The First Law of Thermodynamics (Cambridge International A-Level Physics)

Spec point — 16.2.2

The first law of thermodynamics

• The first law of thermodynamics is based on the principle of conservation of energy
• Heating a gas and doing work on it both increase its internal energy; ∆U is the net result of both:

The increase in internal energy of a system is equal to the sum of the thermal energy added to the system and the work done on it

• The first law of thermodynamics is therefore defined as:

$\Delta U=q+W$

• Where:

  • ∆U = increase in internal energy (J)
  • q = thermal energy added to the system (J)
  • W = work done on the system (J)

• The first law of thermodynamics applies to all thermodynamic systems, not just gases

  • There is an important sign convention used for this equation

• A positive value for internal energy (+∆U) means:

  • The internal energy ∆U of the system increases
  • Heat (+q) is added to the system
  • Work (+W) is done on the system

• A negative value for internal energy (−∆U) means:

  • The internal energy ∆U of the system decreases
  • Heat (-q)* *is taken away from the system
  • Work (-W)* *is done by the system

Worked Example The volume occupied by 1.00 mol of a liquid at 50 °C is 2.4 × 10^-5 m³. When the liquid is vaporised at an atmospheric pressure of 1.03 × 10⁵ Pa, the vapour has a volume of 5.9 × 10^-2 m³. The latent heat to vaporise 1.00 mol of this liquid at 50 °C at atmospheric pressure is 3.48 × 10⁴ J.

Determine, for this change of state, the change in the internal energy of the system.

Answer:

Step 1: Write down the first law of thermodynamics

$\Delta U=q+W$

Step 2: Write the value of heating q of the system

• This is the latent heat, the heat required to vaporise the liquid = 3.48 × 10⁴ J

Step 3: Calculate the work done W

$W=p\Delta V$

• ∆V = final volume − initial volume

∆V = 5.9 × 10^-2 − 2.4 × 10^-5 = 0.058976 m³

• p = atmospheric pressure = 1.03 × 10⁵ Pa
• W = work done

W = (1.03 × 10⁵) × 0.058976 = 6074.528 = 6.07 × 10³ J

• Since the gas is expanding, the work done is negative

W = −6.07 × 10³ J

Step 4: Substitute the values into the first law of thermodynamics

∆U = 3.48 × 10⁴ + (−6.07 × 10³) = 28 730 = 29 000 J (2 s.f.)