Thermodynamics – FMGE Study Notes
(Chemistry – 2nd Year MBBS)
1. Simple Explanation
Thermodynamics is the branch of chemistry that deals with heat, work, energy, and the direction of chemical processes.
- Heat (q) is energy in transit due to temperature difference.
- Work (w) is energy transferred by a force acting through a distance (e.g., pressure × volume).
- Internal Energy (U) is the total microscopic energy of a system.
- Enthalpy (H) = U + PV – convenient for constant‑pressure processes.
- Entropy (S) measures disorder; the second law says that for an isolated system, ΔS ≥ 0.
- Gibbs Free Energy (G) = H – TS – tells whether a reaction is spontaneous (ΔG < 0).
2. Key Points
| Concept | Definition | Important Equations | Typical FMGE Question |
|---|---|---|---|
| First Law | Energy conservation: ΔU = q + w | ΔU = q + w (sign convention: q, w positive when energy flows into system) | “Calculate ΔU for a gas expanding against a constant pressure.” |
| Enthalpy | Heat content at constant pressure | ΔH = q_p | “Find ΔH of a reaction given ΔU and Δ(PV).” |
| Heat Capacity | Energy needed to raise temperature | C = q/ΔT (molar: C_m = q/(nΔT)) | “Calculate heat required to warm 5 g of water from 25°C to 75°C.” |
| Entropy | Measure of randomness | ΔS = q_rev/T (for reversible processes) | “Compute ΔS for melting of ice at 0 °C.” |
| Second Law | ΔS_total ≥ 0 for spontaneous processes | ΔS_total = ΔS_system + ΔS_surroundings | “Is the dissolution of salt in water spontaneous? Explain.” |
| Gibbs Free Energy | ΔG = ΔH – TΔS | ΔG < 0 → spontaneous; ΔG = 0 → equilibrium | “Predict if a reaction is spontaneous at 298 K given ΔH and ΔS.” |
| Kinetic vs Thermodynamic Control | Kinetic: reaction rate; Thermodynamic: ΔG | – | “Which product is favoured at high temperature?” |
| Phase Diagrams | Show stable phases vs P–T conditions | Triple point, critical point | “Explain the meaning of the triple point on a P–T diagram.” |
| Le Chatelier’s Principle | System shifts to counteract changes | – | “Effect of increasing temperature on an exothermic reaction.” |
Quick‑Reference Formulas
| Quantity | Formula | Units | Notes |
|---|---|---|---|
| Internal Energy | ΔU = q + w | J | q in, w in |
| Enthalpy | ΔH = ΔU + Δ(PV) = q_p | J | q_p is heat at constant pressure |
| Heat Capacity (molar) | C_m = q/(nΔT) | J mol⁻¹ K⁻¹ | n = moles |
| Entropy change (reversible) | ΔS = q_rev/T | J K⁻¹ | T in K |
| Gibbs Energy | ΔG = ΔH – TΔS | J | Use ΔG = -RT ln K |
| Equilibrium constant | K = exp(-ΔG°/RT) | – | ΔG° at 298 K |
3. Worked Example
Problem (FMGE style):
A 10 g sample of a pure solid is heated from 25 °C to 150 °C. The molar heat capacity of the solid is 75 J mol⁻¹ K⁻¹. Calculate the heat required (q) and the change in entropy (ΔS) of the solid. (Molar mass of the solid = 100 g mol⁻¹)
Solution
-
Calculate moles (n):
[ n = \frac{m}{M} = \frac{10\ \text{g}}{100\ \text{g mol}^{-1}} = 0.10\ \text{mol} ] -
Heat required (q):
[ q = n,C_m,\Delta T = 0.10,\text{mol} \times 75,\frac{\text{J}}{\text{mol K}} \times (150-25),\text{K} ] [ q = 0.10 \times 75 \times 125 = 937.5\ \text{J} ] (Sign convention: heat absorbed by the solid → q = +937.5 J) -
Entropy change (ΔS):
For a constant‑heat‑capacity process, [ \Delta S = n,C_m,\ln!\left(\frac{T_2}{T_1}\right) ] Convert temperatures to Kelvin: [ T_1 = 25 + 273.15 = 298.15,\text{K} \ T_2 = 150 + 273.15 = 423.15,\text{K} ] [ \Delta S = 0.10 \times 75 \times \ln!\left(\frac{423.15}{298.15}\right) ] [ \Delta S = 7.5 \times \ln(1.419) \approx 7.5 \times 0.350 = 2.63\ \text{J K}^{-1} ]
Answer:
- Heat required: +937.5 J (absorbed by the solid).
- Entropy change: +2.63 J K⁻¹ (increase in disorder).
4. Quick Revision Summary
| Topic | What to Remember |
|---|---|
| **First Law |