JEE Chemistry: Thermodynamics - First Law, Internal Energy, Enthalpy, Hess's Law

Thermodynamics — First Law: Internal Energy, Enthalpy, Hess's Law

What This Is and Why It Matters for JEE

Thermodynamics is a fundamental topic in JEE, appearing in 2-3 questions every year. It's moderately difficult, with a slightly higher weightage in JEE Advanced. Understanding internal energy, enthalpy, and Hess's Law is crucial for solving problems related to heat transfer, chemical reactions, and thermodynamic cycles.

Prerequisites

You should already know: - First Law of Thermodynamics: ?U = Q - W - Work done: W = P?V (for reversible processes) - Heat transfer: Q = mc?T (for ideal gases)

Quick revision: - Review the first law and work done for reversible processes. - Understand ideal gas behavior.

Core Concepts (Exam-Focused)

• Internal Energy (U): A measure of a system's total energy.
• Enthalpy (H): H = U + PV, a measure of a system's total energy plus its pressure-volume product.
• Hess's Law: ?H = H (for a series of reactions), a statement of the conservation of enthalpy.

Key formulae: - ?U = Q - W - ?H = ?U + ?(PV) - ?H = H (for a series of reactions)

Step-by-Step Problem-Solving Strategy

1. Identify the given information (?U, Q, W, ?H, etc.).
2. Determine the unknown quantity (?U, ?H, etc.).
3. Choose the relevant concept (internal energy, enthalpy, Hess's Law).
4. Set up the correct equation using the chosen concept.
5. Check for any special conditions or assumptions (e.g., reversible processes).

Mistake: Assuming a reversible process when it's not given. Fix: Check if the process is reversible or irreversible.

Important Graphs / Diagrams (if applicable)

No specific graphs are relevant to this topic.

Typical JEE Question Patterns

1. Find the change in internal energy: Identify the given information and set up the equation ?U = Q - W.
2. Compare the enthalpy changes: Use Hess's Law to calculate the total enthalpy change.
3. Determine the heat transfer: Use the first law and the given information to calculate the heat transfer.

Common Mistakes & Exam Traps

1. The mistake: Assuming ?U = ?H for all processes. Why it happens: Misunderstanding the difference between internal energy and enthalpy. How to avoid it: Remember that ?U = ?H only for reversible isothermal processes.
2. The mistake: Failing to account for the pressure-volume product in enthalpy changes. Why it happens: Rushing through the problem or misreading the question. How to avoid it: Always include the pressure-volume product in enthalpy changes.
3. The mistake: Using Hess's Law incorrectly. Why it happens: Misunderstanding the concept or rushing through the problem. How to avoid it: Read the question carefully and apply Hess's Law only when the reactions are independent.

Time-Saving Shortcuts

1. Shortcut: Use the equation ?H = ?U + ?(PV) to simplify enthalpy changes.
2. Shortcut: Apply Hess's Law only when the reactions are independent.

Practice MCQs (Exam-Style)

Question 1: A gas expands from 1 L to 2 L against a constant external pressure of 2 atm. If the initial temperature is 300 K, what is the change in internal energy?

A) -100 J B) 0 J C) 100 J D) 200 J

Answer: B) 0 J Solution: Since the process is isothermal, ?U = 0. Common Wrong Answer: A) -100 J, assuming a decrease in internal energy due to expansion.

Question 2: Two reactions are given: Reaction 1: A-B Reaction 2: B-C If the enthalpy change for Reaction 1 is +100 kJ/mol and for Reaction 2 is -50 kJ/mol, what is the total enthalpy change for the reaction A-C?

A) +50 kJ/mol B) +100 kJ/mol C) -50 kJ/mol D) -150 kJ/mol

Answer: B) +100 kJ/mol Solution: Use Hess's Law to calculate the total enthalpy change. Common Wrong Answer: A) +50 kJ/mol, assuming the enthalpy change for Reaction 2 is negligible.

Question 3: A system undergoes a reversible isothermal expansion from 1 L to 2 L against a constant external pressure of 2 atm. If the initial temperature is 300 K, what is the change in enthalpy?

A) -100 J B) 0 J C) 100 J D) 200 J

Answer: B) 0 J Solution: Since the process is isothermal and reversible, ?H = 0. Common Wrong Answer: A) -100 J, assuming a decrease in enthalpy due to expansion.

Quick Revision Card (60-Second Summary)

• ?U = Q - W
• ?H = ?U + ?(PV)
• ?H = H (for a series of reactions)
• Hess's Law: ?H = H (for a series of reactions)
• Reversible processes: ?U = ?H
• Isothermal processes: ?U = 0, ?H = 0

If You Get Stuck in Exam

• Partial marks strategy: Write down what you know and try to relate it to the question.
• Eliminate distractors: Check each option carefully and eliminate any that are clearly incorrect.
• Skip and return: If you're stuck, move on to the next question and come back to it later.

Related JEE Topics

• Thermodynamic cycles: Study the Carnot cycle, Rankine cycle, and other thermodynamic cycles to understand the application of thermodynamics in real-world systems.
• Heat transfer: Study heat transfer through conduction, convection, and radiation to understand the different modes of heat transfer.
• Chemical equilibrium: Study chemical equilibrium to understand how thermodynamics relates to chemical reactions.