Thermodynamic processes — isothermal, adiabatic, isobaric, isochoric comparison

Question

Compare isothermal, adiabatic, isobaric, and isochoric processes. What quantity is constant in each? How do their PV curves differ?

(JEE Main 2024 asked PV diagram identification; NEET tests first law application for each process)

Solution — Step by Step

Temperature is constant ( $\Delta T = 0$ ). For an ideal gas, this means internal energy does not change ( $\Delta U = 0$ ).

From the first law: $Q = W$ (all heat absorbed is converted to work).

PV curve: rectangular hyperbola ( $PV = nRT = \text{constant}$ ).

No heat enters or leaves ( $Q = 0$ ). The system is perfectly insulated.

From the first law: $\Delta U = -W$ . When the gas expands, it does work at the expense of its internal energy, so it cools down.

PV curve: steeper than isothermal ( $PV^\gamma = \text{constant}$ , where $\gamma = C_p/C_v$ ).

Pressure is constant ( $\Delta P = 0$ ). This happens when a gas is heated in a cylinder with a freely movable piston.

Work done: $W = P\Delta V = nR\Delta T$ . Heat: $Q = nC_p\Delta T$ .

PV curve: horizontal line (P constant, V changes).

Volume is constant ( $\Delta V = 0$ ). This happens in a rigid sealed container.

Work done: $W = 0$ (no volume change, no work). From the first law: $Q = \Delta U = nC_v\Delta T$ .

PV curve: vertical line (V constant, P changes).

flowchart TD
    A[Thermodynamic Process] --> B{What is constant?}
    B -->|Temperature: dT = 0| C["Isothermal<br/>PV = const<br/>dU = 0, Q = W"]
    B -->|Heat exchange: Q = 0| D["Adiabatic<br/>PV^γ = const<br/>dU = −W"]
    B -->|Pressure: dP = 0| E["Isobaric<br/>W = PdV<br/>Q = nCₚdT"]
    B -->|Volume: dV = 0| F["Isochoric<br/>W = 0<br/>Q = nCᵥdT"]

Why This Works

The first law of thermodynamics ( $Q = \Delta U + W$ ) is the master equation. Each process sets one variable to zero or constant, which simplifies the first law differently:

Process	Constant	$Q$	$\Delta U$	$W$
Isothermal	$T$	$= W$	$0$	$nRT\ln(V_2/V_1)$
Adiabatic	(insulated)	$0$	$= -W$	$nC_v(T_1-T_2)$
Isobaric	$P$	$nC_p\Delta T$	$nC_v\Delta T$	$P\Delta V$
Isochoric	$V$	$nC_v\Delta T$	$nC_v\Delta T$	$0$

On a PV diagram, the steepness increases as: isobaric (flat) → isothermal → adiabatic → isochoric (vertical). This ordering is crucial for identifying curves in JEE problems.

Alternative Method

The slope comparison on a PV diagram: adiabatic curve is steeper than isothermal by a factor of $\gamma$ . At any point, $\left(\frac{dP}{dV}\right)_{\text{adiabatic}} = \gamma \left(\frac{dP}{dV}\right)_{\text{isothermal}}$ . Since $\gamma > 1$ , the adiabatic curve always falls faster. This is a one-line answer for “compare isothermal and adiabatic PV curves.”

Common Mistake

Students confuse isothermal with adiabatic. In isothermal, temperature is constant — the system exchanges heat with the surroundings to maintain T. In adiabatic, no heat is exchanged — so temperature changes (decreases during expansion, increases during compression). The temperature behaviour is opposite: isothermal holds T constant, adiabatic lets T change freely.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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