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

What is being claimed?

We will derive all three from scratch, never just quote them.

Derivation from first principles

Step 1 — Apply $Q=0$ to the first law. $0 = nC_V\,dT + P\,dV$ Why this step? For an ideal gas internal energy depends only on temperature, so $dU = nC_V\,dT$ always (even when $V$ changes). Setting $\delta Q=0$ leaves the work to be paid by internal energy.

Step 2 — Kill the explicit $P$ using the ideal gas law. From $PV = nRT$ we get $P = \dfrac{nRT}{V}$. Substitute: $nC_V\,dT + \frac{nRT}{V}\,dV = 0$ Why this step? We have three variables $P,V,T$; the ideal gas law lets us eliminate one. We choose to keep $T$ and $V$ to head toward the $TV$ relation first.

Step 3 — Separate variables. Divide by $nC_V T$: $\frac{dT}{T} + \frac{R}{C_V}\frac{dV}{V} = 0$

Step 4 — Replace $R/C_V$ with $\gamma-1$. For an ideal gas, Mayer's relation $C_P - C_V = R$ holds, so $\frac{R}{C_V} = \frac{C_P - C_V}{C_V} = \frac{C_P}{C_V} - 1 = \gamma - 1.$ Why this step? This is what makes $\gamma$ appear — it is the signature of the gas's degrees of freedom hidden in $C_V$.

$\frac{dT}{T} + (\gamma-1)\frac{dV}{V} = 0$

Step 5 — Integrate. $\ln T + (\gamma-1)\ln V = \text{const} \;\Rightarrow\; \boxed{TV^{\gamma-1} = \text{const}}$ Why this step? Each term integrates to a logarithm; a sum of logs = const means a product is const.

Step 6 — Convert to $PV^\gamma$. Use $T = \dfrac{PV}{nR}$: $\frac{PV}{nR}\,V^{\gamma-1} = \text{const} \;\Rightarrow\; PV^{\gamma} = \text{const}$ Why this step? Multiply $T$ by $V^{\gamma-1}$; the $V$ powers add: $V^{1+(\gamma-1)} = V^\gamma$. Constants like $nR$ absorb into the new const.

Step 7 — The $TP$ relation. From $PV^\gamma=$const and $V = \dfrac{nRT}{P}$: $P\left(\frac{nRT}{P}\right)^{\gamma} = \text{const} \;\Rightarrow\; P^{1-\gamma}T^{\gamma} = \text{const}.$

Adiabatic vs Isothermal (Forecast-then-Verify)

Differentiate each:

• Isothermal: $P\,dV + V\,dP = 0 \Rightarrow \left(\dfrac{dP}{dV}\right)_T = -\dfrac{P}{V}$.
• Adiabatic: $V^\gamma dP + \gamma PV^{\gamma-1}dV = 0 \Rightarrow \left(\dfrac{dP}{dV}\right)_{Q} = -\gamma\dfrac{P}{V}$.

Verify: since $\gamma>1$, the adiabatic slope is $\gamma$ times steeper. ✔️ Why? In adiabatic expansion the gas also cools, so $P$ drops faster than isothermal (where $T$ is held up by incoming heat).

Worked examples

Common mistakes (Steel-manned)

Active-recall flashcards

Connections

• First Law of Thermodynamics — the parent equation $\delta Q = dU+\delta W$.
• Isothermal Process — contrast: $PV=$const, $T$ fixed.
• Mayer's Relation — supplies $C_P-C_V=R$, hence $R/C_V=\gamma-1$.
• Degrees of Freedom and $\gamma$ — why $\gamma=5/3$ (mono), $7/5$ (diatomic).
• Speed of Sound in Gases — uses adiabatic bulk modulus $\gamma P$.
• Work Done in Thermodynamic Processes — $W=nC_V(T_1-T_2)$ for adiabats.

Concept Map

Hinglish (regional understanding)