Question bank — Adiabatic relations — PV^γ = const, TV^(γ−1) = const (derivation)

False — adiabatic only means no heat transfer ($\delta Q=0$); a fast expansion into vacuum is also adiabatic. But the relations $P{V}^{\gamma }=$const need it to be reversible/quasi-static so $P,V,T$ are defined throughout.

False — that describes an Isothermal Process. In adiabatic, since no heat refills the internal energy, doing work forces $T$ to change: expansion cools, compression heats.

False — $\gamma >1$, so the adiabat's slope $-\gamma P/V$ is steeper than the isotherm's $-P/V$ at every point; they only touch, never coincide (see figure s02).

False — for an ideal gas $U=U(T)$ alone, so this holds for any process. The "$V$" subscript records how $C_V$ was measured, not when the formula applies.

False — the first law is $0=dU+\delta W$, so $dU=-\delta W$. When the gas does work, its internal energy drops even with $Q=0$.

False — the derivation used the ideal gas law ($PV=nRT$) and $dU=n{C}_{V}dT$. For real gases or liquids the exponent and even the form change.

False — $\gamma$ depends on the active degrees of freedom, which change with $T$. Cold diatomic gas uses only translation+rotation ($\gamma =7/5$); heat it enough to activate vibration and $C_V$ rises, so $\gamma$ drops toward $9/7$ (see figure s03).

True — a reversible adiabatic process has $dS=\delta Q_{\text{rev}}/T=0$ (no reversible heat enters), so $S$ is unchanged, i.e. it is isentropic; that's the deeper reason the curve is a single well-defined line.

False — it loses internal energy to work done on the surroundings, not to heat. $Q=0$ throughout; the energy leaves as mechanical work, not as heat.

Wrong exponent — Step 5 integrates $dT/T+(\gamma -1)dV/V=0$ to $\ln T+(\gamma -1)\ln V=$const, giving $T{V}^{\gamma -1}=$const. Only $P$ carries the full $\gamma$; $T$ gets $\gamma$ minus one because swapping $P\leftrightarrow T$ via $PV=nRT$ shifts one power of $V$.

Wrong substitution — Step 2 uses the ideal gas law $PV=nRT\Rightarrow P=nRT/V$, not $n{C}_{V}T/V$. Using $C_V$ here would break dimensions and never produce $\gamma -1$.

Double error at Step 4: $R/C_V=(C_P-C_V)/C_V=\gamma -1$ by Mayer's Relation, not $\gamma$. This is exactly why the $TV$ exponent is $\gamma -1$.

Reversed — differentiating each (parent's forecast section) gives adiabat slope $-\gamma P/V$ and isotherm slope $-P/V$; since $\gamma >1$ the adiabat is steeper. In adiabatic expansion $P$ falls faster because $T$ also drops (figure s02).

Contradiction with the defining condition $\delta Q=0$ — it absorbs no heat. The work is paid entirely out of internal energy: $W=-\Delta U$.

Sign flipped — from $W=-\Delta U=-n{C}_V(T_2-T_1)=n{C}_V(T_1-T_2)$. For expansion $T_2<T_1$ giving $W>0$, as it should be for a gas that pushes outward.

No — monatomic ($5/3$) adiabats are steeper than diatomic ($7/5$) since a larger $\gamma$ means a larger slope factor. See Degrees of Freedom and $\gamma$.

Because $\gamma =C_P/C_V$, and $C_V$ counts the degrees of freedom each molecule can store energy in — so the curve's steepness literally reads out the molecule's internal structure.

Because the equation links inexact-looking rates of change, not values; it is a differential relation valid at every instant. Integrating each exact differential $dx/x$ gives $\ln x$, and the sum of two logs equalling a constant of integration means their arguments multiply to a fixed number — that is how $T{V}^{\gamma -1}=$const is born. The constant itself is fixed by the gas's starting state $(T_0,V_0)$, so different initial conditions give different adiabats, one per constant.

Fast compression is nearly adiabatic (no time for heat to leak, so $\delta Q\approx 0$), so the work you do goes entirely into internal energy via $dU=-\delta W$, raising $T$ — exactly $P{V}^{\gamma }=$const at play.

Compressions and rarefactions oscillate too fast for heat to diffuse between them, so each parcel obeys $\delta Q=0$; this is why the Speed of Sound in Gases uses the adiabatic bulk modulus $\gamma P$, not $P$.

In adiabatic expansion the gas also cools, so its pressure drops for two reasons (more volume and less temperature), whereas the isotherm keeps $T$ propped up by incoming heat (figure s02).

Because $U$ is a state function of $T$ alone for an ideal gas, so its exact differential $dU$ is fixed by $dT$ regardless of what $V,P$ do simultaneously — the path does not matter for $U$.

Removing heat doesn't keep $T$ fixed — with no heat inflow the temperature must fall during expansion, and that extra temperature change is what the factor $\gamma$ encodes.

A rising air parcel expands (lower pressure aloft) doing work on its surroundings with almost no heat exchange, so it cools adiabatically — this is the dry adiabatic lapse rate, and it is why mountaintops and high clouds are cold (figure s04).

The slope $-\gamma P/V$ approaches the isothermal slope $-P/V$; a gas that stores lots of energy internally barely cools on expansion, so its adiabat nearly overlaps the isotherm.

No — it is adiabatic ($Q=0$) but irreversible and does no work, so $\Delta U=0$ and $T$ stays constant for an ideal gas. The relation only holds for the reversible case.

Yes, they share that one point ($P,V,T$ all equal there), but their slopes differ by the factor $\gamma$, so they immediately diverge on either side (figure s02).

It predicts $P\propto V^{-\gamma }\to \infty$ and $T\propto V^{-(\gamma -1)}\to \infty$. But the ideal-gas assumption fails long before: molecules have finite size and attract each other, so a van der Waals picture takes over — the gas liquefies and $C_V$ itself changes, so no single $\gamma$ survives (figure s03 hints at why $C_V$ is not fixed).

No — then $P{V}^{\gamma }=PV=$const and $T{V}^{\gamma -1}=T=$const, collapsing both to the isothermal case. But $\gamma >1$ always for a real gas, so this is only a limiting sanity check.

$W=0$ — but with $Q=0$ and $\Delta T=0$ nothing happened at all; a genuine adiabatic process with work must change temperature.