◈ Termodinamica
Gas perfetti, trasformazioni termodinamiche, I e II principio, macchine termiche, cicli di Carnot e Otto, entropia.
Complete Theory
Worked Examples
Example 1Cycle A→B→C→A — Net work in a thermodynamic cycle
Example 2Ice → steam — The stages of heating
Exercises with Solutions
Exercise 1Adiabatic compression — The Diesel ignites by itselfHard
📋 Problem to solve
A diatomic gas (, mol, ) undergoes adiabatic compression from atm, K to atm. Find , and . Explain why the temperature rises so much.
📌 Given data
n=2 molγ=1.4C_v=5R/2P_1=1 atm, T_1=300 KP_2=8 atm
Exercise 2Entropy change in simple processesHard
📋 Problem to solve
Calculate for: (a) isobaric heating of mol diatomic gas () from K to K; (b) isothermal expansion at K with for the same gas.
📌 Given data
n=1 molC_p=7R/2T_1=300 K, T_2=600 KV_2/V_1=2
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Integrative Problems
Problems combining all chapters — exam levelProblem 1The Thermoelectric Power Plant: from Gas to Molecules to EntropyEXTREME
A thermoelectric power plant uses of a diatomic gas (, ) running through the following cycle on a PV diagram:
State A: , . A→B: adiabatic compression to (compression ratio ). B→C: isochoric, heating to (combustion). C→D: adiabatic expansion to (return to original volume). D→A: isochoric, cooling (Otto cycle).
State A: , . A→B: adiabatic compression to (compression ratio ). B→C: isochoric, heating to (combustion). C→D: adiabatic expansion to (return to original volume). D→A: isochoric, cooling (Otto cycle).
📌 Problem data
n = 5\,\mathrm{mol}\gamma = 7/5 = 1.4,\; C_v = 5R/2T_A = 300\,\mathrm{K},\; P_A = 1\,\mathrm{atm}r = V_A/V_B = 8T_C = 2400\,\mathrm{K}
(a)Ideal Gas — Thermodynamic States(b)First Law — Work and Heat for Each Process(c)Cycles — Otto vs Carnot Efficiency(d)Kinetic Theory — Molecules in Motion(e)Entropy — Second Law and Global Balance
