vapor pressure varies with temperature. Substances with lower latent heat of vaporization show a faster increase in vapor pressure with rising temperature. This question applies that insight to compare water with methylcyclohexane at 150 °C, given both have the same vapor pressure at 100 °C.

Given Data / Assumptions:

• At 100 °C: P_sat,water = P_sat,methylcyclohexane = 1 atm.
• Latent heats (at ~100 °C): λ_water ≈ 40.63 kJ/mol; λ_methylcyclohexane ≈ 31.55 kJ/mol.
• At 150 °C: P_sat,water = 4.69 atm (given).
• Assume Clausius–Clapeyron holds approximately over this range and λ is weakly temperature dependent.

Concept / Approach:Clausius–Clapeyron in integrated form suggests ln(P2/P1) ≈ −(λ/R) * (1/T2 − 1/T1). For a given temperature interval, smaller λ gives a larger increase in ln(P) and thus a higher P2.

Step-by-Step Solution:

Verification / Alternative check:A rough calculation with constant λ would show a larger ln(P2/P1) for methylcyclohexane. While exact numbers require integration and real-property data, the qualitative conclusion is robust.

Why Other Options Are Wrong:

• (a) and (b) contradict the λ-based trend.
• (d) We have sufficient comparative data (equal P at 100 °C and λ values) to draw a qualitative conclusion.
• (e) Exact equality at 150 °C is not expected because λ values differ.

Common Pitfalls:Assuming equal vapor pressures at one temperature implies equal curves at all temperatures. Always consider latent heat differences.

Final Answer:Significantly more than 4.69 atm (methylcyclohexane higher than water)