# Work in Thermodynamics

Why does steam make things move? Why will an aerosol explode violently when left in a hot car? The simple answer: Thermodynamics is interesting! Heat as energy is complex enough to need its entire branch of physics just to explain its relationship with other energies! Thermodynamics also describes how heat acts in a system, and how its energy transfers into other forms of energy. Work in thermodynamics is, therefore, an interesting topic too, but just how does it... work? What variables do we need to look out for, and what is a thermodynamic process? Let's find out.

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## The Definition of Work in Thermodynamics

We usually think of work in the mechanical sense, where energy is transferred from one place to another through a force. We can also think of this concept in terms of thermodynamics, where energy in a thermodynamic system is transferred into another thermodynamic system.

Work in the field of physics is defined as the energy that is transferred when a force exerts on an object with mass over a certain distance.

Before we go a bit deeper into work in terms of thermodynamics, let's first look at thermodynamics as a whole, and what exactly it is. Thermodynamics is an entire section of physics that involves heat. Specifically, how heat interacts and transfers between other different kinds of energy, and overall the relationship heat has with all forms of energy. We know that work is energy transference, and in thermodynamics, work is only ever done in the mechanical sense.

The first demonstration of work in a thermodynamic system was performed in an experiment devised by James Joule in 1845. James suspended a paddle wheel in a body of water, with a rope and weight attached outside of the water, with the ability to drop as the wheel spun. The wheel was spun, and James noticed that the increase in water temperature due to the mechanical motion had a direct relationship with the height at which the weight on the rope fell. From this information, James Joule was able to determine what the mechanical equivalent of heat is, at $$4.4\,\mathrm{joules\, per\, calorie}$$, with of course the unit of energy being named after James himself.

A simple diagram of the setup of James Joules' experiment, McGill

## Work in Thermodynamics Processes

So we now know how work relates in terms to thermodynamics, but what is a thermodynamic process? A thermodynamic process is quite simply the kind of way work can be done between thermodynamic systems. A thermodynamic system is a confined space with matter inside in which thermodynamic processes will take place. There are at minimum a handful of variables to take into consideration in a thermodynamic system, even more so in the transference of energy between two or more. The changes in these variables as work is done are what differentiates one thermodynamic process from another. Let's look at a few different thermodynamic processes in the context of a gas as our thermodynamic system.

An adiabatic process occurs when the only energy transferred to or from a system is via work done on or by the system. Energy is transferred outward for gas doing work and energy is transferred into the gas when external work is done on it.

An isobaric process occurs when the pressure of the gas in a system is constant. Most often, chemical reactions will take place at constant pressure, making them isobaric processes.

An isochoric process occurs when the volume of gas in a thermodynamic system is constant. Gases burnt in a car's engine are a good example of an isochoric process. These gases usually change in temperature and pressure but their volume is constant.

An isothermal process, as this pattern goes, will in turn be a system in which the temperature of the gas is constant. Melting ice remains at a constant temperature and is an example of an isothermal process.

## Useful Work in Thermodynamics

Before we cover what useful work in thermodynamics is, let's first go over what useful work is. Useful work is described as the portion of work done that directly transfers useful energy. A contrast to this is the waste work done. In a system, there will always be some energy that is dissipated in another form, that is to say, there will never be a perfect transfer of 100% useful energy. The work done that is associated with this transfer of waste energy is waste work done.

So, useful work in thermodynamics can be quite simply described as the useful output from a thermodynamic system, the difference between the input work in the system, and the waste work.

$$\mathrm{Useful\:work=Total\:work - Waste\:work}$$

A thermodynamic system has a total work output of $$200\,000\,\mathrm{J}$$, and $$5\%$$ of that is waste. How much is useful work?

In this, we can simply take $$5\%$$ of our total work, and subtract that from our total work to get our value for useful work as follows:

$$200\,000 \times 0.05 = 10\,000\,\mathrm{J}$$

$$200\,000 - 10\,000 = 190\,000\,\mathrm{J}$$

This means that $$190\,000\,\mathrm{J}$$ of useful work is done by the system.

