Non-flow processes (A-level only)

What you'll learn in this topic:

What a "non-flow" process is and how to apply the First Law of Thermodynamics to it.
The physical meanings and mathematical equations for four specific changes: constant volume, constant pressure, isothermal, and adiabatic.
How to sketch and interpret these processes on a pressure-volume ( $p - V$ ) diagram.

A non-flow process simply means we are looking at a fixed mass of gas trapped within a closed boundary—like gas sealed inside a cylinder with a piston. No gas flows in, and no gas flows out.

To understand how the gas behaves when we heat it, cool it, squash it, or let it expand, we rely heavily on the First Law of Thermodynamics.

Definition

The First Law of Thermodynamics

The First Law is essentially the principle of conservation of energy applied to a gas. For the AQA specification, the equation is written as:

\Delta U + W

$Q$ is the heat energy supplied to the gas.
$ΔU\Delta U$ is the increase in the internal energy of the gas (which corresponds to an increase in temperature).
$W$ is the work done by the gas when it expands.

If heat is removed from the gas, $Q$ is negative. If the gas is compressed, work is done on the gas, so $W$ is negative. If the gas cools down, $ΔU\Delta U$ is negative.

Let's look at the four distinct ways we can change the state of this trapped gas.

1. Constant volume change (Isochoric)

Imagine heating a gas trapped in a rigid, thick-walled metal box. The walls cannot move, so the volume of the gas remains completely constant.

Because the volume cannot change ( $ΔV=0\Delta V = 0$ ), the gas cannot expand. If it cannot expand, it cannot do any work against its surroundings. Therefore, $W = 0$ .

Applying the First Law ( $\Delta U + W$ ), we get:

\Delta U

All the heat energy you supply to the gas goes directly into increasing its internal energy. The molecules move faster, and the temperature and pressure of the gas rise.

2. Constant pressure change (Isobaric)

Now imagine the gas is in a cylinder with a free-moving, frictionless piston. As you heat the gas, it tries to increase in pressure. However, because the piston is free to slide outwards, the gas expands just enough to keep the pressure inside balanced with the constant pressure outside.

Because the gas expands, it is pushing a force over a distance. It is doing work. We can calculate this work using the constant pressure $p$ and the change in volume $ΔV\Delta V$ :

p\Delta V

Applying the First Law, the heat supplied goes into doing two things at once: raising the internal energy (making it hotter) and doing work to push the piston out:

\Delta U + p\Delta V

Example

Heating a gas at constant pressure

A fixed mass of gas is heated at a constant pressure of $1.0 \times 10^5 \text{ Pa}$ . Its volume expands from $\text{ m}^3$ to $\text{ m}^3$ . The total heat energy supplied to the gas is $\text{ J}$ . Calculate the change in internal energy of the gas.

Calculate the work done by the gas: Use the formula for work done at constant pressure. $J\begin{aligned} W &= p\Delta V \\ W &= 1.0 \times 10^5 \times (0.035 - 0.020) \\ W &= 1.0 \times 10^5 \times 0.015 = 1500 \text{ J} \end{aligned}$
Apply the First Law of Thermodynamics: Substitute your values into $\Delta U + W$ and rearrange to find $ΔU\Delta U$ . $J\begin{aligned} 2500 &= \Delta U + 1500 \\ \Delta U &= 2500 - 1500 = 1000 \text{ J} \end{aligned}$

3. Isothermal change

An isothermal change is one where the temperature of the gas remains perfectly constant throughout.

For an ideal gas, the internal energy depends only on its absolute temperature. If the temperature doesn't change, the internal energy doesn't change. Therefore, $ΔU=0\Delta U = 0$ .

Applying the First Law ( $\Delta U + W$ ), we get:

Q = W

If the gas expands and does work ( $W$ is positive), it must be absorbing an equal amount of heat from its surroundings ( $Q$ is positive) to prevent itself from cooling down. This is usually achieved by letting the gas expand very slowly in a cylinder with thin, highly conductive walls, so heat has plenty of time to transfer in.

Because temperature is constant, we can look at the ideal gas equation ( $p V = n R T$ ). Since $n$ , $R$ , and $T$ are all constants, we get Boyle's Law:

\text{constant}

4. Adiabatic change

An adiabatic change is the exact opposite of an isothermal one in terms of heat flow. In an adiabatic process, no heat enters or leaves the system. Therefore, $Q = 0$ .

This usually happens if the cylinder is heavily insulated, or if the change happens so fast that there is simply no time for heat to transfer (like the rapid compression in a car engine cylinder).

Applying the First Law ( $\Delta U + W$ ) with $Q = 0$ :

\Delta U + W \quad \Rightarrow \quad \Delta U = -W

If the gas expands rapidly, it does work ( $W$ is positive). Because no heat can enter to replace the lost energy, the internal energy drops ( $ΔU\Delta U$ is negative), meaning the gas cools down. Conversely, rapid compression causes the gas to heat up dramatically.

