posted by Chris on .
Methyl alcohol can be used as a fuel instead of, or combined with, gasoline. A sample of methyl alcohol, CH3OH, in a flask at constant volume exerts a pressure of 254 mm Hg at 57 degrees C. The flask is slowly cooled.
a) assuming no condensation, use the ideal gas law to calculate the pressure of the vapor at 35 degrees C.
Now I know using the first set of conditions I can find the moles and then do A that way, but I have a question first. When using the ideal gas law with the first set of conditions at constant volume, would I arbitrarily assign a volume to use in all calculations, or would I essentially leave it out of the ideal gas law?
Use an arbirtrary volume. Call it V, or if feel the urge, use 1 liter, that will but the moles on a per liter basis.
I have a follow-up question.
I calculated the number of moles (.01233) and plugged it into the ideal gas equation to find the pressure at 35 degrees C and 45 degrees C (.312 atm and .322 atm, consecutively). These were the pressures of the vapor.
The next questions state:
b) Compare your answers in a) with the equilibrium vapor pressures of methyl alcohol: 203 mmHg at 35 degrees C, and 325 mm Hg at 45 degrees C.
c) based on your answers from a) and b), predict the pressure exerted by the methyl alcohol in the flask at 35 degrees C, at 45 degrees C.
d) What physical states of methyl alcohol are present at both temperatures?
For b, I converted the mmHg pressures to atm (.2671 atm and .4276 atm consecutively), but I don't really get what the question is asking by "compare".
For c, I understand that I'm now supposed to find the pressure of the methyl alcohol, not the vapor as in a). But what is the difference in finding this? How does b) play a role in finding the pressure?
For d, is there an equation for this or should I just look it up?