pV = \(\dfrac13 N m \langle c^{2}\rangle\).
pV = N k T (and the molar form pV = n R T) and deduce
\(\langle E_{\text{trans}}\rangle = \dfrac32 kT\) and \(U = \dfrac32 n R T\).
\(\Delta U = \dfrac32 n R \Delta T\) for a monatomic ideal gas.
\[
\Delta p = 2 m c_x .
\]
\[
\frac{c_x}{2L}.
\]
\[
F = \frac{\Delta p}{\Delta t}= \frac{2 m cx}{2L/cx}= \frac{m c_x^{2}}{L}.
\]
\[
p = \frac{1}{L^{3}}\,m\sum{i=1}^{N}c{x,i}^{2}
= \frac{N m\langle c_x^{2}\rangle}{V}.
\]
\[
\langle cx^{2}\rangle = \langle cy^{2}\rangle = \langle c_z^{2}\rangle,
\qquad
\langle c^{2}\rangle = \langle cx^{2}\rangle+\langle cy^{2}\rangle+\langle cz^{2}\rangle = 3\langle cx^{2}\rangle .
\]
\[
pV = \frac13 N m \langle c^{2}\rangle .
\]
Both expressions describe the same state of an ideal gas, so their right‑hand sides are equal:
\[
\frac13 N m \langle c^{2}\rangle \;=\; N k T .
\]
Cancel the common factor N and multiply both sides by \(\tfrac32\):
\[
\frac12 m \langle c^{2}\rangle \;=\; \frac32 k T .
\]
The left‑hand side is the average translational kinetic energy of a single molecule, therefore
\[
\boxed{\langle E_{\text{trans}}\rangle = \frac12 m \langle c^{2}\rangle = \frac32 k T } .
\]
For a mole of gas (\(N = n N{A}\) where \(N{A}=6.022\times10^{23}\,\text{mol}^{-1}\)) the total translational kinetic energy (the internal energy of a monatomic ideal gas) is
\[
U = N\langle E_{\text{trans}}\rangle = \frac32 N k T = \frac32 n R T ,
\]
and the change in internal energy on heating or cooling is
\[
\boxed{\Delta U = \frac32 n R \,\Delta T } .
\]
| Symbol | Quantity | Units |
|---|---|---|
| p | Pressure | Pa (N m⁻²) |
| V | Volume | m³ |
| N | Number of molecules | – |
| n | Number of moles | mol |
| N_{A} | Avogadro’s constant | 6.022 × 10²³ mol⁻¹ |
| m | Mass of one molecule | kg |
| c | Speed of a molecule | m s⁻¹ |
| \(\langle c^{2}\rangle\) | Mean‑square speed | (m s⁻¹)² |
| \(\langle E_{\text{trans}}\rangle\) | Average translational kinetic energy per molecule | J |
| k | Boltzmann constant = \(1.38\times10^{-23}\) J K⁻¹ | J K⁻¹ |
| R | Universal gas constant = \(N_{A}k = 8.31\) J mol⁻¹ K⁻¹ | J mol⁻¹ K⁻¹ |
| T | Absolute temperature | K |
\(\displaystyle \langle E_{\text{trans}}\rangle = \frac32 k T\).
\(\displaystyle c_{\text{rms}} = \sqrt{\langle c^{2}\rangle}= \sqrt{\frac{3kT}{m}}\).
\(\displaystyle U = \frac32 n R T\).
\(\displaystyle \Delta U = \frac32 n R \Delta T\) – this formula is used in the thermodynamics section (Section 16) when applying the first law.
The expression \(U = \tfrac32 nRT\) and its differential form \(\Delta U = \tfrac32 nR\Delta T\) are required in Section 16 – Thermodynamics for solving first‑law problems involving ideal gases.
Given: \(N = 2.0\times10^{23}\) helium atoms at \(T = 300\) K.
\[
\langle E_{\text{trans}}\rangle = \frac32 k T
= \frac32 (1.38\times10^{-23})(300)
= 6.21\times10^{-21}\,\text{J}.
\]
\[
U = N\langle E_{\text{trans}}\rangle
= (2.0\times10^{23})(6.21\times10^{-21})
= 1.24\times10^{3}\,\text{J}.
\]
Given: 1 mol O₂ at 298 K. Molar mass \(M = 32.0\) g mol⁻¹.
Molecular mass \(m = \dfrac{M}{N_{A}} = \dfrac{32.0\times10^{-3}}{6.022\times10^{23}} = 5.31\times10^{-26}\) kg.
\[
c_{\text{rms}} = \sqrt{\frac{3kT}{m}}
= \sqrt{\frac{3(1.38\times10^{-23})(298)}{5.31\times10^{-26}}}
= 4.8\times10^{2}\,\text{m s}^{-1}.
\]
Given: 2.0 mol of a monatomic gas is heated from 300 K to 350 K.
\[
\Delta U = \frac32 n R \Delta T
= \frac32 (2.0)(8.31)(350-300)
= 1.25\times10^{3}\,\text{J}.
\]
\(\displaystyle \langle E_{\text{trans}}\rangle = \frac32 k T\) and, per mole, \(U = \frac32 n R T\).
Your generous donation helps us continue providing free Cambridge IGCSE & A-Level resources, past papers, syllabus notes, revision questions, and high-quality online tutoring to students across Kenya.