The relationship between pressure and volume for an adiabatic change is given by:

pV^\gamma = \text{constant}

Definition

The adiabatic constant

The symbol $γ\gamma$ (gamma) is the ratio of the specific heat capacities of the gas. You don't need to calculate it from scratch; the exam board will provide its value in the question if you need it (for air, $γ\gamma$ is usually around 1.4).

Key Idea

Isothermal vs. Adiabatic on a p-V diagram

Both isothermal and adiabatic expansions cause the pressure to drop as volume increases, but they look different on a $p - V$ graph.

The isothermal curve follows $\propto \frac{1}{V}$ .
The adiabatic curve follows $\propto \frac{1}{V^\gamma}$ . Since $γ\gamma$ is greater than 1, the adiabatic curve is noticeably steeper.

p-V diagram comparing isothermal and adiabatic expansion

Example

Adiabatic compression in an engine

Air in a diesel engine cylinder is compressed adiabatically. Its initial pressure is $1.0 \times 10^5 \text{ Pa}$ and its initial volume is $5.0 \times 10^{-4} \text{ m}^3$ . It is rapidly compressed to a volume of $2.5 \times 10^{-5} \text{ m}^3$ .

Calculate the final pressure of the air. (Assume $γ=1.4\gamma = 1.4$ )

State the relationship for an adiabatic change: Because $pVγ=constantpV^\gamma = \text{constant}$ , we can equate the initial and final states:

p_1 V_1^\gamma = p_2 V_2^\gamma

Rearrange to solve for the final pressure ( $p_2$ ):

p_2 = p_1 \left( \frac{V_1}{V_2} \right)^\gamma

Substitute the values: $Pa\begin{aligned} p_2 &= 1.0 \times 10^5 \times \left( \frac{5.0 \times 10^{-4}}{2.5 \times 10^{-5}} \right)^{1.4} \\ p_2 &= 1.0 \times 10^5 \times (20)^{1.4} \\ p_2 &= 1.0 \times 10^5 \times 66.29 = 6.6 \times 10^6 \text{ Pa} \end{aligned}$

Common Mistake

Forgetting the power of gamma

When typing $p1V1γ=p2V2γp_1 V_1^\gamma = p_2 V_2^\gamma$ into your calculator, it is very easy to accidentally calculate $(pV)γ(pV)^\gamma$ instead. Remember that the exponent $γ\gamma$ applies only to the volume $V$ , not the pressure! Grouping the volumes into a fraction $(V1V2)γ\left(\frac{V_1}{V_2}\right)^\gamma$ as shown in the example above is the safest way to avoid this calculator error.

Exam technique

In the exam

Watch out for keywords: If a question says the process happens "slowly", assume it is isothermal. If it says it happens "rapidly", assume it is adiabatic.
Check your signs: When applying $\Delta U + W$ , double-check whether the gas is expanding ( $+ W$ ) or being compressed ( $- W$ ), and whether it's gaining heat ( $+ Q$ ) or losing heat ( $- Q$ ).
Use the formula booklet: You don't need to memorise $pVγ=constantpV^\gamma = \text{constant}$ or $p\Delta V$ , as they are provided in your data sheet. But you do need to know which physical situation each formula applies to.

Self review

Check yourself

Can you state why the work done during a constant volume process is zero?
If a gas is compressed adiabatically, does its temperature increase, decrease, or stay the same?
Which curve on a $p - V$ diagram is steeper: an isothermal expansion or an adiabatic expansion?

Non-flow processes (A-level only)

What you'll learn in this topic:

What a "non-flow" process is and how to apply the First Law of Thermodynamics to it.
The physical meanings and mathematical equations for four specific changes: constant volume, constant pressure, isothermal, and adiabatic.
How to sketch and interpret these processes on a pressure-volume ( $p - V$ ) diagram.

A non-flow process simply means we are looking at a fixed mass of gas trapped within a closed boundary—like gas sealed inside a cylinder with a piston. No gas flows in, and no gas flows out.

To understand how the gas behaves when we heat it, cool it, squash it, or let it expand, we rely heavily on the First Law of Thermodynamics.

Definition

The First Law of Thermodynamics

The First Law is essentially the principle of conservation of energy applied to a gas. For the AQA specification, the equation is written as:

\Delta U + W

$Q$ is the heat energy supplied to the gas.
$ΔU\Delta U$ is the increase in the internal energy of the gas (which corresponds to an increase in temperature).
$W$ is the work done by the gas when it expands.

If heat is removed from the gas, $Q$ is negative. If the gas is compressed, work is done on the gas, so $W$ is negative. If the gas cools down, $ΔU\Delta U$ is negative.

Let's look at the four distinct ways we can change the state of this trapped gas.

1. Constant volume change (Isochoric)

Imagine heating a gas trapped in a rigid, thick-walled metal box. The walls cannot move, so the volume of the gas remains completely constant.

Because the volume cannot change ( $ΔV=0\Delta V = 0$ ), the gas cannot expand. If it cannot expand, it cannot do any work against its surroundings. Therefore, $W = 0$ .

