$\Delta U = Q - W$

where:

• $\Delta U$ = change in internal energy (J)
• $Q$ = heat added to system (J)
• $W$ = work done BY system (J)

Sign Conventions:

Heat (Q):

• $Q > 0$: Heat added to system (absorbed)
• $Q < 0$: Heat removed from system (released)

Work (W):

• $W > 0$: System does work on surroundings (expansion)
• $W < 0$: Work done on system (compression)

Internal Energy (ΔU):

• $\Delta U > 0$: Energy increases, temperature rises
• $\Delta U < 0$: Energy decreases, temperature falls

💡 Key Insight: Energy can enter as heat or leave as work, but total energy is conserved.

Internal Energy

Internal energy is the total kinetic and potential energy of all molecules.

For an ideal gas: $U = \frac{3}{2}nRT$

where:

• $n$ = number of moles
• $R$ = 8.31 J/(mol·K)
• $T$ = absolute temperature (K)

For ideal gas: $\Delta U$ depends ONLY on temperature change! $\Delta U = \frac{3}{2}nR\Delta T$

Work in Thermodynamic Processes

For gas expansion/compression:

$W = P\Delta V$

where:

• $P$ = pressure (Pa)
• $\Delta V = V_f - V_i$ (m³)

On a PV diagram: Work = area under curve

• Expansion ($\Delta V > 0$): $W > 0$ (system does work)
• Compression ($\Delta V < 0$): $W < 0$ (work done on system)

Thermodynamic Processes

1. Isothermal (Constant Temperature)

• $\Delta T = 0$ → $\Delta U = 0$
• First Law: $Q = W$
• All heat becomes work
• For ideal gas: $PV =$ constant

2. Adiabatic (No Heat Transfer)

• $Q = 0$
• First Law: $\Delta U = -W$
• Work comes from/goes to internal energy
• Temperature changes!
• Example: Rapid compression/expansion

3. Isobaric (Constant Pressure)

• $P =$ constant
• $W = P\Delta V$
• $Q = \Delta U + P\Delta V$
• Example: Piston free to move

4. Isochoric (Constant Volume)

• $\Delta V = 0$ → $W = 0$
• First Law: $Q = \Delta U$
• All heat changes internal energy
• Example: Rigid container

PV Diagrams

Pressure-Volume (PV) diagrams show thermodynamic processes:

• Horizontal line: Isobaric (constant P)
• Vertical line: Isochoric (constant V)
• Curve: Isothermal or adiabatic

Work = area under curve

• Clockwise cycle: Net work done BY system (W > 0)
• Counter-clockwise: Net work done ON system (W < 0)

Second Law of Thermodynamics

Heat flows spontaneously from hot to cold, not the reverse (without external work).

Equivalently: Entropy of an isolated system always increases.

Entropy (S): Measure of disorder/randomness

• Natural processes increase total entropy
• Energy becomes more spread out, less useful

Implications:

1. No perfect heat engine: Can't convert all heat to work
2. No perfect refrigerator: Can't transfer heat from cold to hot without work
3. Time's arrow: Explains why time has a direction

Heat Engines

Convert heat into mechanical work:

Efficiency: $e = \frac{W}{Q_H} = \frac{Q_H - Q_C}{Q_H} = 1 - \frac{Q_C}{Q_H}$

where:

• $Q_H$ = heat absorbed from hot reservoir
• $Q_C$ = heat expelled to cold reservoir
• $W = Q_H - Q_C$ = useful work output

Maximum efficiency (Carnot engine): $e_{max} = 1 - \frac{T_C}{T_H}$

Temperatures in Kelvin!

💡 Key: Efficiency always < 100%. Some heat must be expelled.

Refrigerators and Heat Pumps

Move heat from cold to hot (requires work):

Coefficient of Performance (COP): $COP_{refrigerator} = \frac{Q_C}{W}$ $COP_{heat pump} = \frac{Q_H}{W}$

Higher COP = more efficient

Problem-Solving Strategy

1. Identify the process: Isothermal, adiabatic, isobaric, isochoric
2. Apply First Law: $\Delta U = Q - W$
3. Use process-specific relationships:
   • Isothermal: $\Delta U = 0$
   • Adiabatic: $Q = 0$
   • Isobaric: $W = P\Delta V$
   • Isochoric: $W = 0$
4. For ideal gas: $\Delta U = \frac{3}{2}nR\Delta T$
5. Watch signs for Q and W!

Common Mistakes

❌ Wrong sign for Q or W ❌ Using Celsius instead of Kelvin for Carnot efficiency ❌ Forgetting ΔU = 0 for isothermal process ❌ Confusing heat added vs. heat expelled in engines ❌ Assuming all heat can be converted to work (violates 2nd Law)

Find: Change in internal energy $\Delta U$

Solution:

Apply First Law: $\Delta U = Q - W$ $\Delta U = 500 - 300 = 200 \text{ J}$

Answer: Internal energy increases by 200 J

The gas gained 500 J as heat but lost 300 J doing work, net gain of 200 J.

Solution:

Part (a): Change in internal energy

For isothermal process with ideal gas: $\Delta T = 0 \Rightarrow \Delta U = 0$

Part (b): Work done

Apply First Law: $\Delta U = Q - W$ $0 = 800 - W$ $W = 800 \text{ J}$

Answer:

• (a) $\Delta U = 0$ (temperature constant)
• (b) $W = 800$ J (all heat converted to work!)

In isothermal expansion, all absorbed heat becomes work.

Solution:

Part (a): Change in internal energy

For isothermal process with ideal gas: $\Delta T = 0 \Rightarrow \Delta U = 0$

Part (b): Work done

Apply First Law: $\Delta U = Q - W$ $0 = 800 - W$ $W = 800 \text{ J}$

Answer:

• (a) $\Delta U = 0$ (temperature constant)
• (b) $W = 800$ J (all heat converted to work!)

In isothermal expansion, all absorbed heat becomes work.

Solution:

Part (a): Maximum efficiency

For Carnot engine: $e_{max} = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{500}$ $e_{max} = 1 - 0.6 = 0.4 = 40\%$

Part (b): Work output

$e = \frac{W}{Q_H}$ $W = e \cdot Q_H = (0.4)(1000) = 400 \text{ J}$

Part (c): Heat expelled

Energy conservation: $Q_H = W + Q_C$ $Q_C = Q_H - W = 1000 - 400 = 600 \text{ J}$

Verification: $e = 1 - \frac{Q_C}{Q_H} = 1 - \frac{600}{1000} = 0.4$ ✓

Answer:

• (a) Maximum efficiency: 40%
• (b) Work done: 400 J
• (c) Heat expelled: 600 J

Even the perfect Carnot engine can't achieve 100% efficiency!

Solution:

Part (a): Maximum efficiency

For Carnot engine: $e_{max} = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{500}$ $e_{max} = 1 - 0.6 = 0.4 = 40\%$

Part (b): Work output

$e = \frac{W}{Q_H}$ $W = e \cdot Q_H = (0.4)(1000) = 400 \text{ J}$

Part (c): Heat expelled

Energy conservation: $Q_H = W + Q_C$ $Q_C = Q_H - W = 1000 - 400 = 600 \text{ J}$

Verification: $e = 1 - \frac{Q_C}{Q_H} = 1 - \frac{600}{1000} = 0.4$ ✓

Answer:

• (a) Maximum efficiency: 40%
• (b) Work done: 400 J
• (c) Heat expelled: 600 J

Even the perfect Carnot engine can't achieve 100% efficiency!