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Study Guide: General Chemistry 1: Thermochemistry SystemSurroundings Heat q and Work w First Law ΔUqw
Source: https://www.fatskills.com/college-chemistry/chapter/generalchemistry1-general-chemistry-1-thermochemistry-systemsurroundings-heat-q-and-work-w-first-law-%CE%B4uqw

General Chemistry 1: Thermochemistry SystemSurroundings Heat q and Work w First Law ΔUqw

By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.

⏱️ ~6 min read

What Is This?

The First Law of Thermodynamics states that the change in internal energy (ΔU) of a system is equal to the heat (q) added to the system minus the work (w) done by the system. This topic appears in exams to test your understanding of energy conservation and your ability to apply the First Law in various scenarios. Questions typically involve calculating ΔU, q, or w given the other two variables.

Why It Matters

This topic is tested in physics, chemistry, and engineering exams, including AP Chemistry, IB Chemistry, and university-level thermodynamics courses. It appears frequently and can carry up to 10-15% of the total marks. The skill being tested is your ability to apply the First Law to different systems and processes, often requiring algebraic manipulation and unit conversions.

Core Concepts

  • System and Surroundings: Understand the distinction between the system (the part of the universe you're studying) and the surroundings (everything else).
  • Heat (q): Energy transferred due to a temperature difference. Positive q means heat is added to the system.
  • Work (w): Energy transferred via a force acting over a distance. Positive w means work is done by the system.
  • Internal Energy (U): The total energy of a system, including kinetic and potential energy of its molecules.
  • Sign Convention: Heat added to the system and work done by the system are positive; the opposite are negative.

Prerequisites

  • Basic understanding of energy and its forms (kinetic, potential, thermal).
  • Familiarity with algebraic manipulation and unit conversions.
  • Without these, you'll struggle to apply the First Law and make errors in sign conventions.

The Rule-Book (How It Works)

  • Primary Rule: ΔU = q + w
  • Sub-rules:
  • If q is added to the system, q is positive.
  • If w is done by the system, w is positive.
  • ΔU is positive if internal energy increases.
  • Edge Cases:
  • Adiabatic process (q = 0): ΔU = w
  • Isovolumetric process (no volume change, so w = 0): ΔU = q
  • Mnemonic: "Heat in, Work out, Energy changes."

Exam / Job / Audit Weighting

  • Frequency: High
  • Difficulty Rating: Intermediate
  • Question Type: Calculation problems, multiple-choice questions, true/false statements

Difficulty Level

Intermediate

Must-Know Rules, Formulas, Standards, or Principles

  1. First Law of Thermodynamics: ΔU = q + w
  2. Sign Convention: Positive q (heat in), Positive w (work out)
  3. Adiabatic Process: ΔU = w (since q = 0)

Worked Examples (Step-by-Step)


Easy

Question: A system absorbs 50 J of heat and does 20 J of work. Calculate the change in internal energy (ΔU).
Step-by-Step: 1. Identify given values: q = 50 J, w = 20 J 2. Apply the First Law: ΔU = q + w 3. Substitute values: ΔU = 50 J + 20 J = 70 J Answer: ΔU = 70 J

Medium

Question: A gas expands and does 150 J of work on the surroundings while absorbing 250 J of heat. Calculate ΔU.
Step-by-Step: 1. Identify given values: q = 250 J, w = -150 J (work done on surroundings) 2. Apply the First Law: ΔU = q + w 3. Substitute values: ΔU = 250 J - 150 J = 100 J Answer: ΔU = 100 J

Hard

Question: In an adiabatic process, a system does 300 J of work. Calculate the change in internal energy (ΔU).
Step-by-Step: 1. Identify given values: w = 300 J, q = 0 (adiabatic process) 2. Apply the First Law: ΔU = q + w 3. Substitute values: ΔU = 0 + 300 J = -300 J (since work is done by the system) Answer: ΔU = -300 J

Common Exam Traps & Mistakes

  1. Mistake: Confusing the sign of work.
  2. Wrong Answer: ΔU = q - w
  3. Correct Approach: Remember, work done by the system is positive.
  4. Mistake: Forgetting the sign convention for heat.
  5. Wrong Answer: ΔU = -q + w
  6. Correct Approach: Heat added to the system is positive.
  7. Mistake: Not recognizing an adiabatic process.
  8. Wrong Answer: ΔU = q + w (with q ≠ 0)
  9. Correct Approach: In an adiabatic process, q = 0.
  10. Mistake: Incorrect unit conversions.
  11. Wrong Answer: Mixing Joules and calories.
  12. Correct Approach: Convert all units to Joules before calculations.

