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Is there an explanation of the Avogadro's Law that uses the kinetic theory?

BiP

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- Thread starter Bipolarity
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- #1

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Is there an explanation of the Avogadro's Law that uses the kinetic theory?

BiP

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- #3

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I don't see how the analogy is relevant here.

BiP

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mfb

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What do you want to explain? Why some gases are nearly ideal?

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Philip Wood

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In the case of the pressure law, it's not just that the more frequent hits increase the pressure, but that, on average each hit imparts a greater impulse. That's why [itex]\overline{c^2}[/itex], with the 'squared', appears in the kinetic theory formula for pressure:

[tex]pV = \frac{1}{3} Nm\overline{c^2}[/tex].

Now let's look at Avogadro. We need the additional input that**molecules of all ideal gases at the same temperature have the same mean KE of translational motion**, [itex]\frac{1}{2}m \overline{c^2}[/itex]. [Jeans, in *The kinetic Theory of Gases* has a nice justification for this, using the fact that on average there must be no energy exchange in collisions between gas molecules and wall molecules if the gas is in equilibrium with its container walls.]

So for any two gases at the same temperature [tex]m_1\overline{c_1^2} = m_2\overline{c_2^2}[/tex].

So, using the pressure formula above:

[tex]\frac{p_1V_1}{N_1} = \frac{p_2V_2}{N_2}[/tex]

This formula applies for equal temperatures, but if we also impose the conditions that [itex]p_1=p_2[/itex] and [itex]V_1=V_2[/itex], then [itex]N_1=N_2[/itex].

So at the same temperature and pressure, equal volumes of gases contain the same number of molecules!

[You can reach the same conclusion using [itex]\frac{1}{2}m \overline{c^2}=\frac{3}{2}kT[/itex], but this isn't quite as economical because the argument above does not require a specific relationship between temperature and mean KE, merely a knowledge that if two gases have the same mean molecular KE, their temperatures are the same, and the converse.]

[tex]pV = \frac{1}{3} Nm\overline{c^2}[/tex].

Now let's look at Avogadro. We need the additional input that

So for any two gases at the same temperature [tex]m_1\overline{c_1^2} = m_2\overline{c_2^2}[/tex].

So, using the pressure formula above:

[tex]\frac{p_1V_1}{N_1} = \frac{p_2V_2}{N_2}[/tex]

This formula applies for equal temperatures, but if we also impose the conditions that [itex]p_1=p_2[/itex] and [itex]V_1=V_2[/itex], then [itex]N_1=N_2[/itex].

So at the same temperature and pressure, equal volumes of gases contain the same number of molecules!

[You can reach the same conclusion using [itex]\frac{1}{2}m \overline{c^2}=\frac{3}{2}kT[/itex], but this isn't quite as economical because the argument above does not require a specific relationship between temperature and mean KE, merely a knowledge that if two gases have the same mean molecular KE, their temperatures are the same, and the converse.]

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