Thermodynamics

AP Physics 2 thermodynamics: laws of thermodynamics, heat transfer, ideal gas law, PV diagrams, heat engines, and entropy.

# Thermodynamics — AP Physics 2

Thermodynamics connects heat, work, and energy at the macroscopic level. AP Physics 2 covers the ideal gas law, the laws of thermodynamics, PV diagrams, and heat engines. Mastering these topics requires understanding both the mathematical relationships and the conceptual meaning of each process.

Key Concepts

Temperature and Thermal Energy

  • Temperature: measure of average kinetic energy of particles.
  • Average KE per molecule: KE=32kBT\overline{KE} = \frac{3}{2}k_B T (for ideal monatomic gas).
  • kB=1.38×1023 J/Kk_B = 1.38 \times 10^{-23}\ \text{J/K}.

Ideal Gas Law

PV=nRT=NkBTPV = nRT = Nk_B T where nn = moles, R=8.314 J/(mol\cdotpK)R = 8.314\ \text{J/(mol·K)}, NN = number of molecules.

First Law of Thermodynamics

ΔU=QW\Delta U = Q - W

  • ΔU\Delta U: change in internal energy
  • QQ: heat added to the system (positive = heat in)
  • WW: work done by the system (positive = expansion)

Thermodynamic Processes

Process Constant Consequence
Isothermal TT ΔU=0\Delta U = 0, Q=WQ = W
Isobaric PP W=PΔVW = P\Delta V
Isochoric (Isovolumetric) VV W=0W = 0, ΔU=Q\Delta U = Q
Adiabatic Q=0Q = 0 ΔU=W\Delta U = -W

PV Diagrams

  • Work done by gas = area under the curve on a PV diagram.
  • Clockwise cycle: net positive work (heat engine).
  • Counterclockwise cycle: net negative work (refrigerator).

Second Law of Thermodynamics

  • Heat flows spontaneously from hot to cold.
  • Entropy of an isolated system never decreases.
  • No heat engine can be 100% efficient.

Heat Engine Efficiency

e=WnetQH=1QCQHe = \frac{W_{\text{net}}}{Q_H} = 1 - \frac{Q_C}{Q_H}

Carnot (maximum) efficiency: eCarnot=1TCTHe_{\text{Carnot}} = 1 - \frac{T_C}{T_H}

Worked Example

Problem: An ideal gas in a sealed container (V=0.01 m3V = 0.01\ \text{m}^3) at 300 K300\ \text{K} and 1 atm1\ \text{atm} is heated to 600 K600\ \text{K} at constant volume. What is the final pressure?

Solution:

At constant volume: P1/T1=P2/T2P_1/T_1 = P_2/T_2 P2=P1T2T1=1 atm600300=2 atmP_2 = P_1 \cdot \frac{T_2}{T_1} = 1\ \text{atm} \cdot \frac{600}{300} = 2\ \text{atm}

Practice Questions

  1. 1. How much work does a gas do when it expands from 0.02 m30.02\ \text{m}^3 to 0.05 m30.05\ \text{m}^3 at a constant pressure of 200 kPa200\ \text{kPa}?

    W=PΔV=200,000×0.03=6000 JW = P\Delta V = 200{,}000 \times 0.03 = 6000\ \text{J}.

    2. A gas absorbs 500 J500\ \text{J} of heat and does 200 J200\ \text{J} of work. What is ΔU\Delta U?

    ΔU=QW=500200=300 J\Delta U = Q - W = 500 - 200 = 300\ \text{J}.

    3. A heat engine operates between 600 K600\ \text{K} and 300 K300\ \text{K}. What is its maximum efficiency?

    eCarnot=1300/600=0.5=50%e_{\text{Carnot}} = 1 - 300/600 = 0.5 = 50\%.

    4. During an adiabatic expansion, what happens to the temperature of an ideal gas?

    It decreases. Since Q=0Q = 0, the gas does work at the expense of internal energy, so ΔU<0\Delta U < 0 and temperature drops.

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Summary

  • The ideal gas law (PV=nRTPV = nRT) describes gas behavior.
  • First law: ΔU=QW\Delta U = Q - W — energy is conserved.
  • PV diagram area = work done by the gas.
  • Carnot efficiency sets the upper limit for heat engines.

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