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Size of a cube for a molecule of ideal gas

  1. Jan 15, 2015 #1
    1. The problem statement, all variables and given/known data
    The temperature of an ideal gas is 00C and the pressure is 1[atm]. imagine every molecule is enclosed in a cube, what's it's side length?

    2. Relevant equations
    PV=nRT
    Avogadro's number: 6.023E23

    3. The attempt at a solution
    I assume volume of i liter:
    $$1[atm]\cdot 1[liter]=n\cdot 0.08208\cdot 273\rightarrow n=0.0446[mole]$$
    Molecules per 1 liter:
    $$0.0446\cdot 6.023\cdot 10^{23}=2.687\cdot 10^{22}$$
    How many molecules are on one side?
    $$\sqrt[3]{2.687\cdot 10^{22}}=29951774$$
    The length of a side:
    $$\frac{10[cm]}{2995177}=3.34\cdot 10^{-7}[cm]$$
    The answer in the book: 3E-7[cm]
     
  2. jcsd
  3. Jan 15, 2015 #2

    Bystander

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    No specification of significant figures anywhere in the problem statement? Haven't checked your arithmetic in detail, but the set-up and execution looks great.
     
  4. Jan 15, 2015 #3
    Thanks, that's what i asked
     
  5. Jan 15, 2015 #4

    Quantum Defect

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    I always like to draw a picture of what the molecule in the box looks like when I teach this, to give students a sense of scale.

    The box is about 30 Angstrom on a side. How big is a typical molecule? How big is the typical box for a liquid?
     
  6. Jan 16, 2015 #5
    A bi atomic molecule is about 1[A] am i right? then the side is bigger 30 times more, but in the book it's written that's it's only 10 times larger.
    The volume of one mole of water is 18[cm3]. Molecular weight 18:
    $$\sqrt[3]{6.023\times 10^{23}}=84450901,\ \sqrt[3]{18}=2.62[cm]$$
    $$\frac{2.62}{84450901}=31\times10^{-9}[cm]=31\times 10^{-11}[m]=3.1\times 10^{-10}[m]=3.1[Angstram]$$
    Water molecule's size is about 1.5[A]
     
  7. Jan 16, 2015 #6

    BvU

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    Water molecule is a bit bigger (diameter 2.75 A here); O-H centers are .94 Angstrom apart.

    And here are a few other molecules. 3 to 4 A appears to be a good estimate
     
    Last edited: Jan 16, 2015
  8. Jan 16, 2015 #7
    Thanks
     
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