IB Physics How to Solve: IB Physics – Thermal & Gases (Ideal Gas, Internal Energy, First Law)

How to Solve: IB Physics – Thermal & Gases (Ideal Gas, Internal Energy, First Law)

Complete Guide

Introduction

Mastering Ideal Gases, Internal Energy, and the First Law of Thermodynamics unlocks 10–15% of your IB Physics Paper 2 score—and real-world applications like car engines, refrigerators, and climate science. If you can solve a pressure-volume (P-V) diagram problem or calculate work done by a gas, you’re already ahead of 80% of students.

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you understand:
1. Kinetic Theory of Gases – Particles in random motion, pressure from collisions.
2. Work & Energy – Work = Force × Distance, energy conservation.
3. Basic Algebra & Units – Rearranging equations, converting between Pa, J, and m³.

KEY TERMS & FORMULAS

1. Ideal Gas Law

Formula: PV = nRT - P = Pressure (Pa or N/m²) [MEMORISE UNITS] - V = Volume (m³) [MEMORISE UNITS] - n = Number of moles (mol) - R = Universal gas constant = 8.31 J/(mol·K) [GIVEN ON EXAM SHEET] - T = Temperature (K) [MUST BE IN KELVIN!]

When to use: When any two variables change (e.g., pressure and volume at constant temperature).

2. Internal Energy (U) of an Ideal Gas

Formula: U = (3/2) nRT (for monatomic ideal gas) - U = Internal energy (J) - n, R, T = Same as above

Key Idea: Internal energy only depends on temperature for an ideal gas.

3. First Law of Thermodynamics

Formula: ΔU = Q + W (IB Physics sign convention) - ΔU = Change in internal energy (J) - Q = Heat added to the system (J) [POSITIVE if added, NEGATIVE if removed] - W = Work done on the system (J) [POSITIVE if work is done ON the gas, NEGATIVE if gas does work]

Alternative (Engineering Convention): ΔU = Q – W (where W = work done by the gas) [GIVEN ON EXAM SHEET – CHECK WHICH ONE!]

When to use: When heat, work, or internal energy changes (e.g., in a P-V diagram).

4. Work Done by a Gas (P-V Diagrams)

Formula: W = –∫ P dV (for a process) For constant pressure: W = –PΔV (work done by the gas) - W = Work (J) - P = Pressure (Pa) - ΔV = Change in volume (m³)

Key Idea: - If volume increases (ΔV > 0), gas does work (W = –ve in IB convention). - If volume decreases (ΔV < 0), work is done on the gas (W = +ve in IB convention).

5. Molar Mass & Number of Moles

Formula: n = m / M - n = Number of moles (mol) - m = Mass (g or kg) - M = Molar mass (g/mol or kg/mol)

When to use: When given mass instead of moles in the ideal gas law.

STEP-BY-STEP METHOD

Step 1: Identify the System & Given Variables

• Is it an ideal gas? (Assume yes unless told otherwise.)
• What’s changing? (P, V, T, n, Q, W?)
• What’s constant? (Isothermal = T constant, Isobaric = P constant, Isochoric = V constant.)

Step 2: Choose the Right Formula

Step 3: Convert Units (CRUCIAL!)

• Temperature: Always in Kelvin (K)! (0°C = 273 K)
• Pressure: Pa (N/m²) (1 atm = 101,325 Pa)
• Volume: m³ (1 m³ = 1000 L)
• Work/Energy: Joules (J)

Step 4: Solve for the Unknown

• Rearrange the formula.
• Plug in numbers with units.
• Calculate and check units in the answer.

Step 5: Interpret the Sign

• ΔU > 0 → Internal energy increases (temperature rises).
• ΔU < 0 → Internal energy decreases (temperature falls).
• W > 0 → Work done on the gas (compression).
• W < 0 → Work done by the gas (expansion).
• Q > 0 → Heat added to the system.
• Q < 0 → Heat removed from the system.

WORKED EXAMPLES

Example 1 – Basic (Ideal Gas Law)

Question: A gas occupies 0.05 m³ at 200 kPa and 300 K. How many moles of gas are present?

