Thermodynamics problem solving — which formula for which process

Question

2 moles of an ideal gas at 300 K expand isothermally and reversibly from 10 L to 20 L. Calculate the work done by the gas. Also determine $\Delta U$ and $q$ .

(JEE Main / NEET — Chemical Thermodynamics)

Thermodynamics Formula Selection

flowchart TD
    A["Thermodynamics Problem"] --> B{Identify the process}
    B -->|"Temperature constant"| C["Isothermal"]
    B -->|"Pressure constant"| D["Isobaric"]
    B -->|"Volume constant"| E["Isochoric"]
    B -->|"No heat exchange"| F["Adiabatic"]
    C --> C1["w = -nRT ln(V2/V1)"]
    C --> C2["delta U = 0 (ideal gas)"]
    D --> D1["w = -P_ext x delta V"]
    D --> D2["q = n Cp delta T"]
    E --> E1["w = 0"]
    E --> E2["q = n Cv delta T = delta U"]
    F --> F1["q = 0"]
    F --> F2["delta U = w = n Cv delta T"]

Solution — Step by Step

The problem says “isothermal” (constant temperature) and “reversible.” For an isothermal reversible expansion of an ideal gas:

w = -nRT\ln\frac{V_2}{V_1}

The negative sign convention: work done by the gas is negative (system loses energy to surroundings) in the chemistry convention. Some textbooks use the physics convention where work done by the gas is positive — check which your exam uses.

$n = 2$ mol, $R = 8.314$ J/(mol K), $T = 300$ K, $V_1 = 10$ L, $V_2 = 20$ L

w = -2 \times 8.314 \times 300 \times \ln\frac{20}{10}

w = -2 \times 8.314 \times 300 \times \ln 2

w = -2 \times 8.314 \times 300 \times 0.693

\boxed{w = -3457.2 \text{ J} \approx -3.46 \text{ kJ}}

The gas does 3.46 kJ of work on the surroundings (expansion).

For an ideal gas in an isothermal process, internal energy depends only on temperature. Since $T$ is constant:

\Delta U = 0

By the first law of thermodynamics: $\Delta U = q + w$

0 = q + (-3457.2)

\boxed{q = +3457.2 \text{ J} = +3.46 \text{ kJ}}

The gas absorbs 3.46 kJ of heat from the surroundings, and all of it is used to do work. No energy is stored internally.

Why This Works

The first law of thermodynamics ( $\Delta U = q + w$ ) is the energy conservation law for thermodynamics. For an ideal gas, internal energy depends only on temperature (not on volume or pressure). So in an isothermal process, $\Delta U = 0$ regardless of what happens to volume or pressure. All heat absorbed goes directly into doing work.

The $\ln(V_2/V_1)$ term appears because reversible work is calculated by integrating $P\,dV$ with $P = nRT/V$ (ideal gas law).

Alternative Method — Quick Process Summary

Process	Constant	w	q	delta U
Isothermal	T	$-nRT\ln(V_2/V_1)$	$-w$	0
Isobaric	P	$-P\Delta V$	$nC_p\Delta T$	$nC_v\Delta T$
Isochoric	V	0	$nC_v\Delta T$	$nC_v\Delta T$
Adiabatic	q = 0	$nC_v\Delta T$	0	$nC_v\Delta T$

For JEE, remember: isothermal means “use ln,” adiabatic means “use gamma ( $C_p/C_v$ ),” isochoric means “work = 0,” isobaric means “easy $P\Delta V$ calculation.” The first step in any thermodynamics numerical is identifying the process — everything follows from that.

Common Mistake

The sign convention trap: in chemistry (IUPAC), work done ON the system is positive ( $w > 0$ ), and work done BY the system is negative ( $w < 0$ ). In physics, it is the opposite. JEE uses the chemistry convention. If you get a positive answer for expansion work, your sign is wrong — expansion means the gas pushes outward, doing work on the surroundings, so $w$ is negative (chemistry convention).