Question Vapor Pressure of Ethanol ( Physics Help and Math Help  Physics Forums Other Sciences ) Updated: 20080226 15:05:39 (2) 

Vapor Pressure of Ethanol
A 4.7L sealed bottle containing 0.33 g of liquid ethanol, C2H6O, is placed in a refrigerator and reaches equilibrium with its vapor at 11 degrees C.
a) What is the mass of ethanol present in the vapor?
b) When the container is removed and warmed to room temperature, 20 degrees C, will all the ethanol vaporize?
c) How much liquid ethanol would be present at 0 degrees C?
The vapor pressure of ethanol is 10 torr at 2.3 degrees C and 40 torr at 19 degrees C.

My attempt:
a) I have no idea how to approach this.
b) No, because there will be condensation occurring at the same rate as vaporization, once it reaches equilibrium again. (?)
c) I think I need to use PV=nRT for this... but I don't know how to relate the 2 different temperatures I'm getting (0 C and 19 C)
Any suggestions would be appreciated, thank you!!


Answers: Vapor Pressure of Ethanol ( Physics Help and Math Help  Physics Forums Other Sciences ) 

Vapor Pressure of Ethanol
To attack all three problems here, you need to calculate the vapour
pressure of ethanol as a function of the absolute temperature T.
Use the Clausius Clapyron equation for this:
log p = A/T +B,
where p is the vapour pressure, T the absolute temperature and A and B are
constant. You can easily get A and B from the given vapour pressures at
T=270.85 K and T=292.15 K, which will enable you to calculate the vapour pressure
at 262.15 K, 273.15 K and 293.15 K, or indeed any other temperature.
To find the mass of ethanol in the vapour phase, use the ideal gas
equation to get the number of moles, n:
n = pV/(RT), since V is given and p is the vapour pressure.
From n and the molecular weight of ethanol (C2H5OH), you can calculate the
mass of ethanol in the vapour. Since the vessel is sealed, the amount of
ethanol in the liquid and vapour phases together remains constant.
The answers to (a) and (c) should be easy. The answer to (b) is a bit
trickier because it could be that there is not enough ethanol to produce
the required partial pressure in the vapour. In this case, all of the
liquid will evaporate and your answer to (b) is incorrect.
You will have to do the calculation to see what comes out.
I assume you can take it from here, being careful to choose a consistent
set of units.
pkleinod


Vapor Pressure of Ethanol
To attack all three problems here, you need to calculate the vapour
pressure of ethanol as a function of the absolute temperature T.
Use the Clausius Clapyron equation for this:
log p = A/T +B,
where p is the vapour pressure, T the absolute temperature and A and B are
constant. You can easily get A and B from the given vapour pressures at
T=270.85 K and T=292.15 K, which will enable you to calculate the vapour pressure
at 262.15 K, 273.15 K and 293.15 K, or indeed any other temperature.
To find the mass of ethanol in the vapour phase, use the ideal gas
equation to get the number of moles, n:
n = pV/(RT), since V is given and p is the vapour pressure.
From n and the molecular weight of ethanol (C2H5OH), you can calculate the
mass of ethanol in the vapour. Since the vessel is sealed, the amount of
ethanol in the liquid and vapour phases together remains constant.
The answers to (a) and (c) should be easy. The answer to (b) is a bit
trickier because it could be that there is not enough ethanol to produce
the required partial pressure in the vapour. In this case, all of the
liquid will evaporate and your answer to (b) is incorrect.
You will have to do the calculation to see what comes out.
I assume you can take it from here, being careful to choose a consistent
set of units.
pkleinod


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