How Does Temperature Affect Ethanol Vapor Pressure and Mass?

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SUMMARY

This discussion focuses on the relationship between temperature and the vapor pressure of ethanol (C2H6O) in a sealed environment. The Clausius-Clapeyron equation is essential for calculating vapor pressure as a function of temperature, allowing for the determination of ethanol mass in vapor form at various temperatures. The ideal gas law is utilized to find the number of moles of ethanol vapor, which is crucial for solving the posed questions regarding ethanol mass at -11°C, 0°C, and 20°C. Key constants include the vapor pressures of ethanol at -2.3°C (10 torr) and 19°C (40 torr).

PREREQUISITES
  • Understanding of the Clausius-Clapeyron equation
  • Familiarity with the ideal gas law (PV=nRT)
  • Knowledge of vapor pressure concepts
  • Basic thermodynamics principles
NEXT STEPS
  • Calculate vapor pressure using the Clausius-Clapeyron equation for ethanol at various temperatures
  • Apply the ideal gas law to determine the mass of ethanol vapor at -11°C and 20°C
  • Explore the relationship between temperature and vapor pressure for other substances
  • Investigate the implications of condensation and vaporization equilibrium
USEFUL FOR

Chemistry students, researchers in thermodynamics, and professionals in chemical engineering will benefit from this discussion, particularly those interested in phase changes and vapor pressure calculations.

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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!
 
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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
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 gave a excellent answer

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
 

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