Thermodynamics – Study Notes
(For ITI Admission Test – Chemistry)
1. Simple Explanation
Thermodynamics is the branch of chemistry (and physics) that deals with energy, heat, and how energy is transferred and transformed in chemical and physical processes.
Key idea: Energy can’t be created or destroyed; it can only change form – this is the First Law. The Second Law tells us that natural processes increase the overall disorder (entropy) of the universe.
2. Key Points
| Concept | What it Means | Important Formula/Statement |
|---|---|---|
| First Law of Thermodynamics | Energy balance in a closed system. | ΔU = Q – W (ΔU = change in internal energy, Q = heat added to system, W = work done by system) |
| Second Law of Thermodynamics | Entropy of an isolated system never decreases. | ΔS_universe = ΔS_system + ΔS_surroundings ≥ 0 |
| Third Law (optional) | Entropy of a perfect crystal at 0 K is zero. | N/A for exam level |
| Thermodynamic Processes | Types of changes in a system. | <ul><li>Isothermal (T constant)</li><li>Adiabatic (Q = 0)</li><li>Isobaric (P constant)</li><li>Isochoric (V constant)</li></ul> |
| Work in Different Processes | Work depends on type of process. | W = –∫P dV (general); For isothermal ideal gas: W = –nRT ln(V₂/V₁) |
| Enthalpy (H) | Useful for chemistry because it is constant at constant pressure. | H = U + PV → ΔH = ΔU + Δ(PV) |
| Entropy (S) | Measure of disorder or randomness. | S = k ln Ω (statistical definition); ΔS = Q_rev / T (thermodynamic definition) |
| Gibbs Free Energy (G) | Predicts spontaneity at constant T & P. | G = H – TS → ΔG = ΔH – TΔS |
| Spontaneity Criterion | Process is spontaneous if ΔG < 0. |
|
| Heat Capacity (C) | Heat needed to change temperature by 1 K. | C = dQ/dT (at constant P or V) |
| Carnot Efficiency | Maximum possible efficiency of a heat engine. | η = 1 – T_cold / T_hot |
3. Worked Example
Problem
Calculate the change in enthalpy (ΔH) for the combustion of methane:
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
Given the standard enthalpies of formation (ΔH_f°) at 25 °C:
| Species | ΔH_f° (kJ mol⁻¹) |
|---|---|
| CH4(g) | –74.8 |
| O2(g) | 0 |
| CO2(g) | –393.5 |
| H2O(g) | –241.8 |
Solution
-
Write ΔH° for the reaction
[ \Delta H^\circ = \sum \nu_i \Delta H_f^\circ(\text{products}) - \sum \nu_j \Delta H_f^\circ(\text{reactants}) ] -
Insert stoichiometric coefficients (ν)
[ \Delta H^\circ = \big[1(-393.5) + 2(-241.8)\big] - \big[1(-74.8) + 2(0)\big] ]
-
Calculate
[ \Delta H^\circ = \big[-393.5 - 483.6\big] - \big[-74.8\big] = (-877.1) + 74.8 = -802.3 \text{ kJ} ]
Answer: The combustion of methane releases 802.3 kJ mol⁻¹ of heat (exothermic).
4. Quick Revision Summary
- First Law: Energy conservation →
ΔU = Q – W. - Second Law: Entropy of an isolated system increases →
ΔS_universe ≥ 0. - Enthalpy (H):
ΔH = ΔU + Δ(PV); constant‑pressure heat. - Entropy (S):
ΔS = Q_rev / T; measure of disorder. - Gibbs Free Energy (G):
ΔG = ΔH – TΔS;ΔG < 0⇒ spontaneous at constant T & P. - Work in Processes
- Isothermal ideal gas:
W = –nRT ln(V₂/V₁) - Adiabatic:
Q = 0→W = –ΔU - Isochoric:
W = 0→Q = ΔU
- Isothermal ideal gas:
- Heat Capacity:
C = dQ/dT. - Carnot Cycle Efficiency:
η = 1 – T_cold/T_hot. - Spontaneity Check:
ΔG < 0→ spontaneous; `