Thermodynamic Processes (DP IB Physics)

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Katie M

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17 December 2024

Thermodynamic Processes

The four main thermodynamic processes are
- Isovolumetric $(W = 0)$
- Isobaric $(∆ p = 0)$
- Isothermal $(∆ T = 0)$
- Adiabatic $(∆ Q = 0)$

Constant pressure (isobaric)

An isobaric 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

Representing an isobaric process on a p-V diagram

Constant volume (isovolumetric)

An isovolumetric 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:

$Q = ∆ U + W = ∆ U + 0$

$Q = ∆ U$

Representing an isovolumetric process 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:

$Q = ∆ U + W = 0 + W$

$Q = W$

Representing an isothermal process 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

Representing an adiabatic process on a p-V diagram

Entropy in Thermodynamic Processes

At a constant temperature T, the change in entropy is related to heat by

$∆ S = \frac{∆ Q}{T}$

When heat is gained by a system $(∆ Q > 0)$ , entropy increases $(∆ S > 0)$
When heat is lost from a system $(∆ Q < 0)$ , entropy decreases $(∆ S < 0)$
For a reversible process $(∆ Q = 0)$ that returns the system to its original state $(∆ S = 0)$

Process	Heat gained or lost, ΔQ	Change in entropy, ΔS
Isothermal	Expansion	$∆ Q > 0$ Heat gained = work done by gas	$∆ S > 0$ Increases
Compression	$∆ Q < 0$ Heat lost = work done on gas	$∆ S < 0$ Decreases
Isobaric	Expansion	$∆ Q > 0$ Heat gained = increase in internal energy + work done by gas	$∆ S > 0$ Increases
Compression	$∆ Q < 0$ Heat lost = decrease in internal energy + work done on gas	$∆ S < 0$ Decreases
Isovolumetric	Pressure rise	$∆ Q > 0$ Heat gained due to temperature rise	$∆ S > 0$ Increases
Pressure drop	$∆ Q < 0$ Heat lost due to temperature drop	$∆ S < 0$ Decreases
Adiabatic	Expansion	$∆ Q = 0$ Pressure & temperature decrease with no heat gained or lost	$∆ S = 0$ No change
Compression	$∆ Q = 0$ Pressure & temperature increase with no heat gained or lost	$∆ S = 0$ No change

Process

Heat gained or lost, ΔQ

Change in entropy, ΔS

Isothermal

Expansion

$∆ Q > 0$

Heat gained = work done by gas

$∆ S > 0$

Increases

Compression

$∆ Q < 0$

Heat lost = work done on gas

$∆ S < 0$

Decreases

Isobaric

Expansion

$∆ Q > 0$

Heat gained = increase in internal energy + work done by gas

$∆ S > 0$

Increases

Compression

$∆ Q < 0$

Heat lost = decrease in internal energy + work done on gas

$∆ S < 0$

Decreases

Isovolumetric

Pressure rise

$∆ Q > 0$

Heat gained due to temperature rise

$∆ S > 0$

Increases

Pressure drop

$∆ Q < 0$

Heat lost due to temperature drop

$∆ S < 0$

Decreases

Adiabatic

Expansion

$∆ Q = 0$

Pressure & temperature decrease with no heat gained or lost

$∆ S = 0$

No change

Compression

$∆ Q = 0$

Pressure & temperature increase with no heat gained or lost

$∆ S = 0$

No change

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.

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

an isovolumetric change from state B to state C
an adiabatic compression from state C to state A

(b) State and explain at which point in the cycle ABCA the entropy of the gas is the largest.

Answer:

(a)

Isovolumetric = 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

(b)

Entropy and heat (at a constant T) are related by

$∆ S = \frac{∆ Q}{T}$

From state A to state B:
- In an isothermal expansion, entropy increases
- Because T = constant but the volume increases so work is done by gas, ΔQ > 0 so ΔS > 0

From state B to state C:
- In an isovolumetric change where pressure decreases, entropy decreases
- Because temperature decreases, so energy has been removed, ΔQ < 0 so ΔS < 0

From state C to state A:
- In an adiabatic compression, entropy is constant
- Because it is an adiabatic process, ΔQ = 0 so ΔS = 0

Therefore, entropy is greatest at B

Adiabatic Processes

Adiabatic processes in monatomic ideal gases can be modelled by the equation

$p V^{\frac{5}{3}} = c o n s t a n t$

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 for monatomic ideal gases

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

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

Worked Example

An ideal monatomic gas 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

Worked Example

An ideal monatomic gas is compressed adiabatically from a state with volume 3.1 × 10⁻³ m³ and temperature 590 K to a state of volume 2.1 × 10⁻³ m³.

Calculate the new temperature of the gas.

Answer:

For an ideal monatomic gas undergoing an adiabatic change:

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

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

From the ideal gas law:

$p V = n R T \Rightarrow p = \frac{n R T}{V}$

$(\frac{n R T_{1}}{V_{1}}) {V_{1}}^{\frac{5}{3}} = (\frac{n R T_{2}}{V_{2}}) {V_{2}}^{\frac{5}{3}}$

$T_{1} {V_{1}}^{\frac{2}{3}} = T_{2} {V_{2}}^{\frac{2}{3}}$

Where:
- Initial temperature, $T_{1}$ = 590 K
- Final temperature = $T_{2}$
- Initial volume, $V_{1}$ = 3.1 × 10⁻³ m³
- Final volume, $V_{2}$ = 2.1 × 10⁻³ m³

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

$T_{2} = 590 \times {(\frac{3.1 \times 10^{- 3}}{2.1 \times 10^{- 3}})}^{\frac{2}{3}}$

New temperature: $T_{2}$ = 765 K

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Previous:Second Law of ThermodynamicsNext:Heat Engines

Thermodynamic Processes (DP IB Physics)

Revision Note

Thermodynamic Processes

Constant pressure (isobaric)

Constant volume (isovolumetric)

Constant temperature (isothermal)

Constant thermal energy (adiabatic)

Entropy in Thermodynamic Processes

Adiabatic Processes

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