Applying the First Law ( $\Delta U + W$ ), we get:

\Delta U

All the heat energy you supply to the gas goes directly into increasing its internal energy. The molecules move faster, and the temperature and pressure of the gas rise.

2. Constant pressure change (Isobaric)

Because the gas expands, it is pushing a force over a distance. It is doing work. We can calculate this work using the constant pressure $p$ and the change in volume $ΔV\Delta V$ :

p\Delta V

Applying the First Law, the heat supplied goes into doing two things at once: raising the internal energy (making it hotter) and doing work to push the piston out:

\Delta U + p\Delta V

Example

Heating a gas at constant pressure

Calculate the work done by the gas: Use the formula for work done at constant pressure. $J\begin{aligned} W &= p\Delta V \\ W &= 1.0 \times 10^5 \times (0.035 - 0.020) \\ W &= 1.0 \times 10^5 \times 0.015 = 1500 \text{ J} \end{aligned}$
Apply the First Law of Thermodynamics: Substitute your values into $\Delta U + W$ and rearrange to find $ΔU\Delta U$ . $J\begin{aligned} 2500 &= \Delta U + 1500 \\ \Delta U &= 2500 - 1500 = 1000 \text{ J} \end{aligned}$

3. Isothermal change

An isothermal change is one where the temperature of the gas remains perfectly constant throughout.

For an ideal gas, the internal energy depends only on its absolute temperature. If the temperature doesn't change, the internal energy doesn't change. Therefore, $ΔU=0\Delta U = 0$ .

Applying the First Law ( $\Delta U + W$ ), we get:

Q = W

Because temperature is constant, we can look at the ideal gas equation ( $p V = n R T$ ). Since $n$ , $R$ , and $T$ are all constants, we get Boyle's Law:

\text{constant}

4. Adiabatic change

An adiabatic change is the exact opposite of an isothermal one in terms of heat flow. In an adiabatic process, no heat enters or leaves the system. Therefore, $Q = 0$ .

Applying the First Law ( $\Delta U + W$ ) with $Q = 0$ :

\Delta U + W \quad \Rightarrow \quad \Delta U = -W

The relationship between pressure and volume for an adiabatic change is given by:

pV^\gamma = \text{constant}

Definition

The adiabatic constant

Key Idea

Isothermal vs. Adiabatic on a p-V diagram

Both isothermal and adiabatic expansions cause the pressure to drop as volume increases, but they look different on a $p - V$ graph.

The isothermal curve follows $\propto \frac{1}{V}$ .
The adiabatic curve follows $\propto \frac{1}{V^\gamma}$ . Since $γ\gamma$ is greater than 1, the adiabatic curve is noticeably steeper.

p-V diagram comparing isothermal and adiabatic expansion

Example

Adiabatic compression in an engine

Calculate the final pressure of the air. (Assume $γ=1.4\gamma = 1.4$ )

State the relationship for an adiabatic change: Because $pVγ=constantpV^\gamma = \text{constant}$ , we can equate the initial and final states:

p_1 V_1^\gamma = p_2 V_2^\gamma

Rearrange to solve for the final pressure ( $p_2$ ):

p_2 = p_1 \left( \frac{V_1}{V_2} \right)^\gamma

Substitute the values: $Pa\begin{aligned} p_2 &= 1.0 \times 10^5 \times \left( \frac{5.0 \times 10^{-4}}{2.5 \times 10^{-5}} \right)^{1.4} \\ p_2 &= 1.0 \times 10^5 \times (20)^{1.4} \\ p_2 &= 1.0 \times 10^5 \times 66.29 = 6.6 \times 10^6 \text{ Pa} \end{aligned}$

Common Mistake

Forgetting the power of gamma

Exam technique

In the exam

Watch out for keywords: If a question says the process happens "slowly", assume it is isothermal. If it says it happens "rapidly", assume it is adiabatic.
Check your signs: When applying $\Delta U + W$ , double-check whether the gas is expanding ( $+ W$ ) or being compressed ( $- W$ ), and whether it's gaining heat ( $+ Q$ ) or losing heat ( $- Q$ ).
Use the formula booklet: You don't need to memorise $pVγ=constantpV^\gamma = \text{constant}$ or $p\Delta V$ , as they are provided in your data sheet. But you do need to know which physical situation each formula applies to.

Self review

Check yourself

Can you state why the work done during a constant volume process is zero?
If a gas is compressed adiabatically, does its temperature increase, decrease, or stay the same?
Which curve on a $p - V$ diagram is steeper: an isothermal expansion or an adiabatic expansion?

Non-flow processes (A-level only)

1. Constant volume change (Isochoric)

2. Constant pressure change (Isobaric)

3. Isothermal change

4. Adiabatic change

Non-flow processes (A-level only) Revision Guide

Non-flow processes (A-level only)

1. Constant volume change (Isochoric)

2. Constant pressure change (Isobaric)

3. Isothermal change

4. Adiabatic change

Non-flow processes (A-level only) Revision Guide