\(\langle cx^{2}\rangle =\langle cy^{2}\rangle =\langle c_z^{2}\rangle\).
Consider a cubic container of side L (so \(V=L^{3}\)). A single molecule of mass m approaches the wall perpendicular to the x‑axis with velocity component \(c_{x}>0\).
\[
\Delta p{x}=(-m c{x})-(+m c{x})=-2m c{x}.
\]
The magnitude of the impulse is \(|\Delta p{x}|=2m|c{x}|\).
\[
\Delta t=\frac{2L}{|c_{x}|}.
\]
\[
F{x}= \frac{|\Delta p{x}|}{\Delta t}= \frac{2m c{x}^{2}}{2L}= \frac{m c{x}^{2}}{L}.
\]
\[
p{x}= \frac{F{x}}{A}= \frac{m c{x}^{2}}{L^{3}}= \frac{m c{x}^{2}}{V}.
\]
Summing the contributions of every molecule in the gas gives
\[
p = \frac{m}{V}\sum{i=1}^{N}c{x,i}^{2}
= \frac{Nm}{V}\,\langle c_{x}^{2}\rangle,
\]
where
\[
\langle c{x}^{2}\rangle =\frac{1}{N}\sum{i=1}^{N}c_{x,i}^{2}
\]
is the mean‑square value of the x‑component of velocity.
(The sum runs over all molecules, not just a representative subset.)
Because the velocity distribution is isotropic,
\[
\langle c{x}^{2}\rangle =\langle c{y}^{2}\rangle =\langle c_{z}^{2}\rangle .
\]
The total mean‑square speed is the sum of the three orthogonal components:
\[
\langle c^{2}\rangle = \langle c{x}^{2}\rangle+\langle c{y}^{2}\rangle+\langle c_{z}^{2}\rangle
= 3\langle c_{x}^{2}\rangle,
\qquad\text{so}\qquad
\langle c_{x}^{2}\rangle = \frac{\langle c^{2}\rangle}{3}.
\]
Substituting into the pressure expression yields the fundamental kinetic‑theory relation
\[
\boxed{\,pV = \frac{1}{3}\,Nm\langle c^{2}\rangle\,}.
\]
\[
N = nN_{\!A},
\]
where \(n\) is the number of moles and \(N_{\!A}=6.022\times10^{23}\,\text{mol}^{-1}\).
\[
pV = \frac{1}{3}\,nM\langle c^{2}\rangle .
\]
\[
pV = nRT,
\]
with \(R = 8.314\;\text{J mol}^{-1}\text{K}^{-1}\).
Using the Boltzmann constant \(k{\!B}=R/N{\!A}=1.38\times10^{-23}\;\text{J K}^{-1}\) we can write
\[
pV = nN{\!A}k{\!B}T .
\]
\[
\frac{1}{3}\,nM\langle c^{2}\rangle = nN{\!A}k{\!B}T
\;\;\Longrightarrow\;\;
\frac{1}{2}m\langle c^{2}\rangle = \frac{3}{2}k_{\!B}T .
\]
This is the average translational kinetic energy per molecule. Multiplying by \(N_{\!A}\) (or by \(n\)) yields the per‑mole form
\[
\frac{1}{2}M\langle c^{2}\rangle = \frac{3}{2}RT .
\]
Thus the kinetic‑theory expression provides the bridge between microscopic motion, temperature, and the macroscopic ideal‑gas law.
\[
U = \frac{3}{2}nRT .
\]
\[
C{V,m}= \left(\frac{\partial U}{\partial T}\right){V}= \frac{3}{2}R .
\]
\[
C_{V,m}= \frac{5}{2}R ,\qquad
C{p,m}= C{V,m}+R = \frac{7}{2}R .
\]
These results follow directly from the kinetic‑theory expression \(\tfrac12 m\langle c^{2}\rangle = \tfrac32 k_{\!B}T\) together with the equipartition theorem (a point often examined in the exam).
| ❌ “Only molecules moving *directly* towards a wall create pressure.” | ✅ All molecules contribute; only the component of their velocity normal to the wall matters, and the isotropic distribution ensures the same average contribution from every molecule. |
| ❌ “Pressure comes from the *weight* of the gas.” | ✅ Pressure is the rate of momentum transfer to the walls, independent of gravity. |
| ❌ “Internal collisions change the net force on the walls.” | ✅ Internal collisions merely redistribute momentum among molecules; the total momentum transferred to the walls per unit time remains unchanged. |
Given:
\(N = 2.5\times10^{23}\) molecules of O\(_2\) (molar mass \(M = 32\;\text{g mol}^{-1}\) → \(m = 5.31\times10^{-26}\;\text{kg}\)),
\(\langle c^{2}\rangle = 4.5\times10^{4}\;\text{m}^{2}\text{s}^{-2}\),
\(V = 2.0\times10^{-2}\;\text{m}^{3}\).
Solution:
\[
p = \frac{Nm\langle c^{2}\rangle}{3V}
= \frac{(2.5\times10^{23})(5.31\times10^{-26})(4.5\times10^{4})}{3(2.0\times10^{-2})}
\approx 1.0\times10^{4}\;\text{Pa}.
\]
Result: The gas exerts a pressure of about 10 kPa.
Given:
A sample of helium contains \(n = 0.250\;\text{mol}\) and occupies \(V = 5.00\times10^{-3}\;\text{m}^{3}\).
The pressure measured is \(p = 2.00\times10^{5}\;\text{Pa}\).
Helium is monatomic, so \(m = M/N_{\!A}\) with \(M = 4.00\times10^{-3}\;\text{kg mol}^{-1}\).
Find: the absolute temperature \(T\) and the mean‑square speed \(\langle c^{2}\rangle\).
Solution:
\[
T = \frac{pV}{nR}
= \frac{(2.00\times10^{5})(5.00\times10^{-3})}{0.250\times8.314}
\approx 4.82\times10^{2}\;\text{K}.
\]
\[
\langle c^{2}\rangle = \frac{3k_{\!B}T}{m}
= \frac{3(1.38\times10^{-23})(4.82\times10^{2})}{4.00\times10^{-3}/6.022\times10^{23}}
\approx 7.9\times10^{5}\;\text{m}^{2}\text{s}^{-2}.
\]
Result: \(T \approx 4.8\times10^{2}\;\text{K}\) and \(\langle c^{2}\rangle \approx 7.9\times10^{5}\;\text{m}^{2}\text{s}^{-2}\).
| Symbol | Quantity | Units | Notes |
|---|---|---|---|
| p | Pressure | Pa (N m⁻²) | Force per unit area on the container walls |
| V | Volume | m³ | Space occupied by the gas |
| N | Number of molecules | – | Exact count of particles |
| n | Amount of substance | mol | Number of moles ( \(N=nN_{\!A}\) ) |
| m | Mass of one molecule | kg | \(m = M/N_{\!A}\) |
| M | Molar mass | kg mol⁻¹ | Mass of one mole of the gas |
| \(\langle c^{2}\rangle\) | Mean‑square speed | m² s⁻² | Average of \(c^{2}\) over all molecules |
| \(k{\!B}\) | Boltzmann constant | J K⁻¹ | \(k{\!B}=R/N_{\!A}=1.38\times10^{-23}\) |
| R | Universal gas constant | J mol⁻¹ K⁻¹ | 8.314 |
| \(C{V,m},\,C{p,m}\) | Molar heat capacities | J mol⁻¹ K⁻¹ | Derived from degrees of freedom |
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