recall and use the first law of thermodynamics ∆U = q + W expressed in terms of the increase in internal energy, the heating of the system (energy transferred to the system by heating) and the work done on the system

Cambridge A‑Level Physics 9702 – First Law of Thermodynamics

1. Internal Energy ( U )

• Definition: the total microscopic energy of a system – the sum of the translational, rotational and vibrational kinetic energies of all particles plus the intermolecular potential energy.
• Closed system: for a given composition, U is a state function; it depends only on the state variables that describe the system.
• Ideal gas (syllabus 16.1): the intermolecular potential energy is negligible, so

  \$U = U(T) \qquad\text{and}\qquad \Delta U = nC_{V}\Delta T\$

  where n is the amount of gas (mol) and C_V is the molar heat capacity at constant volume (units J mol⁻¹ K⁻¹).

• Real gas: the internal energy may also depend on volume or pressure because of intermolecular forces. This nuance is mentioned in the syllabus but is not required for most A‑Level calculations.
• Because C_V > 0, any increase in temperature always means an increase in internal energy.

2. First‑Law Statement (Syllabus 16.2)

For a closed system (no mass crosses the boundary) the conservation of energy is expressed as

\$\boxed{\Delta U = q + W}\$

• q – heat transferred to the system (positive when heat flows into the system).
• W – work done on the system (positive when work is done on the system).
• For an open system the law is written in terms of enthalpy, but the A‑Level syllabus only requires the closed‑system form.

2.1 Special cases (useful shortcuts)

3. Sign Conventions (Cambridge A‑Level)

4. Work Terms Required for 9702

1. Pressure–volume work (p V work) – for a quasi‑static change of a gas

   \$W{pV}= -\int{Vi}^{Vf} p\,\mathrm{d}V\$

   • Expansion (ΔV > 0) → W < 0 (system does work).
   • Compression (ΔV < 0) → W > 0 (work done on system).

2. Surface work (constant external pressure)

   \$W{\text{surf}} = -p{\text{ext}}\Delta V\$

3. Electrical work (current I through a potential difference V)

   \$W{\text{elec}} = -\int V I \,\mathrm{d}t = -q{\text{elec}}V\$

   (negative when the system delivers electrical energy).

4. Other work forms (e.g. stretching a wire, magnetic work) are rarely examined but obey the same sign rule: work on the system is positive.

5. Heat‑Transfer Expressions

• Sensible heating (no phase change)

  \$q = mc\Delta T\$

  m = mass (kg), c = specific heat capacity (J kg⁻¹ K⁻¹).

• Phase change

  \$q = mL\$

  where L is the latent heat of fusion or vapourisation (J kg⁻¹).

• Ideal‑gas heating at constant pressure

  \$q = nC_{P}\Delta T\$

• Ideal‑gas heating at constant volume

  \$q = nC_{V}\Delta T\$

6. Solving Problems – Step‑by‑Step (AO2)

1. Define the system (closed, open, cyclic) and draw a clear schematic if helpful.
2. Identify the process (isothermal, adiabatic, isochoric, etc.) – this tells you which terms are zero.
3. Calculate heat, q using the appropriate formula (mcΔT, mL, nCΔT …) and assign the correct sign according to the convention.
4. Calculate work, W – choose the correct work type, evaluate the integral or use the simple expression, then apply the sign rule.
5. Apply the first law ΔU = q + W to obtain the change in internal energy.
6. If required, relate ΔU to temperature (e.g. ΔU = nC_VΔT for an ideal gas) to find ΔT or any other unknown.
7. Finally, check units and signs; a quick “+” or “–” next to each quantity helps avoid common errors.

7. Worked Examples

Example 1 – Heating a metal block while it is compressed

Given: 0.50 kg aluminium, c = 900 J kg⁻¹ K⁻¹, ΔT = 60 K, work done on the block = +20 J.

Solution:

1. Heat added: \$q = mc\Delta T = (0.50)(900)(60) = 27\,000\ \text{J} \;(+)\$
2. Work on the system: \$W = +20\ \text{J}\$
3. First law: \$\Delta U = q + W = 27\,000 + 20 = 27\,020\ \text{J}\$

Example 2 – Adiabatic expansion of a monatomic ideal gas

Data: n = 1.00 mol, γ = 5/3, V₁ = 10 L, V₂ = 30 L, T₁ = 300 K.

Find: (a) final temperature T₂, (b) work done on the gas.

Solution:

1. Adiabatic ⇒ q = 0, so ΔU = W.
2. Adiabatic relation for an ideal gas:

   \$T V^{\gamma-1} = \text{constant}\$

   \$\$T{2}=T{1}\left(\frac{V{1}}{V{2}}\right)^{\gamma-1}

   =300\left(\frac{10}{30}\right)^{2/3}\approx 1.58\times10^{2}\ \text{K}\$\$

3. Change in internal energy:

   \$\$\Delta U = nC{V}(T{2}-T_{1})

   = (1.00)\left(\frac{3}{2}R\right)(158-300)\$\$

   \$\approx \frac{3}{2}(8.314)(-142) = -1.77\times10^{3}\ \text{J}\$

4. Since ΔU = W, \$W = -1.77\ \text{kJ}\$

   (negative → work done by the gas).

Example 3 – Heat engine completing a cycle

Given: during one cycle the engine absorbs 500 J of heat from a hot reservoir and does 300 J of work on the surroundings.

Find: net heat released to the cold reservoir and ΔU of the working substance.

Solution:

1. For a cyclic process, ΔU = 0.
2. First law: \$0 = q{\text{in}} + q{\text{out}} + W\$

   with q_in = +500 J, W = +300 J.

3. Re‑arrange: \$q{\text{out}} = -(q{\text{in}} + W) = -(500 + 300) = -800\ \text{J}\$

   The negative sign indicates 800 J of heat is released to the cold reservoir.

8. Common Pitfalls (AO2)

• Sign errors – always write “+” or “–” next to each quantity before substituting into ΔU = q + W.
• Omitting work when a volume change or electrical device is involved.
• Using q = mcΔT for a phase change – replace with q = mL.
• Assuming ΔU = 0 for any isothermal process – only true for an ideal gas; real gases may have a small ΔU.
• Confusing C_P and C_V – remember C_P > C_V and they appear in constant‑pressure and constant‑volume situations respectively.
• For cyclic processes forgetting that ΔU = 0, so heat and work must balance exactly.

9. Key Points to Remember (AO1)

• The first law links heat, work and internal energy: ΔU = q + W.
• Positive q = heat added to the system; positive W = work done on the system.
• For p V work, W = –∫p dV (negative for expansion, positive for compression).
• Internal energy of an ideal gas depends only on temperature: ΔU = nC_VΔT (C_V is a molar quantity, J mol⁻¹ K⁻¹).
• Special cases to memorise:

  • Adiabatic q = 0 → ΔU = W
  • Isothermal (ideal gas) ΔU = 0 → q = –W
  • Cyclic ΔU = 0 → q = –W
  • Isochoric W = 0 → ΔU = q

• Always state the system, choose the correct sign convention, and use the appropriate formula for q and W before applying the first‑law equation.

10. Syllabus Checklist – How the Notes Meet the Requirements