Why is $\Delta H = 0$ for an isothermal expansion of an ideal gas even if pressure changes?

For an ideal gas, internal energy $U$ and enthalpy $H$ depend strictly on temperature $T$. Since temperature is constant in an isothermal process ($T_1 = T_2$), $\Delta T = 0$, which mathematically yields $\Delta U = n C_v \Delta T = 0$ and $\Delta H = n C_p \Delta T = 0$ regardless of pressure or volume variations.

How do I easily remember the sign of $\Delta G$ for spontaneity?

Remember that nature favors minimum energy and maximum disorder. $\Delta G = \Delta H - T \Delta S$. For spontaneity, $\Delta G$ must be negative ($\Delta G 0$) are always spontaneous at all temperatures because both terms make $\Delta G$ negative.

What is the key operational difference between reversible and irreversible work calculations in numericals?

For reversible processes, external pressure equals internal gas pressure at every point ($P_{ext} = P_{int} \pm dP$), requiring integration ($w_{rev} = -\int P_{int} dV$). For irreversible processes, external pressure is constant ($P_{ext} = \text{constant}$), simplifying the work formula to $w_{irrev} = -P_{ext} \Delta V$.

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Solved High-Yield Sample Questions

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Question #1NEET 2024

An ideal gas expands isothermally from $2.0\text{ L}$ to $10.0\text{ L}$ at $298\text{ K}$ against a constant external pressure of $2.0\text{ atm}$ . Calculate the work done $w$ on the gas in Joules according to IUPAC convention. (Given: $1\text{ L}\cdot\text{atm} = 101.3\text{ J}$ )

Solution Locked

Step-by-Step Solution & Concept

Step 1

Identify the process and use the formula for irreversible work against a constant external pressure: $w = -P_{ext} (V_2 - V_1)$ .

Step 2

Substitute given values: $w = -(2.0\text{ atm}) \times (10.0\text{ L} - 2.0\text{ L}) = -16.0\text{ L}\cdot\text{atm}$ .

Step 3

Convert $\text{L}\cdot\text{atm}$ to Joules: $w = -16.0 \times 101.3\text{ J} = -1620.8\text{ J}$ .

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Question #2JEE Main 2023

$1\text{ mole}$ of an ideal monoatomic gas ( $\gamma = \frac{5}{3}$ ) expands reversibly and adiabatically from an initial volume of $1\text{ L}$ at $300\text{ K}$ to a final volume of $8\text{ L}$ . The final temperature $T_2$ of the gas is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

For a reversible adiabatic process of an ideal gas, use the temperature-volume relation $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$ .

Step 2

Calculate exponent $\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$ . Substitute parameters: $300 \times (1)^{\frac{2}{3}} = T_2 \times (8)^{\frac{2}{3}}$ .

Step 3

Simplify $(8)^{\frac{2}{3}} = (2^3)^{\frac{2}{3}} = 2^2 = 4$ . Therefore, $300 = T_2 \times 4 \implies T_2 = \frac{300}{4} = 75\text{ K}$ .

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Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

First Law of Thermodynamics (IUPAC Convention)

Enthalpy and Internal Energy Relation

Reversible Isothermal Work (Ideal Gas)

Reversible Adiabatic Expansion Relations

Gibbs Free Energy & Spontaneity

Standard Free Energy & Equilibrium Constant

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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