## The Formula for Work in Thermodynamics

Work occurs in many different senses, as we have gone over. This means that many different equations are used to calculate for work. In a thermodynamic system, however, this formula is the most used and most widely applicable:

$$W=\Delta U-Q$$

Where $$W$$ is the work done on or by a system, $$Q$$ is the heat that enters or leaves a system, and $$\Delta U$$ is the change in the internal energy of a system. The formula above is the mathematical expression of the first law of thermodynamics. This equation defines all of these variables and their relationship to one another in a thermodynamic system. For example, we now know that if we wish for thermodynamic work to remain constant, and the internal energy declines, we must then increase the amount of heat entering the system.

An arguably equally as important concept in terms of thermodynamic work is the relationship between pressure and volume in a thermodynamic system. These variables are used directly in solving for work, as shown here:

$$W=-P\Delta V$$

Where $$P$$ is the pressure externally acting on the system, and $$\Delta V$$ is the change in volume as it is doing work. If the system does work externally $$W<0$$ and if an external force does work on the system then $$W>0$$. Note that this method only works for an isobaric process with constant pressure $$P$$. Essentially, the work done on a system at constant pressure, in which the volume is changing, is the area under a $$P-V$$ graph. This is illustrated by the graph below.

The area under a graph of P vs V is the work done by a thermodynamic system at constant pressure, Wikimedia Commons CC BY-SA 4.0

## Work in Thermodynamics Examples

Now that we've seen what equations are used to calculate work done in a thermodynamic system, let's try putting them into practice.

Consider a thermodynamic system of a gas contained in a cylinder by a piston as shown in the figure below. The piston is moved inward, compressing the gas in the process. The volume of the gas decreases by $$50\,\mathrm{m^{3}}$$ as the piston applies a constant pressure of $$10\,\mathrm{N/m^{2}}$$. What is the work done on the gas?

Gas contained in a cylinder by a moveable piston, adapted from image by Dpumroy CC BY-SA 3.0

To solve this, we first have to look at our variables and see which equations have these two. We know we have pressure and volume, so we need our second equation. We also know that work is done on the gas so it gains energy and $$W>0$$. Putting our numbers in the equation gives us this:

\begin{align} W&=-P\Delta V \\ &=-(10\,\mathrm{N\,m^{2}})\times(-50\,\mathrm{m^{3}}) \\ &=+\,500\,\mathrm{N\,m} \\ &=+\,500\,\mathrm{J} \end{align}

Notice that our change in volume is negative since the gas is being compressed. Our final answer is $$+\,500\,\mathrm{J}$$ which makes sense as work is being done on the gas.

The internal energy of a thermodynamic system increases by $$50 \,\mathrm{J}$$, with the work done on the system being equal to $$25\,\mathrm{J}$$. How much heat has entered the system in Kelvin?

Once again, we need to look at the equations available to us and determine which one uses the variables we have available. Upon doing this, we can see the first equation does, but we're going to need to do some tweaking. In this equation, work done is the focus, so we need to rearrange it to solve for temperature, giving us this equation instead:

$$Q=\Delta U-W$$

The work done is positive since work is done on the system and the change in internal energy is positive. After this, it's as simple as our previous example, plug in the numbers we have:

$$Q=50\,\mathrm{J}-25\,\mathrm{J}=25\,\mathrm{J}$$

Which added together gives us a heat of $$25\,\mathrm{J}$$ entering the system.

## Work in Thermodynamics - Key takeaways

• Work in thermodynamics is the energy transferred from one thermodynamic system into another thermodynamic system, often a transfer from heat energy to another form of energy.
• There are many different kinds of thermodynamic processes, including adiabatic, isobaric, isochoric, and isothermal.
• The formulas for work done in a thermodynamic system are $$W=\Delta U-Q$$ and $$W=-P\Delta V$$.

#### Flashcards inWork in Thermodynamics 15

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What does work mean in thermodynamics?

Work in thermodynamics means the energy that is used in the transference from one thermodynamic system to another.

How to calculate work done in thermodynamics

The equations used to calculate work done are  and  .

What is an example of work in thermodynamics?

An example of work in thermodynamics would occur in an adiabatic thermodynamic process, where no heat loss or gain occurs in the system. This means the work done is only a product of the physical changes in the system, such as changes in volume and pressure.

What are different types of processes in thermodynamics?

The different types of processes in thermodynamics are adiabatic, isobaric, isometric, and isothermic.

## Test your knowledge with multiple choice flashcards

Who first demonstrated work in a thermodynamic system?

When was work first demonstrated in a thermodynamic system?

Which of the following equations is correct regarding the relationship between the work done, the heat transfer and internal energy of a thermodynamic system?

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