Explain why internal energy \(U\) is a state function and show that it can be expressed as the sum of the microscopic kinetic and potential energies of the molecules that constitute the system.
For a system containing \(N\) molecules
\[
U = \sum{i=1}^{N}\Bigl(Ki^{\text{trans}}+Ki^{\text{rot}}+Ki^{\text{vib}}+V_i\Bigr)
\]
For an ideal gas the potential term is taken as zero; for real gases, liquids and solids it can be large, especially during phase changes.
| Gas / Phase | Degrees of Freedom (DoF) | Kinetic‑energy contribution to \(U\) (per mole) | Potential‑energy contribution to \(U\) |
|---|---|---|---|
| Monatomic ideal gas | 3 translational (no rotation, no vibration) | \(\displaystyle U_{\text{kin}} = \frac{3}{2}RT\) | Negligible (ideal‑gas assumption) |
| Linear diatomic ideal gas | 3 translational + 2 rotational = 5 DoF Vibrational modes frozen when \(T \ll \Theta_{\text{vib}}\) | \(\displaystyle U{\text{kin}} = \frac{5}{2}RT\) (vibration frozen) When one vibrational mode is active: \(\displaystyle U{\text{kin}} = \frac{5}{2}RT + \frac{1}{2}RT = 3RT\) | Zero when vibration frozen; each active vibrational mode adds \(\tfrac{1}{2}RT\) as potential energy. |
| Non‑linear polyatomic ideal gas | 3 translational + 3 rotational = 6 DoF \(f = 3N_{\text{atoms}}-6\) vibrational modes | \(\displaystyle U{\text{kin}} = 3RT\) (vibrations frozen) With \(f\) active vibrational modes: \(\displaystyle U{\text{kin}} = 3RT + \frac{f}{2}RT\) | Zero for frozen vibrations; each active mode contributes an additional \(\tfrac{1}{2}RT\) as potential energy. |
| Real gas / liquid / solid | All translational, rotational and vibrational DoF are present; strong intermolecular forces. | Depends on temperature; vibrational modes become progressively excited as \(T\) rises. | Significant – accounts for cohesion, latent heat of melting/boiling, and the large heat capacities of condensed phases. |
\[
U = \frac{f}{2}\,nRT
\]
where \(f\) is the total number of active kinetic degrees of freedom (translational + rotational) and \(n\) is the amount of substance in moles.
\[
\Delta U = CV\,\Delta T\qquad\text{with}\qquad CV = \left(\frac{\partial U}{\partial T}\right)_V = \frac{f}{2}nR.
\]
\(f = 3\) → \(C_V = \tfrac{3}{2}nR = 3R\).
\(\Delta U = C_V\Delta T = 3R(50\;\text{K}) = 3(8.314)(50) \approx 1.25\times10^3\;\text{J}.\)
The Cambridge syllabus uses the convention
\[
\Delta U = q + w
\]
Note: Some textbooks write \(\Delta U = Q - W\) (where \(W\) is work done by the system). Both are equivalent; the syllabus adopts the \(q + w\) form.
Two different processes that start and finish at the same temperature and volume give the same \(\Delta U\):
Both routes lead to the same change in internal energy, confirming that \(U\) is a state function.
\[
U = nC_VT = \frac{f}{2}nRT.
\]
Example: 1 mol of a linear diatomic gas at 300 K (vibration frozen) → \(U = \tfrac{5}{2}R(300) \approx 3.11\times10^3\;\text{J}\).
\[
U \approx 3nRT,\qquad C_V \approx 3nR.
\]
\[
CV = \left(\frac{\partial U}{\partial T}\right)V,\qquad Cp = \left(\frac{\partial H}{\partial T}\right)p.
\]
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