Thermodynamics and kinetic theory of gases
Thermodynamics and kinetic theory
This domain tests the ideal-gas law, kinetic theory, the first law, thermodynamic processes, heat engines and the Carnot limit. Expect problems that combine a P–V diagram with energy accounting, and conceptual questions about temperature and internal energy.
Core rules
- Ideal gas: PV=nRT=NkT. Temperature must be in kelvin. At constant T, P1V1=P2V2; at constant V, P/T is constant; at constant P, V/T is constant.
- Kinetic theory: average translational kinetic energy per molecule is (3/2)kT; rms speed is v_rms=√(3RT/M). For a monatomic ideal gas, U=(3/2)nRT.
- First law: ΔU=Q−W, where W is work done by the gas. W=∫P dV; on a P–V diagram it is the area under the curve.
- Processes: isochoric W=0; isobaric W=PΔV; isothermal ΔU=0; adiabatic Q=0. In an adiabatic expansion the gas cools.
- Heat engines: efficiency η=W/Q_in=1−Q_out/Q_in. Carnot efficiency is 1−T_c/T_h, the maximum possible.
- Second law: entropy of an isolated system does not decrease; heat flows spontaneously from hot to cold.
Worked example
Two moles of monatomic ideal gas are heated at constant volume from 300 K to 400 K. Find Q. Since V is constant, W=0, so Q=ΔU=(3/2)nRΔT=(3/2)(2)(8.31)(100)=2.49×10³ J. The pressure rises by a factor 400/300=1.33. If the same gas were heated at constant pressure, Q would be (5/2)nRΔT=4.16×10³ J, because some heat would become work.
Traps
Using Celsius in PV=nRT; using the wrong sign for work done on the gas; assuming an isothermal process has Q=0 when it has ΔU=0; treating internal energy as depending on pressure for an ideal gas; using Carnot efficiency for a real engine; forgetting that efficiency must be less than 1.