Thermodynamic Processes

The four main thermodynamic processes are
- Constant volume $(W = 0)$
- Constant pressure $(∆ p = 0)$
- Isothermal $(∆ T = 0)$
- Adiabatic $(∆ Q = 0)$

Constant pressure

An isobaric (constant pressure) process is defined as:

A process in which no change in pressure occurs

This occurs when gases are allowed to expand or contract freely during a change in temperature
When there is a change in volume ΔV at a constant pressure p, work done W is equal to

$W = p ∆ V$

From the first law of thermodynamics:

$Q = ∆ U + W$

$Q = ∆ U \pm p ∆ V$

The ± sign reflects whether work has been done on or by the gas as a result of the change in volume

2-4-6-isobaric-pv-diagram

The solid blue line represents an isobaric process at constant pressure on a p-V diagram

Constant volume

An isovolumetric (constant volume) process is defined as:

A process where no change in volume occurs and the system does no work

If there is no change in volume, then there is no work done on or by the gas, so $W = 0$
Therefore, from the first law of thermodynamics:

$Q = ∆ U + W = ∆ U + 0$

$Q = ∆ U$

2-4-6-isovolumetric-pv-diagram

The solid blue line represents an isovolumetric process at constant volume on a p-V diagram

Constant temperature (isothermal)

An isothermal process is defined as:

A process in which no change in temperature occurs

If the temperature does not change, then the internal energy of the gas will not change, so $∆ U = 0$
Therefore, from the first law of thermodynamics:

$Q = ∆ U + W = 0 + W$

$Q = W$

2-4-6-isothermal-pv-diagram

The solid blue line represents an isothermal process with constant temperature on a p-V diagram

Constant thermal energy (adiabatic)

An adiabatic process is defined as:

A process where no heat is transferred into or out of the system

If there is no heat entering or leaving the system then $Q = 0$
Therefore, from the first law of thermodynamics:

$Q = ∆ U + W = 0$

$W = - ∆ U$

This means that all the work done is at the expense of the system's internal energy
Hence, an adiabatic process will usually be accompanied by a change in temperature

2-4-6-adiabatic-pv-diagram

The solid blue line represents an adiabatic process with constant thermal energy on a p-V diagram

Adiabatic Processes

Adiabatic processes in ideal gases can be modelled by the equation

$p V^{γ} = c o n s t a n t$

Where:
- p = pressure of the gas (Pa)
- V = volume occupied by the gas (m³)

This equation can be used for calculating changes in pressure, volume and temperature, e.g. for monatomic ideal gases, where $γ = \frac{5}{3}$

$p_{1} {V_{1}}^{\frac{5}{3}} = p_{2} {V_{2}}^{\frac{5}{3}}$

Where:
- $p_{1}$ = initial pressure (Pa)
- $p_{2}$ = final pressure (Pa)
- $V_{1}$ = initial volume (m³)
- $V_{2}$ = final volume (m³)

Worked example

A quantity of energy Q is supplied to three ideal gases X, Y and Z.

Gas X absorbs Q isothermally, gas Y isovolumetrically and gas Z isobarically.

Complete the table by inserting the words ‘positive’, ‘zero’ or ‘negative’ for the work done W, the change in internal energy ΔU and the temperature change ΔT for each gas.

	$W$	$∆ U$	$∆ T$
X
Y
Z

Answer:

X: Isothermal = constant temperature, no change in internal energy
- Temperature: $∆ T = 0$
- Internal energy: $∆ T \propto ∆ U$ , so, $∆ U = 0$
- Work done: $Q = ∆ U + W \Rightarrow Q = + W$

Y: Isovolumetric = constant volume, no work done
- Work done: $W \propto ∆ V$ , so, $W = 0$
- Internal energy: $Q = ∆ U + W \Rightarrow Q = + ∆ U$
- Temperature: $∆ T \propto ∆ U$ , so, $∆ T > 0$

Z: Isobaric = constant pressure
- Work done: $∆ p = 0$ , so $W = p ∆ V$ , so $W > 0$
- Internal energy: $Q = ∆ U + W$ , so $∆ U > 0$
- Temperature: $∆ T \propto ∆ U$ , so $∆ T > 0$

	$W$	$∆ U$	$∆ T$
X	positive	0	0
Y	0	positive	positive
Z	positive	positive	positive

Worked example

A heat engine operates on the cycle shown in the pressure-volume diagram. One step in the cycle consists of an isothermal expansion of an ideal gas from state A of volume V to state B of volume 2V.

2-4-6-entropy-in-a-heat-engine-worked-example

On the graph, complete the cycle ABCA by drawing curves to show

a change at constant volume from state B to state C
an adiabatic compression from state C to state A

Answer:

Constant volume = no work done
Next step is a compression (where pressure increases), so this step should involve a pressure drop
- Hence, B to C: line drawn vertically down

Adiabatic = no heat supplied or removed, compression = work is done on the gas, volume decreases
- Hence, C to A: line curves up to meet A

2-4-6-entropy-in-a-heat-engine-worked-example-ma

Worked example

An ideal monatomic gas $(γ = \frac{5}{3})$ expands adiabatically from a state with pressure 7.5 × 10⁵ Pa and volume 1.8 × 10⁻³ m³ to a state of volume 4.2 × 10⁻³ m³.

Calculate the new pressure of the gas.

Answer:

For an ideal monatomic gas undergoing an adiabatic change:

$p V^{\frac{5}{3}} = C$

$p_{1} {V_{1}}^{\frac{5}{3}} = p_{2} {V_{2}}^{\frac{5}{3}}$

Where:
- Initial pressure, $p_{1}$ = 7.5 × 10⁵ Pa
- Final pressure = $p_{2}$
- Initial volume, $V_{1}$ = 1.8 × 10⁻³ m³
- Final volume, $V_{2}$ = 4.2 × 10⁻³ m³

$p_{2} = p_{1} {(\frac{V_{1}}{V_{2}})}^{\frac{5}{3}}$

$p_{2} = (7.5 \times 10^{5}) \times {(\frac{1.8 \times 10^{- 3}}{4.2 \times 10^{- 3}})}^{\frac{5}{3}}$

New pressure: $p_{2}$ = 1.8 × 10⁵ Pa

Thermodynamic Processes (AQA A Level Physics)

Revision Note