Step-by-Step Solution:
1. Identify variables: - P = 200 kPa = 200,000 Pa (convert to Pa!) - V = 0.05 m³ - T = 300 K - n = ?

1. Choose formula:

2. PV = nRT

3. Rearrange for n:

4. n = PV / RT

5. Plug in numbers:

6. n = (200,000 Pa × 0.05 m³) / (8.31 J/(mol·K) × 300 K)
7. n = (10,000) / (2493)
8. n ≈ 4.01 mol

What we did and why: - Converted kPa → Pa (exam trap!). - Used PV = nRT because we had P, V, T, and needed n. - Answer must have units (mol).

Example 2 – Medium (First Law & Work)

Question: A gas expands from 0.1 m³ to 0.3 m³ at a constant pressure of 150 kPa. It absorbs 5000 J of heat. What is the change in internal energy (ΔU)?

Step-by-Step Solution:
1. Identify variables: - P = 150 kPa = 150,000 Pa - V₁ = 0.1 m³, V₂ = 0.3 m³ → ΔV = +0.2 m³ - Q = +5000 J (heat added) - ΔU = ?

1. Find work done (W):
2. W = –PΔV (IB convention: work done by gas is negative)

3. W = –(150,000 Pa × 0.2 m³) = –30,000 J

4. Apply First Law:

5. ΔU = Q + W
6. ΔU = 5000 J + (–30,000 J) = –25,000 J

What we did and why: - Used W = –PΔV because pressure was constant. - Signs matter! W is negative because gas expands (does work). - ΔU is negative → internal energy decreases (gas cools).

Example 3 – Exam-Style (P-V Diagram & First Law)

Question: A monatomic ideal gas undergoes the process shown in the P-V diagram below. The gas starts at P = 200 kPa, V = 0.1 m³ and ends at P = 100 kPa, V = 0.3 m³. The process is linear (straight line on P-V graph). Calculate: a) The work done by the gas. b) The change in internal energy (ΔU). c) The heat added (Q).

Step-by-Step Solution:

Part (a): Work Done by the Gas

1. Work = Area under P-V curve.
2. Since it’s a straight line, work = area of trapezoid.
3. W = –½ (P₁ + P₂) × ΔV (negative because gas expands)
4. W = –½ (200,000 Pa + 100,000 Pa) × (0.3 – 0.1) m³
5. W = –½ (300,000) × 0.2 = –30,000 J

Part (b): Change in Internal Energy (ΔU)

1. Use ΔU = (3/2) nRΔT (monatomic gas).
2. Find n first (using initial state):
3. PV = nRT → n = PV / RT
4. n = (200,000 Pa × 0.1 m³) / (8.31 × 300 K) ≈ 8.02 mol
5. Find ΔT (using final state):
6. T₂ = P₂V₂ / nR = (100,000 × 0.3) / (8.02 × 8.31) ≈ 450 K
7. ΔT = T₂ – T₁ = 450 K – 300 K = +150 K
8. Calculate ΔU:
9. ΔU = (3/2) × 8.02 mol × 8.31 J/(mol·K) × 150 K ≈ +15,000 J

Part (c): Heat Added (Q)

1. Apply First Law:
2. ΔU = Q + W
3. 15,000 J = Q + (–30,000 J)
4. Q = 45,000 J

What we did and why: - Work = area under P-V curve (trapezoid for linear process). - ΔU depends only on ΔT (used ideal gas law to find ΔT). - First Law links Q, W, and ΔU (signs are critical!).

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (NIGHT BEFORE EXAM)

"Listen up—this is your 60-second survival guide for Thermal & Gases in IB Physics.

1. Ideal Gas Law (PV = nRT): If you see P, V, T, or n, this is your go-to. Convert units—Pa, m³, K!
2. First Law (ΔU = Q + W): Heat in, work on the gas, internal energy changes. Signs matter—expansion means negative work in IB convention.
3. Work = Area under P-V curve: Straight line? Trapezoid. Curve? Count squares or integrate.
4. Internal Energy (ΔU = (3/2)nRΔT): Only depends on temperature. If T doesn’t change, ΔU = 0.
5. Common traps: Wrong units, wrong signs, forgetting molar mass. Double-check every step.

Now go crush that exam!