A Level Physics - How to Solve: Thermal Physics (Ideal Gas Equation, Internal Energy, First Law of Thermodynamics)

How to Solve: Thermal Physics (Ideal Gas Equation, Internal Energy, First Law of Thermodynamics)

Complete Guide For GCSE/A-Level Physics, Chemistry, and Biology

Introduction

"Mastering the Ideal Gas Equation and the First Law of Thermodynamics doesn’t just get you marks—it unlocks real-world problems like how car engines work, why balloons expand in heat, and even how your lungs breathe. On your exam, this topic is worth 10-15% of your thermal physics marks, so let’s make sure you nail it."

WHAT YOU NEED TO KNOW FIRST

Before diving in, you must already understand:
1. Pressure, volume, and temperature relationships (Boyle’s Law, Charles’s Law, Pressure Law).
2. Kinetic theory of gases (particles in random motion, collisions with walls = pressure).
3. Energy conservation (energy cannot be created or destroyed, only transferred).

If any of these are shaky, pause and review them first.

KEY TERMS & FORMULAS

Key Terms

Formulas

1. Ideal Gas Equation

Formula: pV = nRT - p = pressure (Pa or N/m²) [MEMORISE] - V = volume (m³) [MEMORISE] - n = number of moles (mol) [MEMORISE] - R = universal gas constant = 8.31 J/mol·K (given on exam sheet) - T = temperature (K, not °C!) [MEMORISE]

When to use: When you have pressure, volume, temperature, or moles and need to find a missing variable.

2. Internal Energy of an Ideal Gas

Formula: U = (3/2) nRT (for a monatomic ideal gas) - U = internal energy (J) [MEMORISE] - n = number of moles (mol) - R = 8.31 J/mol·K - T = temperature (K)

For diatomic gases (e.g., O₂, N₂): U = (5/2) nRT (given on exam sheet if needed)

When to use: When asked for internal energy or temperature change in an ideal gas.

3. First Law of Thermodynamics

Formula: ΔU = Q + W (work done on the gas) OR ΔU = Q - W (work done by the gas)

• ΔU = change in internal energy (J) [MEMORISE]
• Q = heat added to the system (J) [MEMORISE]
• W = work done (J) [MEMORISE]

Sign conventions (CRUCIAL!): - Q > 0 → Heat added to the system. - Q < 0 → Heat removed from the system. - W > 0 → Work done on the gas (compression). - W < 0 → Work done by the gas (expansion).

When to use: When given heat, work, or internal energy changes and asked to find a missing variable.

4. Work Done by a Gas (Isobaric Process)

Formula: W = pΔV - W = work done (J) [MEMORISE] - p = pressure (Pa) - ΔV = change in volume (m³)

When to use: When pressure is constant and volume changes.

5. Molar Mass & Number of Moles

Formula: n = m / M - n = number of moles (mol) [MEMORISE] - m = mass (g or kg) - M = molar mass (g/mol or kg/mol)

When to use: When given mass and need to find moles (e.g., for the Ideal Gas Equation).

STEP-BY-STEP METHOD

How to Solve Any Thermal Physics Problem

Follow these 5 steps for every question:

1. Identify the process (isothermal, adiabatic, isobaric, isochoric).
2. List knowns and unknowns (write down all given values).
3. Choose the right formula (Ideal Gas Equation, First Law, etc.).
4. Check units (convert to Pa, m³, K, J if needed).
5. Solve step-by-step (show all working—examiners love this!).

Worked Example Using the Steps

Question: A gas expands from 0.02 m³ to 0.05 m³ at a constant pressure of 150 kPa. Calculate the work done by the gas.

Step 1: Identify the process - Isobaric (pressure is constant).

Step 2: List knowns and unknowns - p = 150 kPa (must convert to Pa) - V₁ = 0.02 m³ - V₂ = 0.05 m³ - W = ? (work done by the gas)

Step 3: Choose the right formula - W = pΔV (work done by the gas in an isobaric process).

Step 4: Check units - 150 kPa = 150,000 Pa (1 kPa = 1000 Pa) - Volume is already in m³.

