The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.4 atm? (∆Hvap = 28.5 kJ/mol) What is the vapor pressure of a liquid at 313.83 K if its ∆Hvap = 28.9 kJ/mol and its normal boiling point is 341.88 K?
Added by Dennis S.
Step 1
First, we need to convert the given values into the correct units. The given ∆Hvap is in kJ/mol, but we need it in J/mol for our calculations. So, we multiply it by 1000: 28.9 kJ/mol * 1000 = 28900 J/mol. Show more…
Show all steps
Your feedback will help us improve your experience
Ankur S and 92 other Chemistry 101 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.365 atm? (ΔHvap = 28.5 kJ/mol)
Dr. Satish I.
The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.765 atm? (∆Hvap = 28.5 kJ/mol)
Adi S.
A liquid has a $\Delta_{\text {vap }} H$ of $38.7 \mathrm{~kJ} / \mathrm{mol}$ and a boiling point of $110^{\circ} \mathrm{C}$ at 1 atm pressure. Calculate the vapor pressure of the liquid at $97^{\circ} \mathrm{C}$.
Recommended Textbooks
Chemistry: Structure and Properties
Chemistry The Central Science
Chemistry
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD