I'm having trouble getting the right answer for a problem, I'm hoping that by posting my work someone can help me figure out where I've gone wrong.

A two-liter soft drink bottle can withstand a pressure of 5 atm. Half a cup [approximately 120 mL) of ethyl alcohol C2H5OH, (d= 0.789 g/mL) is poured into a soft drink bottle at room temperature. The bottle is then heated to 100 degrees C [3 sig figs], changing the liquid alcohol to a gas. Will the soft drink bottle withstand the pressure, or will it explode?

My work:
P= nRT/V

to find n:
.789 g/mL x 120 mL = 94.68 g = xmol x 46.08 g/mol
x= 2.055 mol

(2.055 mol x .08206 L x atm/mol x K x 373.15 K) / 2 L
= 31.46 atm

This number seems high to me. Am I setting up the problem correctly?

You have done it correctly, if the assumption that all of the ethyl alcohol vaporizes is valid. The way they worded the question, you cannot say for sure if they expect you to assume that it all evaporates. If it does, it is ok to use the ideal gas law, as you have done.

However, you need to consider the vapor pressure of alcohol as a function of temperature to see if it can all evaporate. Extrapolating from vapor pressure data I can find, and using a log P = a + b/T fit, it is appears that the vapor pressure of C2H5OH is about 2 atm at 100 C. (It is 1 atm at 78.5 C.) This means that the vapor pressure in the bottle will rise to a temperature insufficient to break the bottle, and that most of the alcohol will remain in liquid form.

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  1. You may have copied the question wrong because I had the same one for homework, only d=0.789g/L instead of g/mL. That error would account for your unusually high number

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    posted by Rachael
  2. xe kia k3

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    posted by Jens

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