Step 5: Solve step-by-step - ΔV = V₂ - V₁ = 0.05 - 0.02 = 0.03 m³ - W = pΔV = 150,000 × 0.03 = 4500 J

Answer: The work done by the gas is 4500 J.

What we did and why: - We recognised it was an isobaric process, so W = pΔV applies. - We converted kPa to Pa to match units. - We calculated ΔV first, then multiplied by pressure to find work.

WORKED EXAMPLES

Example 1 – Basic (Ideal Gas Equation)

Question: A container holds 0.5 moles of an ideal gas at 300 K and 200 kPa. What is its volume?

Solution:
1. Process: Unknown, but we have p, n, T → use pV = nRT.
2. Knowns: - n = 0.5 mol - T = 300 K - p = 200 kPa = 200,000 Pa - R = 8.31 J/mol·K
3. Formula: pV = nRT
4. Rearrange: V = nRT / p
5. Plug in values: - V = (0.5 × 8.31 × 300) / 200,000 - V = (1246.5) / 200,000 - V = 0.00623 m³ = 6.23 × 10⁻³ m³

Answer: The volume is 6.23 × 10⁻³ m³.

What we did and why: - We used the Ideal Gas Equation because we had p, n, T. - We converted kPa to Pa to match units with R. - We rearranged the formula to solve for V.

Example 2 – Medium (First Law of Thermodynamics)

Question: A gas absorbs 500 J of heat and does 200 J of work on its surroundings. What is the change in internal energy?

Solution:
1. Process: Not specified, but we have Q and W → use First Law.
2. Knowns: - Q = +500 J (heat added to the system) - W = -200 J (work done by the gas → negative in ΔU = Q + W)
3. Formula: ΔU = Q + W
4. Plug in values: - ΔU = 500 + (-200) = 300 J

Answer: The change in internal energy is +300 J.

What we did and why: - We used the First Law because we had heat and work. - We applied the correct sign convention (work done by the gas is negative). - We added Q and W to find ΔU.

Example 3 – Exam-Style (Combined Ideal Gas & First Law)

Question: A 2.0 mol sample of an ideal monatomic gas is compressed isothermally from 0.04 m³ to 0.02 m³ at 350 K. Calculate: (a) The work done on the gas. (b) The change in internal energy. (c) The heat transferred to the gas.

Solution (a): Work done on the gas
1. Process: Isothermal (T = constant).
2. Knowns: - V₁ = 0.04 m³ - V₂ = 0.02 m³ - T = 350 K - n = 2.0 mol
3. For isothermal process, work done on gas: - W = nRT ln(V₂/V₁) - (Note: ln = natural logarithm, given on formula sheet)
4. Plug in values: - W = 2.0 × 8.31 × 350 × ln(0.02/0.04) - W = 5817 × ln(0.5) - ln(0.5) ≈ -0.693 - W = 5817 × (-0.693) ≈ -4030 J
5. Work done on gas = +4030 J (negative sign means work is done by the gas, so we take the positive value for work done on the gas).

Answer (a): The work done on the gas is +4030 J.

Solution (b): Change in internal energy
1. For an isothermal process, ΔU = 0 (temperature doesn’t change).
2. Answer (b): The change in internal energy is 0 J.

Solution (c): Heat transferred
1. First Law: ΔU = Q + W
2. ΔU = 0 (from part b)
3. 0 = Q + W → Q = -W
4. W = +4030 J (work done on the gas)
5. Q = -4030 J (negative means heat is removed from the gas).

Answer (c): 4030 J of heat is removed from the gas.

What we did and why: - We recognised it was an isothermal process, so ΔU = 0. - We used W = nRT ln(V₂/V₁) for work in an isothermal process. - We applied the First Law to find Q after finding W and ΔU.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP

"Okay, let’s crush this in 60 seconds. The Ideal Gas Equation is pV = nRT—pressure times volume equals moles times R times temperature in Kelvin. Always convert units! For internal energy, monatomic gases use U = (3/2)nRT, diatomic use U = (5/2)nRT. The First Law is ΔU = Q + W—heat added plus work done on the gas. Signs matter! Work done on the gas is positive, by the gas is negative. For isothermal processes, ΔU = 0. For adiabatic, Q = 0. For isobaric, W = pΔV. And always check units—Pa, m³, K, J. You’ve got this!"