Use the Ideal Gas Law to predict the approximate boiling/condensing temperature

Question

Use the Ideal Gas Law to predict the approximate boiling/condensing temperature of water at one atmosphere of pressure, proceeding from the assumption that we will start with 10^24 molecules of water vapor at room temperature and one atmosphere of pressure, and end up with the molecules as tightly packed together as they can be (in other words, liquefied), but the temperature is now so high that they want to automatically boil. Assume each molecule is a cube measuring 0.5 nm on each side and, for simplicity, the Boltzmann constant is 1 times 10^-23 J/K, one atmosphere is 1 times 10^5 Pa, and room temperature is 300 K. (Obviously I will know you did not think about this at all if you just write 373 K because this method doesn't produce very good results.)

Explanation / Answer

initial conditions:

pressure=1 atm=10^5 Pa

number of molecules=10^24 molecules

number of moles=10^24/(6.022*10^23)=1.6606

then volume=1.6606*22.4 ltr=37.1969 ltr=37.1969*10^(-3) m^3

temperature=300 K

as pressure remains constant,

volume/temperature=constant

final volume=volume of one cube*number of molecules

=(0.5*10^(-9))^3*10^24

=125*10^(-6) m^3

then if final temperature is T,

125*10^(-6)/300=37.1969*10^(-3)/T

==>T=37.1969*10^(-3)*300/(125*10^(-6))=8927.3 K

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