HSC Physics 2 Initial Notes

Part III Physics — Second Paper 43

Chapter 15 Thermodynamics 15.1 Key Concepts & Theoretical Foundations Thermodynamics studies heat, work and internal energy in systems of many particles. A system is the region of interest; everything else is the surroundings. Internal energy & the first law Internal energy U is the total microscopic kinetic + potential energy of a system. The first law is energy conservation: ∆U = Q −W, where Q is heat added to the system and W is work done by the system. Four standard processes Isothermal (T const): ∆U = 0, so Q = W = nRT ln(V2/V1). Adiabatic (Q = 0): ∆U = −W and PV γ = const. Isobaric (P const): W = P ∆V . Isochoric (V const): W = 0, so Q = ∆U. For an ideal gas the molar heat capacities satisfy Mayer’s relation Cp −Cv = R, and γ = Cp/Cv. The second law forbids complete conversion of heat into work in a cycle. A heat engine takes Q1 from a hot reservoir, rejects Q2 to a cold one, and does W = Q1 −Q2 with efficiency η = W/Q1. The ideal Carnot engine has the maximum possible efficiency η = 1 −Tc/Th (temperatures in kelvin). Entropy S measures disorder; for a reversible exchange ∆S = Q/T, and total entropy never decreases. 15.2 Master Formula Sheet Master Formula Sheet ∆U = Q −W, W

HSC 2027 Science — Teacher’s Guide 45 15.3 Tables, Diagrams & Visual Layouts Process Constant Q W ∆U Isothermal T nRT ln V2 V1 = Q 0 Adiabatic Q = 0 0 −∆U nCv∆T Isobaric P nCp∆T P∆V nCv∆T Isochoric V nCv∆T 0 = Q V P Carnot cycle Th Tc Figure 1.1 — Carnot cycle: two isotherms + two adiabatics on a P–V diagram. 15.4 Worked Mathematical Problems Worked Problem 15.1 (First law) 500 J of heat is added to a gas which does 200 J of work. Find the change in internal energy. Solution. ∆U = Q −W = 500 −200 = 300 J. The gas warms even though it expands. ∆U = 300 J . Worked Problem 15.2 (Isothermal work) Two moles of an ideal gas expand isothermally at 300 K from 10 L to 20 L. Find the work done. (R = 8.31.) Solution. W = nRT ln V2 V1 = 2(8.31)(300) ln 2 = 4986(0.693) = 3455 J. Since T is constant, Q = W = 3455 J. W ≈3.46 kJ . Worked Problem 15.3 (Adiabatic relation) A gas (γ = 1.4) at 1 atm and 300 K is compressed adiabatically to half its volume. Find the final temperature. Solution. TV γ−1 = const ⇒T2 = T1 V1 V2 γ−1 = 300(2)0.4 = 300(1.320) = 396 K. T2 ≈396 K (heats on compression). Worked Problem 15.4 (Carnot efficiency) A Carnot engine operates between 500 K and 300 K. Find its efficien