Thermodynamics – Study Notes
Program: State Polytechnic Entrance
Subject: Chemistry
Topic: Thermodynamics
1. Simple Explanation
Thermodynamics is the branch of chemistry that deals with energy changes, heat, work, and the direction of chemical processes. It tells us what will happen (e.g., a reaction will occur or not) and how much energy will be involved.
| Concept | What It Means | Practical Example |
|---|---|---|
| System | The part of the universe we study. | A beaker of reacting chemicals. |
| Surroundings | Everything else. | The laboratory bench and air. |
| Energy | The ability to do work or transfer heat. | Heat from a burner. |
| Heat (q) | Energy in transit due to temperature difference. | Warm water cooling to room temp. |
| Work (w) | Energy transfer by physical means (pressure–volume). | A piston compressing a gas. |
2. Key Points (Core Principles)
2.1 Laws of Thermodynamics
| Law | Symbolic Statement | What It Tells Us |
|---|---|---|
| 0th Law | If A ≈ B and B ≈ C → A ≈ C | Temperature is a state property. |
| 1st Law | ΔU = q + w | Energy conservation: change in internal energy = heat added + work done. |
| 2nd Law | ΔS_univ ≥ 0 | Spontaneous processes increase total entropy. |
| 3rd Law | S → 0 as T → 0 K for perfect crystal | Absolute entropy at 0 K is zero. |
2.2 State Functions
- Internal Energy (U)
- Enthalpy (H = U + PV)
- Entropy (S)
- Gibbs Free Energy (G = H – TS)
Only state functions depend on the state, not on the path taken.
2.3 Heat Capacities
| Quantity | Symbol | Definition | Units |
|---|---|---|---|
| Molar Heat Capacity (C<sub>m</sub>) | C<sub>m</sub> | Heat needed to raise 1 mol by 1 K | J mol⁻¹ K⁻¹ |
| Constant‑Pressure Heat Capacity (C<sub>p</sub>) | C<sub>p</sub> | Heat at constant pressure | J mol⁻¹ K⁻¹ |
| Constant‑Volume Heat Capacity (C<sub>v</sub>) | C<sub>v</sub> | Heat at constant volume | J mol⁻¹ K⁻¹ |
For ideal gases:
- (C_p = C_v + R)
- (R = 8.314,\text{J mol}^{-1}\text{K}^{-1})
2.4 Calorimetry (Measuring Heat)
- Constant‑pressure calorimeter (coffee‑cup): (q_p = C_p \Delta T).
- Constant‑volume calorimeter (bomb): (q_v = C_v \Delta T).
2.5 Phase Transitions
| Transition | Symbol | Energy Change |
|---|---|---|
| Fusion | (L_f) | Heat of fusion |
| Vaporisation | (L_v) | Heat of vaporisation |
| Sublimation | (L_s) | Heat of sublimation |
During phase change, temperature remains constant while energy changes phase.
2.6 Gibbs Free Energy and Spontaneity
- ΔG = ΔH – TΔS
- ΔG < 0 → spontaneous at constant T, P.
- ΔG = 0 → equilibrium.
- ΔG > 0 → non‑spontaneous.
3. Worked Example
Problem:
A 0.500 L beaker contains 0.200 mol of water at 25 °C. We heat it to 75 °C in a constant‑pressure calorimeter with a known heat capacity of 2.50 kJ K⁻¹. Calculate the heat supplied and the change in enthalpy of the water.
Step‑by‑Step
-
Find ΔT
[ \Delta T = 75^\circ\text{C} - 25^\circ\text{C} = 50,\text{K} ] -
Heat supplied (q<sub>p</sub>)
[ q_p = C_{\text{cal}} \Delta T = (2.50,\text{kJ K}^{-1})(50,\text{K}) = 125,\text{kJ} ] -
Heat absorbed by water
The calorimeter is insulated, so all heat goes to the water.
[ q_{\text{water}} = +125,\text{kJ} ] (positive because heat is absorbed by the system). -
Enthalpy change of water
For a liquid at constant pressure,
[ \Delta H = q_p ] Therefore,
[ \Delta H_{\text{water}} = +125,\text{kJ} ] -
Verification using molar heat capacity
[ \Delta H = n C_{p,\text{water}} \Delta T ] Water’s (C_{p} = 4.18,\text{J g}^{-1}\text{K}^{-1}) (≈ 75.4 J mol⁻¹ K⁻¹).
[ \Delta H = 0.200 \times 75.4 \times 50 \approx 754,\text{J} \approx 0.754,\text{kJ} ] The discrepancy arises because the calorimeter’s heat capacity dominates; the water’s own heat capacity contributes only ~0.754 kJ.
Answer:
- Heat supplied: 125 kJ
- Enthalpy change of