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According to Avogadro's law equal volume of all gases at constant temperature and pressure contain equal number of molecules. But how is this possible ??

Imagine helium and radon gas.

In this case is $m_\text{Rn} \gt m_\text{He}$. As pressure is directly proportional to density of gas, if the volume is constant or equal for both the gases then mass should also be same in order to have same or constant pressure.

So, how the number of molecules will be same if $m_\text{Rn} \gt m_\text{He}$, there should be $N_\text{Rn} \lt N_\text{He}$.

So, how they will have same number of molecules at equal volume, constant temp and pressure ?

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  • $\begingroup$ Think of a mixture of helium and radon, 50% of each atom type. Temperature is obviously constant. The partial pressure due to helium is the same as the partial pressure due to radon. The kinetic energy distribution of the two types of atoms is the same. Thus for a given kinetic energy the helium atoms are traveling faster than the radon atoms. $\endgroup$
    – MaxW
    Commented Nov 4, 2023 at 5:40
  • $\begingroup$ It is radon, not Raedon. Element names are not written capitalized. Proper text punctation is essential for text readability. $\endgroup$
    – Poutnik
    Commented Nov 4, 2023 at 8:23
  • $\begingroup$ Chem+Math Expression formatting reference: MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial // MathJax is preferred not to be used in CH SE Q titles. $\endgroup$
    – Poutnik
    Commented Nov 4, 2023 at 8:36

2 Answers 2

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The gas hydrostatic pressure is proportional to the mass of molecules. The kinetic pressure is not.

Evaluating an ideal gas at given temperature, volume and number of molecules, the mean translational kinetic energy of molecules is $E_\text{k}=\frac 32 k_\text{B} T$, where $k_\text{B} \approx \pu{1.38E-23 J K-1}$ is the Bolzmann constant

The mean quadratic speed of such molecules is then $v = \sqrt{\frac{2 E_\text{k}}{m}}=\sqrt{\frac{3k_\text{B}T}{m}}=\sqrt{\frac{3RT}{M}}$

and the mean impulse is

$$p=mv=\sqrt{3k_\text{B}mT}$$

Pressure due molecular kinetics is then:

$$p=\frac{\text{|d}\vec {F}|}{|\text{d}\vec {A}|}=\frac{\text{|d}\vec {p}|}{|\text{d}\vec {A}|\cdot \text{d}t}$$

As the pressure is proportional to both the speed and impulse:

$$p \propto |\vec {v}| \cdot |\vec {p}| = \sqrt{\frac{3k_\text{B}T}{m}} \cdot \sqrt{3k_\text{B}mT} = 3k_\text{B}T$$

Therefore, the pressure for given volume, temperature and number of molecules of an ideal gas does not depend on mass of a molecule.

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In the ideal gas or perfect gas model, the gas atoms/molecules are considered as point masses. Further the law is only valid when the intermolecular distances are large and only elastic collisions occur. Thus, when the intermolecular distances are large, the size of individual atoms, and corresponding masses do not affect the system, through some form of interactions.

Van der Waals equation, amongst others, accounts for the sizes of particles and intermolecular interactions.

In short, masses and sizes do matter but not in the ideal gas or perfect gas model and Avogadro law is valid for such systems.

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