The vapor pressure of liquid propanol, C3H7OH, is 40.0 mm Hg at 309 K. A sample
ID: 991530 • Letter: T
Question
The vapor pressure of liquid propanol, C3H7OH, is 40.0 mm Hg at 309 K. A sample of C3H7OH is placed in a closed, evacuated container of constant volume at a temperature of 494 K. It is found that all of the C3H7OH is in the vapor phase and that the pressure is 60.0 mm Hg. If the temperature in the container is reduced to 309 K, which of the following statements are correct?
Choose all that apply.
The pressure in the container will be 40.0 mm Hg.
Some of the vapor initially present will condense.
Only propanol vapor will be present.
Liquid propanol will be present.
No condensation will occur.
Explanation / Answer
At constant volume the temperature and pressure are related to each other as follows:
P1/T1= P2/T2
Here P1 = 60.0 mmHg; T1=494K
P2= ? and T2= 309 K
P2= P1/T1*T2
P2 =60/494*309
P2 = 37.53 mmHg
Hence the correct answer is Some of the vapor initially present will condense.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.