91Ó°ÊÓ

The thermal conductivity and other macroscopic properties measured for an ideal gas can be used to generate a rough estimate of Avogadro's number. The formula for thermal conductivity is:

$k_t=\frac{C_V}{2V}\mathrm{ℓ}\stackrel{¯}{v}$ Let be equation ($1$)

Assume we've configured the system so that we have a volume box $V$and cross-sectional area $A$. We know that the average speed $\stackrel{¯}{v}$is approximately the RMS speed, which is:

$\stackrel{¯}{v}≈v_{rms}=\sqrt{\frac{3kT}{m}}$

multiply with $\frac{\sqrt{N}}{\sqrt{N}}$, so:

Let the following be Equation ($2$)

localid="1650414641595" $$\begin{gathered} \stackrel{¯}{v}=\sqrt{\frac{3NkT}{Nm}} \\ =\sqrt{\frac{3NkT}{M}} \\ =\sqrt{\frac{3PV}{M}} \end{gathered}$$

where $ M$is the total mass of the gas. Substitute from ($2$) into ($1$), so:

localid="1650414665465" $$\begin{gathered} k_t=\frac{C_V}{2V}\mathrm{ℓ}\stackrel{¯}{v} \\ =\frac{C_V}{2V}\mathrm{ℓ}\sqrt{\frac{3PV}{M}} \\ =\frac{C_V}{2}\mathrm{ℓ}\sqrt{\frac{3P}{MV}} \end{gathered}$$

$\mathrm{â„“}=\frac{2}{C_V}{k}_t\sqrt{\frac{MV}{3P}}$

Schroeder's expression for $\mathrm{â„“}$is based on the idea that the mean path length is equal to the length of a cylinder of radius equal to the diameter and volume of the molecule equal to the average volume per molecule $\frac{V}{N}$, so that:

$\mathrm{â„“}=\frac{1}{4\mathrm{Ï€}{r}^2}\left(\frac{N}{V}\right)$

where $r$the radius of the molecule, To get $N$from this formula, we need to know $r$, but this is a microscopic quantity that we're assuming we don't know. I can't see any way of progressing from here unless we take a different value localid="1650296672007" $\mathrm{â„“}$. Since we're after only a rough approximation of Avogadro's number, we can take $\mathrm{â„“}$Instead of the average distance between collisions, it should be the average distance between molecules. That is to say:

Let be equation ($4$)

$\mathrm{ℓ}≈{\left[\frac{V}{N}\right]}^{\frac{1}{3}}$

We can now combine equations ($4$) and ($3$), so:

${\left[\frac{V}{N}\right]}^{\frac{1}{3}}=\frac{2}{C_V}{k}_t\sqrt{\frac{MV}{3P}}$

apply the power $3$on both sides:

$\frac{V}{N}≈{\left(\frac{2}{C_V}{k}_t\right)}^3{\left(\frac{MV}{3P}\right)}^{\frac{3}{2}}$

solve for $N$, so:

$\frac{N}{V}≈{\left(\frac{C_V}{2k_t}\right)}^3{\left(\frac{3P}{MV}\right)}^{\frac{3}{2}}→N≈{\left(\frac{C_V}{2k_t}\right)}^3{\left(\frac{3P}{MV}\right)}^{\frac{3}{2}}V$

$N≈{\left(\frac{C_V}{2k_t}\right)}^3{\left(\frac{3P}{M}\right)}^{\frac{3}{2}}\frac{1}{\sqrt{V}}$ let be equation ($5$)

Assuming we know the gas constant $R$, The ideal gas law can be used to calculate the number of moles:

$n=\frac{PV}{RT}$

And Avogadro number is given by:

$N_A=\frac{N}{n}=\frac{RT}{PV}N$

substitute from equation ($5$) so Avogadro's number is roughly:

$N_A≈\frac{RT}{PV}\left[{\left(\frac{C_V}{2k_t}\right)}^3{\left(\frac{3P}{M}\right)}^{\frac{3}{2}}\frac{1}{\sqrt{V}}\right]$

$N_A≈RT\sqrt{P}\left[{\left(\frac{C_V}{2k_t}\right)}^3{\left(\frac{3}{MV}\right)}^{\frac{3}{2}}\right]$