nitrogen

logistic_guy

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What is the average distance between nitrogen molecules at STP\displaystyle \text{STP}?
 
We start with the ideal gas law.

PV=NkT\displaystyle PV = NkT

We will rearrange the equation so that we get moleculesvolume\displaystyle \frac{\text{molecules}}{\text{volume}}.

NV=PkT\displaystyle \frac{N}{V} = \frac{P}{kT}

At STP\displaystyle \text{STP}, the pressure is 101325 Pa\displaystyle 101325 \ \text{Pa} and the temperature is 273.15 K\displaystyle 273.15 \ \text{K}, and k\displaystyle k is just the Boltzmann\displaystyle \text{Boltzmann} constant.

Then,

NV=101325 Pa1.38×1023 J/K(273.15 K)=101325 Pa1.38×1023 J(273.15)\displaystyle \frac{N}{V} = \frac{101325 \ \text{Pa}}{1.38 \times 10^{-23} \ \text{J/K}(273.15 \ \text{K})} = \frac{101325 \ \text{Pa}}{1.38 \times 10^{-23} \ \text{J}(273.15)}

Our goal at this point is to write the units in terms of m3,cm3,or mm3\displaystyle \text{m}^3, \text{cm}^3, \text{or} \ \text{mm}^3.

PaN/m2\displaystyle \text{Pa} \rightarrow \text{N/}\text{m}^2
JNm\displaystyle \text{J} \rightarrow \text{Nm}

This gives:

NV=1013251.38×1023 m3(273.15)=2.69×1025 molecules/m3\displaystyle \frac{N}{V} = \frac{101325}{1.38 \times 10^{-23} \ \text{m}^3(273.15)} = 2.69 \times 10^{25} \ \text{molecules/}\text{m}^3

Let us assume that the volume of one molecule is a3\displaystyle a^3 and it is a cube. Then, the volume of all molecules is:

V=Na3\displaystyle V = Na^3

Then, the average distance is the length of one side of one cube. That is:

a=VN3=1NV3=12.69×10253=3.34×109 m=3.34×107 cm\displaystyle a = \sqrt[3]{\frac{V}{N}} = \sqrt[3]{\frac{1}{\frac{N}{V}}} = \sqrt[3]{\frac{1}{2.69 \times 10^{25}}} = 3.34 \times 10^{-9} \ \text{m} = \textcolor{blue}{3.34 \times 10^{-7} \ \text{cm}}
 
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