Thermodynamics Domain Guide

Every ideal-gas question is really the question 'what is being held fixed?'. Isothermal fixes $T$, so $\Delta U = 0$ and $Q = W$. Adiabatic fixes $Q = 0$, so $W = -\Delta U$ and the gas cools as it expands. Isobaric fixes $P$, so $W = P\Delta V$. Isochoric fixes $V$, so $W = 0$ and $Q = \Delta U$. Four sentences, one of which is always the intended shortcut — and each of them kills a term in the first law before you write it.

The second law is a counting statement, not a mystical one. $S = k_B\ln\Omega$ says entropy measures how many microscopic arrangements produce the same macroscopic state, and processes run toward the overwhelmingly more numerous arrangements. That is why heat flows from hot to cold with no force pushing it, and why a Carnot engine — which passes heat through infinitesimal temperature differences — is the only kind that generates no entropy and therefore extracts the maximum work.

Temperature is not heat and neither is internal energy. Temperature measures the average energy per degree of freedom; internal energy is the total; heat is energy in transit because of a temperature difference. A large iceberg has more internal energy than a cup of boiling water and a lower temperature, and heat still flows from the cup to the iceberg. Keeping these three separate resolves most conceptual items in the domain without any calculation.

ETS lists thermodynamics and statistical mechanics as one content area, and our learning model groups the laws, the ideal gas, kinetic theory, engines, calorimetry, heat transfer, and thermal expansion beneath it. Start with the sign convention, because a large share of wrong answers here are sign errors rather than physics errors. In the form used above, $\Delta U = Q - W$, heat added to the gas is positive and work done by the gas is positive. Some texts write $\Delta U = Q + W$ with $W$ meaning work done on the gas; both are correct and they differ only in bookkeeping, so decide which one you use and never mix them mid-problem.

The ideal gas law, $PV = nRT = Nk_BT$, links the state variables, and the named processes fix one of them. Isothermal: $\Delta U = 0$, $W = nRT\ln(V_2/V_1)$, and $Q = W$. Isochoric: $W = 0$ and $Q = nC_V\Delta T$. Isobaric: $W = P\Delta V$ and $Q = nC_P\Delta T$, with $C_P = C_V + R$ — a relation that holds for any ideal gas because expanding at constant pressure requires the extra work $P\Delta V = nR\Delta T$. Adiabatic: $Q = 0$, $PV^{\gamma}$ and $TV^{\gamma-1}$ are constant, and $W = (P_1V_1 - P_2V_2)/(\gamma-1)$. Throughout, remember that $\Delta U = nC_V\Delta T$ is valid for every one of these, because $U$ depends only on $T$.

Kinetic theory supplies the microscopic picture and three speeds that are routinely confused. From $\tfrac{1}{2}m\langle v^{2}\rangle = \tfrac{3}{2}k_BT$ comes $v_{\text{rms}} = \sqrt{3k_BT/m} = \sqrt{3RT/M}$ with $M$ the molar mass in kilograms per mole. The mean speed is $\sqrt{8k_BT/\pi m}$ and the most probable speed is $\sqrt{2k_BT/m}$, so the ordering is always $v_{\text{mp}} < v_{\text{avg}} < v_{\text{rms}}$. Equipartition generalises this: each quadratic degree of freedom holds $\tfrac{1}{2}k_BT$ on average, which gives $C_V = \tfrac{3}{2}R$ for a monatomic gas, $\tfrac{5}{2}R$ for a rigid diatomic at ordinary temperatures, and the Dulong-Petit value $3R$ for a classical solid with three vibrational modes each contributing kinetic and potential terms.

Engines convert the first and second laws into a number. Over a complete cycle $\Delta U = 0$, so $W = Q_H - Q_C$ and the efficiency is $e = W/Q_H = 1 - Q_C/Q_H$. The Carnot bound, $e_{\text{Carnot}} = 1 - T_C/T_H$ in absolute temperature, is the ceiling for any engine between the same two reservoirs, and it is achieved only by a reversible cycle running infinitely slowly. Refrigerators and heat pumps are the same cycle read backwards, with $\mathrm{COP}_{\text{fridge}} = Q_C/W \le T_C/(T_H-T_C)$ and $\mathrm{COP}_{\text{pump}} = Q_H/W \le T_H/(T_H-T_C)$. A claimed efficiency above the Carnot value is not an approximation error; it is an impossibility, and on a multiple-choice paper it is a free elimination.

Entropy makes the second law quantitative. For a reservoir at fixed $T$ exchanging heat $Q$, $\Delta S = Q/T$. For an ideal gas between two states, $\Delta S = nC_V\ln(T_2/T_1) + nR\ln(V_2/V_1)$, which reduces to $nR\ln(V_2/V_1)$ for an isothermal change and to zero for a reversible adiabatic (hence 'isentropic'). Because $S$ is a state function, an irreversible process such as free expansion into a vacuum — where $Q = 0$ and $W = 0$ and therefore $\Delta T = 0$ for an ideal gas — still has $\Delta S = nR\ln(V_2/V_1) > 0$, computed along an imaginary reversible isothermal path. That single example contains the whole method: never compute $\Delta S$ along the actual irreversible path.

Calorimetry and transport close the domain. Heating without a phase change costs $Q = mc\Delta T$; a phase change costs $Q = mL$ at constant temperature, and forgetting the latent term is the classic mixing-problem error. Steady conduction through a slab gives $P = kA\Delta T/L$, so series slabs add thermal resistances $L/kA$ exactly as resistors add. Radiation follows the Stefan-Boltzmann law $P = \sigma\varepsilon AT^{4}$ with $\sigma = 5.67\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$, and net exchange with surroundings at $T_0$ is $\sigma\varepsilon A(T^{4}-T_0^{4})$ — always in kelvin, where the fourth power makes a small temperature change matter enormously. Thermal expansion is $\Delta L = \alpha L_0\Delta T$ with $\beta \approx 3\alpha$ for volume, and the counter-intuitive consequence worth remembering is that a hole in a heated plate gets larger, because every linear dimension scales together.