Discussion
Both the vapour pressure at 25°C and normal boiling point of toluene are higher than that of water, which of the following explains these observations ?
Answer: C
The explanation is that toluene has a lower heat of vaporization than water.
The Clausius-Clapeyron equation links how fast vapour pressure rises with temperature to the enthalpy of vaporization:
\[ \ln P = -\frac{\Delta H_{vap}}{RT} + C \]
A small ΔHvap means a shallow curve. Toluene starts with a higher vapour pressure than water at 25 °C, yet reaches 1 atm (its boiling point) at a higher temperature, 110.6 °C against 100 °C. Its vapour pressure must therefore rise more slowly with temperature, which requires the smaller ΔHvap: about 33 kJ/mol for toluene against about 41 kJ/mol for hydrogen-bonded water.
The first option states the opposite of the usual trend (a higher vapour pressure normally goes with a lower boiling point). A higher molar mass would lower the vapour pressure, not raise it, and the density of the vapour does not govern evaporation.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development