Thermodynamic processes — isothermal, adiabatic, isobaric, isochoric comparison

Question

Compare the four thermodynamic processes — isothermal, adiabatic, isobaric, and isochoric. How do work, heat, and internal energy change in each?

Solution — Step by Step

Condition: $\Delta T = 0$ , which means $\Delta U = 0$ (for ideal gas, internal energy depends only on temperature).

From the first law: $q = -w$ (heat absorbed = work done by the gas).

Work done (reversible): $w = -nRT \ln\left(\frac{V_2}{V_1}\right)$

On a PV diagram, the curve follows $PV = \text{constant}$ (a hyperbola). It is less steep than an adiabatic curve.

Condition: $q = 0$ (perfectly insulated system).

From the first law: $\Delta U = w$ (all work comes from internal energy).

If the gas expands adiabatically, it cools down (does work at the expense of its own energy). If compressed, it heats up.

Equation: $PV^\gamma = \text{constant}$ , where $\gamma = C_p/C_v$ .

The adiabatic curve on a PV diagram is steeper than the isothermal curve (because the gas cools as it expands, reducing pressure faster).

Condition: $\Delta P = 0$ .

Work done: $w = -P\Delta V = -P(V_2 - V_1)$ (simplest formula).

Heat: $q = nC_p\Delta T$ (at constant pressure, we use $C_p$ ).

On a PV diagram, this is a horizontal line. Heating at constant pressure causes the gas to expand.

Condition: $\Delta V = 0$ , so $w = 0$ (no work done — the gas cannot expand).

From the first law: $\Delta U = q$ (all heat goes into changing internal energy).

Heat: $q = nC_v\Delta T$ (at constant volume, we use $C_v$ ).

On a PV diagram, this is a vertical line. Heating at constant volume increases pressure.

graph TD
    A[Thermodynamic Process] --> B{What is held constant?}
    B -->|Temperature: dT=0| C[Isothermal]
    B -->|Heat: q=0| D[Adiabatic]
    B -->|Pressure: dP=0| E[Isobaric]
    B -->|Volume: dV=0| F[Isochoric]
    C --> C1["dU=0, q=-w"]
    D --> D1["q=0, dU=w"]
    E --> E1["w=-PdV, q=nCpdT"]
    F --> F1["w=0, dU=q=nCvdT"]

Why This Works

Process	Constant	$\Delta U$	$q$	$w$	PV curve
Isothermal	$T$	0	$-w$	$-nRT\ln(V_2/V_1)$	Hyperbola ( $PV = c$ )
Adiabatic	$q=0$	$w$	0	$nC_v\Delta T$	Steeper hyperbola ( $PV^\gamma = c$ )
Isobaric	$P$	$q + w$	$nC_p\Delta T$	$-P\Delta V$	Horizontal line
Isochoric	$V$	$q$	$nC_v\Delta T$	0	Vertical line

The first law of thermodynamics ( $\Delta U = q + w$ , with $w$ as work done on the gas) governs all four. Each process sets one variable to zero or constant, simplifying the equation differently.

For PV diagrams, remember: adiabatic is steeper than isothermal. During adiabatic expansion, both $P$ and $T$ decrease (so $P$ drops faster than in isothermal where $T$ stays constant).

Alternative Method

For JEE questions comparing work done in different processes between the same initial and final states:

w_{\text{isobaric}} > w_{\text{isothermal}} > w_{\text{adiabatic}}

This is because on a PV diagram, the area under the curve (which represents work) is largest for isobaric (horizontal line at constant P) and smallest for adiabatic (steepest curve).

Common Mistake

Students often confuse the sign convention. In chemistry, $w = -P_{ext}\Delta V$ (work done ON the gas is positive). In physics, $W = P\Delta V$ (work done BY the gas is positive). JEE questions may use either convention. Always check which sign convention the question uses before calculating. Writing the wrong sign is the fastest way to lose marks in thermodynamics.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next