Thermodynamic processes comparison — work, heat, internal energy for each type

Question

Compare isothermal, adiabatic, isobaric, and isochoric processes. For each, what happens to work done, heat exchanged, and internal energy?

(CBSE 11 + JEE Main + NEET — appears almost every year)

Solution — Step by Step

Process	Constant quantity	Equation	Key feature
Isothermal	Temperature ( $T$ )	$PV = \text{constant}$	Slow process, system in thermal equilibrium with surroundings
Adiabatic	No heat exchange ( $Q = 0$ )	$PV^\gamma = \text{constant}$	Fast process or insulated walls
Isobaric	Pressure ( $P$ )	$V/T = \text{constant}$	Heating gas in open container or with movable piston
Isochoric	Volume ( $V$ )	$P/T = \text{constant}$	Heating gas in rigid container

The first law: $Q = \Delta U + W$

Process	$W$ (work done by gas)	$Q$ (heat supplied)	$\Delta U$ (change in internal energy)
Isothermal	$nRT \ln(V_f/V_i)$	$W$ (since $\Delta U = 0$ )	$0$ (temperature constant)
Adiabatic	$\frac{P_iV_i - P_fV_f}{\gamma - 1}$	$0$ (by definition)	$-W$
Isobaric	$P\Delta V = nR\Delta T$	$nC_p\Delta T$	$nC_v\Delta T$
Isochoric	$0$ (no volume change)	$nC_v\Delta T$	$Q$ (all heat goes to $\Delta U$ )

For the same initial and final volumes, the work done follows this order:

W_{isobaric} > W_{isothermal} > W_{adiabatic}

This is because the PV curve is steepest for adiabatic (drops fastest), so the area under the curve is smallest.

An ideal gas at $P_1 = 2$ atm, $V_1 = 10$ L is expanded isothermally to $V_2 = 20$ L.

W = nRT\ln\frac{V_2}{V_1} = P_1V_1\ln\frac{V_2}{V_1} = 2 \times 10 \times \ln 2 = 20 \times 0.693 = \mathbf{13.86 \text{ L}\cdot\text{atm}}

Since it is isothermal: $\Delta U = 0$ and $Q = W = 13.86$ L-atm.

flowchart TD
    A["Identify the thermodynamic process"] --> B{"What is held constant?"}
    B -- "Temperature T" --> C["ISOTHERMAL: ΔU = 0, Q = W"]
    B -- "No heat exchange" --> D["ADIABATIC: Q = 0, ΔU = -W"]
    B -- "Pressure P" --> E["ISOBARIC: W = PΔV, Q = nCpΔT"]
    B -- "Volume V" --> F["ISOCHORIC: W = 0, Q = ΔU"]
    C --> G["PV = const, gentle curve"]
    D --> H["PV^γ = const, steep curve"]
    E --> I["Horizontal line on PV diagram"]
    F --> J["Vertical line on PV diagram"]

Why This Works

The first law of thermodynamics ( $Q = \Delta U + W$ ) is conservation of energy for thermal systems. Heat added either increases the internal energy (temperature rises) or does work (volume expands) or both. Each process constrains one variable, which fixes the relationship between the other two.

For an ideal gas, internal energy depends ONLY on temperature. So in an isothermal process, $\Delta U = 0$ regardless of pressure or volume changes. In adiabatic, all work comes from internal energy — the gas cools when it expands.

Alternative Method

For JEE MCQs, remember the slopes of PV curves: adiabatic is steeper than isothermal (by factor $\gamma$ ). This means adiabatic compression raises pressure MORE than isothermal compression for the same volume change. When asked “which process does more work?” — compare the areas under the curves.

Common Mistake

Students confuse “isothermal” with “adiabatic.” Isothermal means constant temperature (heat CAN flow to maintain it). Adiabatic means no heat flow (temperature DOES change). A slow process in a conducting container is isothermal. A fast process in an insulated container is adiabatic. Many students mistakenly think “no heat” means “no temperature change” — it is exactly the opposite.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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