Thermodynamic processes — isothermal (T const), isochoric (V const), isobaric (P const), adiabatic (Q = 0)

0. The foundation we never leave: First Law & Work

1. Isochoric (V = const) — "iso-CHORic = no CHaR (no room) to move"

Since $V$ is constant, $dV = 0$ : $W = \int P\,dV = 0$ $\Delta U = Q - W = Q \quad\Rightarrow\quad \boxed{Q = nC_V\,\Delta T}$

On a $P$ – $V$ diagram this is a vertical line.

2. Isobaric (P = const) — "iso-BARic = constant BARometric pressure"

$P$ is constant, so it pulls out of the integral: $W = \int_{V_1}^{V_2} P\,dV = P(V_2 - V_1) = P\Delta V = nR\Delta T$ Why this step? $PV=nRT$ at constant $P$ gives $P\Delta V = nR\Delta T$ .

$Q = nC_P\,\Delta T,\qquad \Delta U = nC_V\Delta T$ On a $P$ – $V$ diagram this is a horizontal line; $W$ = rectangle area.

3. Isothermal (T = const) — "iso-THERMal = constant THERMometer"

$\Delta T = 0 \Rightarrow \Delta U = 0 \Rightarrow Q = W$ . Now find $W$ . With $T$ const, $P = nRT/V$ : $W = \int_{V_1}^{V_2} \frac{nRT}{V}\,dV = nRT\int_{V_1}^{V_2}\frac{dV}{V} = nRT\ln\frac{V_2}{V_1}$ $\boxed{W = Q = nRT\ln\frac{V_2}{V_1} = nRT\ln\frac{P_1}{P_2}}$ Why this step? $nRT$ is constant so it leaves the integral; $\int dV/V = \ln V$ . Used $P_1V_1=P_2V_2 \Rightarrow V_2/V_1 = P_1/P_2$ .

The curve is a hyperbola $PV=\text{const}$ .

4. Adiabatic (Q = 0) — "a-DIA-batic = no DIArrhea of heat (heat can't pass)"

$Q=0 \Rightarrow W = -\Delta U = -nC_V\Delta T$ . To get the curve, start from First Law differential form: $dQ = 0 = dU + dW = nC_V\,dT + P\,dV$ Use $PV=nRT \Rightarrow nR\,dT = P\,dV + V\,dP$ , so $nC_V\,dT = \frac{C_V}{R}(P\,dV+V\,dP)$ : $\frac{C_V}{R}(P\,dV+V\,dP) + P\,dV = 0$ Multiply by $R/C_V$ and group, with $\gamma \equiv C_P/C_V$ and $R/C_V = \gamma-1$ : $(P\,dV + V\,dP) + (\gamma-1)P\,dV = 0 \;\Rightarrow\; \gamma P\,dV + V\,dP = 0$ $\gamma\frac{dV}{V} + \frac{dP}{P} = 0 \;\xrightarrow{\text{integrate}}\; \gamma\ln V + \ln P = \text{const}$ $\boxed{PV^\gamma = \text{const}};\quad TV^{\gamma-1}=\text{const};\quad P^{1-\gamma}T^{\gamma}=\text{const}$

Because $\gamma>1$ , the adiabat is steeper than the isotherm on a $P$ – $V$ plot.

Figure — Thermodynamic processes — isothermal (T const), isochoric (V const), isobaric (P const), adiabatic (Q = 0)

5. One-glance summary table

Process	Constant	$W$	$Q$	$\Delta U$	Curve
Isochoric	$V$	$0$	$nC_V\Delta T$	$nC_V\Delta T$	vertical
Isobaric	$P$	$P\Delta V=nR\Delta T$	$nC_P\Delta T$	$nC_V\Delta T$	horizontal
Isothermal	$T$	$nRT\ln\frac{V_2}{V_1}$	$=W$	$0$	hyperbola
Adiabatic	$Q=0$	$\frac{P_1V_1-P_2V_2}{\gamma-1}$	$0$	$-W$	steep ( $PV^\gamma$ )

Common mistakes (Steel-manned)

In which process is work done by the gas zero, and why?

Isochoric —

dV=0

W=\int P\,dV=0

For an ideal gas,

\Delta U

equals what in ALL processes?

nC_V\Delta T

(since

U

depends only on

T

Isothermal work formula?

