# How Does Temperature Affect Ethanol Vapor Pressure and Mass?

• fk378
In summary: K, so you can just use thatIn summary, at -11 degrees Celsius, a 4.7-L sealed bottle containing 0.33 g of liquid ethanol, C2H6O, will reach equilibrium with its vapor at -11 degrees Celsius. At 20 degrees Celsius, all of the ethanol vaporized. There would be 0.33 g of ethanol present at 0 degrees Celsius.
fk378
A 4.7-L 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!

A sketch of the solution.

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
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.

there is another form of the Clausius Clapeyron equation

Ln(P1/P2)=ΔHVap/(1/T2-1/T1)

which might be easier for your specific question as you can plug in all the variables and not have to solve for C

the R value is a constant, 8.31457 J/K

## What is vapor pressure?

Vapor pressure is the pressure exerted by the vapor of a substance in equilibrium with its liquid or solid phase at a given temperature.

## How is vapor pressure measured?

Vapor pressure can be measured using a variety of methods, including the isopiestic method, the static method, and the dynamic method.

## What factors affect the vapor pressure of ethanol?

The vapor pressure of ethanol is affected by temperature, pressure, and the presence of other substances (such as water or other solvents) in the surrounding environment.

## Why is the vapor pressure of ethanol important?

The vapor pressure of ethanol is important in various industries, such as the production of alcoholic beverages, fuel production, and in the chemical and pharmaceutical industries. It is also important in understanding the behavior of mixtures of ethanol and other substances.

## What is the relationship between vapor pressure and boiling point?

Vapor pressure and boiling point are inversely related - as vapor pressure increases, the boiling point decreases. This is because at a higher vapor pressure, more molecules are in the gas phase and therefore less energy is required to break the bonds and reach the boiling point.

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