Shortcut Strategies & Exam Hacks

  • Memory Aid: "Heat in, Work out, Energy changes."
  • Elimination Strategy: If a question involves no heat transfer, eliminate options with q ≠ 0.
  • Pattern Recognition: Identify adiabatic processes quickly by looking for q = 0.

Question-Type Taxonomy

  1. Calculation Problems: Direct application of ΔU = q + w.
  2. Mini-Example: A system absorbs 100 J of heat and does 50 J of work. Calculate ΔU.
  3. Favored Exams: AP Chemistry, IB Chemistry
  4. Multiple-Choice Questions: Identify the correct application of the First Law.
  5. Mini-Example: If a system does 200 J of work in an adiabatic process, what is ΔU?
  6. Favored Exams: University-level thermodynamics
  7. True/False Statements: Verify the correctness of statements about the First Law.
  8. Mini-Example: True or False: In an isovolumetric process, ΔU = q.
  9. Favored Exams: Engineering thermodynamics

Practice Set (MCQs)


Question 1

Question: A system absorbs 150 J of heat and does 75 J of work. What is the change in internal energy (ΔU)? Options: A. 75 J B. 225 J C. -75 J D. -225 J Correct Answer: B. 225 J Explanation: ΔU = q + w = 150 J + 75 J = 225 J Why the Distractors Are Tempting: - A: Confuses the sign of work.
- C: Incorrectly subtracts work.
- D: Misapplies the sign convention.

Question 2

Question: In an adiabatic process, a system does 400 J of work. What is ΔU? Options: A. 400 J B. -400 J C. 0 J D. 800 J Correct Answer: B. -400 J Explanation: ΔU = q + w = 0 + 400 J = -400 J (work done by the system) Why the Distractors Are Tempting: - A: Forgets the sign of work.
- C: Incorrectly assumes no change in internal energy.
- D: Doubles the work done.

Question 3

Question: A gas absorbs 300 J of heat and does no work. What is ΔU? Options: A. 300 J B. -300 J C. 0 J D. 600 J Correct Answer: A. 300 J Explanation: ΔU = q + w = 300 J + 0 = 300 J Why the Distractors Are Tempting: - B: Incorrectly subtracts heat.
- C: Assumes no change in internal energy.
- D: Doubles the heat absorbed.

Question 4

Question: A system does 50 J of work on the surroundings while absorbing 100 J of heat. What is ΔU? Options: A. 150 J B. 50 J C. -50 J D. -150 J Correct Answer: B. 50 J Explanation: ΔU = q + w = 100 J - 50 J = 50 J Why the Distractors Are Tempting: - A: Adds work instead of subtracting.
- C: Incorrectly subtracts heat.
- D: Misapplies the sign convention.

Question 5

Question: In an isovolumetric process, a system absorbs 200 J of heat. What is ΔU? Options: A. 200 J B. -200 J C. 0 J D. 400 J Correct Answer: A. 200 J Explanation: ΔU = q + w = 200 J + 0 = 200 J Why the Distractors Are Tempting: - B: Incorrectly subtracts heat.
- C: Assumes no change in internal energy.
- D: Doubles the heat absorbed.

30-Second Cheat Sheet

  • First Law: ΔU = q + w
  • Sign Convention: Positive q (heat in), Positive w (work out)
  • Adiabatic Process: ΔU = w (q = 0)
  • Isovolumetric Process: ΔU = q (w = 0)
  • Memory Aid: "Heat in, Work out, Energy changes."

Learning Path

  1. Beginner Foundation: Understand energy forms and basic thermodynamics.
  2. Core Rules: Memorize the First Law and sign conventions.
  3. Practice: Solve calculation problems and multiple-choice questions.
  4. Timed Drills: Practice under exam conditions.
  5. Mock Tests: Take full-length practice exams.

Related Topics

  1. Second Law of Thermodynamics: Explains entropy and the direction of heat transfer.
  2. Enthalpy (H): Relates to heat transfer at constant pressure.
  3. Heat Engines and Refrigerators: Applications of the First Law in engineering.


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