W=nRT\ln(V_2/V_1)

, and

Q=W

\Delta U=0

Why is

C_P>C_V

At constant

P

, extra heat pays for expansion work; Mayer:

C_P=C_V+R

Adiabatic relation between

P

and

V

PV^\gamma=

const,

\gamma=C_P/C_V

Adiabatic relation between

T

and

V

TV^{\gamma-1}=

const.

In an adiabatic expansion does the gas heat or cool? Why?

Cools —

Q=0

, so work comes from internal energy, lowering

T

Which is steeper on PV diagram, isotherm or adiabat?

Adiabat (slope factor

\gamma>1

Isobaric work formula?

W=P\Delta V=nR\Delta T

What constraint defines an adiabatic process?

Q=0

(no heat exchanged), often due to insulation or rapidity.

Heat added in isochoric process?

Q=nC_V\Delta T

(all into internal energy).

Adiabatic work in terms of states?

W=(P_1V_1-P_2V_2)/(\gamma-1)=nC_V(T_1-T_2)

Recall Feynman: explain to a 12-year-old

Imagine a bike pump with gas inside.

Sealed & heated (isochoric): you can't push the handle. All the heat just makes the gas angrier (hotter), no movement, no work.
Push slowly while keeping it cool (isothermal): every bit of energy you'd give goes straight back out as gentle work — temperature never budges.
Push with a fixed weight on top (isobaric): pressure stays the same, gas grows and heats; energy splits between getting hot and lifting the weight.
Pump super fast with insulation (adiabatic): no time for heat to escape, so the gas heats up purely from being squeezed — that's why a bike pump gets warm!

Connections

First Law of Thermodynamics — the parent equation $\Delta U=Q-W$ .
Ideal Gas Law — $PV=nRT$ used in every derivation.
Heat Capacities Cv and Cp — Mayer's relation $C_P=C_V+R$ .
Carnot Cycle — built from 2 isothermals + 2 adiabatics.
Entropy — adiabatic-reversible = isentropic ( $\Delta S=0$ ).
Work done by gas — PV diagrams — area interpretation.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho yaar, thermodynamic processes basically gas ko ek state se doosri state tak le jaane ke chaar special tareeke hain. Har baar hum ek cheez ko fix (constant) kar dete hain taaki First Law $\Delta U = Q - W$ ka hisaab aasaan ho jaaye. Ek baat hamesha yaad rakho: ideal gas ke liye internal energy sirf temperature pe depend karti hai, isliye $\Delta U = nC_V\Delta T$ har process me valid hai — chahe volume change ho ya na ho.

Isochoric (V constant): piston hil hi nahi sakta, toh gas koi work nahi karti ( $W=0$ ), saara heat seedha temperature badhane me chala jaata hai. Isobaric (P constant): upar weight fixed hai, gas garam hoke phailti hai, toh heat do kaam karta hai — thoda temperature, thoda expansion ka work ( $W=P\Delta V$ ). Isothermal (T constant): gas ko huge reservoir se touch karaya, dheere-dheere process chalta hai, temperature nahi badalta, isliye $\Delta U=0$ aur saara heat work ban jaata hai ( $Q=W=nRT\ln(V_2/V_1)$ ).

Adiabatic (Q=0): yahan heat ka aana-jaana band — ya toh insulation hai ya process itna fast hai ki heat ko time hi nahi milta. Toh gas apni hi internal energy se work karti hai, isliye expand karne pe thandi ho jaati hai (cloud banna, diesel engine isi pe chalte hain). Iska curve $PV^\gamma=$ const hota hai, jo isotherm se zyada steep girta hai kyunki $\gamma>1$ .

Sabse common galti: bachche samajh lete hain "isothermal matlab $Q=0$ " — galat! Isothermal me temperature constant, $Q=0$ wala adiabatic hai. Aur dhyaan rakho compression me work ka sign negative hota hai (gas pe work hota hai). Bas yeh table aur $P$ – $V$ diagram ratt mat lo — derive karke samjho, exam me kabhi nahi bhuloge.

Thermodynamic processes — isothermal (T const), isochoric (V const), isobaric (P const), adiabatic (Q = 0)

0. The foundation we never leave: First Law & Work

1. Isochoric (V = const) — "iso-CHORic = no CHaR (no room) to move"

2. Isobaric (P = const) — "iso-BARic = constant BARometric pressure"

3. Isothermal (T = const) — "iso-THERMal = constant THERMometer"

4. Adiabatic (Q = 0) — "a-DIA-batic = no DIArrhea of heat (heat can't pass)"

5. One-glance summary table

Common mistakes (Steel-manned)

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Thermodynamics

Connections