# First Law of Thermodynamics: Definition & Example

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The laws of thermodynamics are some of the most important laws in all of physics, and understanding how to apply each one of them is a crucial skill for any physics student.

The first law of thermodynamics is essentially a statement of the conservation of energy, but there are many uses for this specific formulation you’ll need to understand if you want to solve problems involving things like heat engines.

Learning what adiabatic, isobaric, isochoric and isothermal processes are, and how to apply the first law of thermodynamics in these situations, helps you mathematically describe the behavior of a thermodynamic system as it evolves in time.

## Internal Energy, Work and Heat

The first law of thermodynamics – like the other laws of thermodynamics – requires an understanding of some key terms. The internal energy of a system is a measure of the total kinetic energy and potential energy of an isolated system of molecules; intuitively, this just quantifies the amount of energy contained in the system.

Thermodynamic work is the amount of work a system does on the environment, for example, by the heat-induced expansion of a gas pushing a piston outwards. This is an example of how heat energy in a thermodynamic process can be converted into mechanical energy, and it is the core principle behind the operation of many engines.

In turn, heat or thermal energy is the thermodynamic energy transfer between two systems. When two thermodynamic systems are in contact (not separated by an insulator) and are at different temperatures, heat transfer occurs in this way, from the hotter body towards the colder one. All of these three quantities are forms of energy, and so are measured in joules.

## The First Law of Thermodynamics

The first law of thermodynamics states that the heat added to the system adds to its internal energy, while the work done by the system reduces the internal energy. In symbols, you use ∆U to denote the change in internal energy, Q to stand for heat transfer and W for the work done by the system, and so the first law of thermodynamics is:

∆U = Q - W

The first law of thermodynamics therefore relates the internal energy of the system to two forms of energy transfer that can take place, and as such it’s best thought of as a statement of the law of conservation of energy.

Any changes to the internal energy of the system come from either heat transfer or work done, with heat transfer to the system and work done on the system increasing internal energy, and heat transfer from the system and work done by it reducing the internal energy. The expression itself is easy to use and understand, but finding valid expressions for the heat transfer and work done to use in the equation can be challenging in some cases.

## Example of the First Law of Thermodynamics

Heat engines are a common type of thermodynamic system that can be used to understand the basics of the first law of thermodynamics. Heat engines essentially convert heat transfer into usable work through a four-step process that involves heat being added to a reservoir of gas to increase its pressure, it expanding in volume as a result, the pressure reducing as heat is extracted from the gas and finally the gas being compressed (i.e., reduced in volume) as work is done on it to bring it back into the original state of the system and start the process over again.

This same system is often idealized as a Carnot cycle, in which all of the processes are reversible and involve no change in entropy, with a stage of isothermal (i.e., at the same temperature) expansion, a stage of adiabatic expansion (with no heat transfer), a stage of isothermal compression and a stage of adiabatic compression to bring it back to the original state.

Both of these processes (the idealized Carnot cycle and the heat engine cycle) are usually plotted on a PV diagram (also called a pressure-volume plot), and these two quantities are related by the ideal gas law, which states:

PV = nRT

Where P = pressure, V = volume, n = the number of moles of the gas, R = the universal gas constant = 8.314 J mol−1 K−1 and T = temperature. In combination with the first law of thermodynamics, this law can be used to describe the stages of a heat engine cycle. Another useful expression gives the internal energy U for an ideal gas:

U = \frac{3}{2}nRT

## The Heat Engine Cycle

A simple approach to analyzing the heat engine cycle is to imagine the process taking place on a straight-sided box in the PV plot, with each stage either taking place at a constant pressure (an isobaric process) or a constant volume (an isochoric process).

First, starting from V1, heat is added and the pressure rises from P1 to P2, and since the volume remains constant, you know that the work done is zero. To tackle this stage of the problem, you make two versions of the ideal gas law for the first and second state (remembering that V and n are constant): P1V1 = nRT1 and P2V1 = nRT2, and then subtract the first from the second to get:

V_1 (P_2-P_1) = nR (T_2 -T_1)

Solving for the change in temperature gives:

(T_2 - T_1) = \frac{ V_1 (P_2 - P_1)}{nR}

If you look for the change in internal energy, you can then insert this into the expression for internal energy U to get:

\begin{aligned} ∆U &= \frac{3}{2}nR∆T \\ \\ &=\frac{3}{2} nR \bigg(\frac{ V_1 (P_2 - P_1)}{nR}\bigg) \\ \\ &=\frac{3}{2} V_1 (P_2 -P_1) \end{aligned}

For the second stage in the cycle, the volume of the gas expands (and so the gas does work) and more heat is added in the process (to maintain a constant temperature). In this case, the work W done by the gas is simply the change in volume multiplied by the pressure P2, which gives:

W = P_2 (V_2 -V_1)

And the change in temperature is found with the ideal gas law, as before (except keeping P2 as a constant and remembering that the volume changes), to be:

T_2 - T_1 = \frac{ P_2 (V_2 - V_1)}{nR}

If you want to find out the exact amount of heat added, you can use the specific heat equation at a constant pressure to find it. However, you can directly calculate the internal energy of the system at this point as before:

\begin{aligned} ∆U &= \frac{3}{2}nR∆T \\ \\ &=\frac{3}{2}nR\bigg(\frac{ P_2 (V_2 – V_1)}{nR}\bigg) \\ \\ &=\frac{3}{2} P_2 (V_2 – V_1) \end{aligned}

The third stage is essentially the reverse of the first stage, so the pressure decreases at a constant volume (this time V2), and heat is extracted from the gas. You can work through the same process based on the ideal gas law and the equation for the internal energy of the system to get:

∆U = -\frac{3}{2} V_2 (P_2 - P_1)

Note the leading minus sign this time because the temperature (and therefore the energy) has decreased.

Finally, the last stage sees the volume decrease as work is done on the gas and heat extracted in an isobaric process, producing a very similar expression to last time for the work, except with a leading minus sign:

W = -P_1 (V_2 -V_1)

The same calculation gives the change in internal energy as:

∆U = -\frac{3}{2} P_1 (V_2 - V_1)

## Other Laws of Thermodynamics

The first law of thermodynamics is arguably the most practically useful for a physicist, but the other three major laws are worth a brief mention too (although they’re covered in more detail in other articles). The zeroth law of thermodynamics states that if system A is in thermal equilibrium with system B, and system B is in equilibrium with system C, then system A is in equilibrium with system C.

The second law of thermodynamics states that the entropy of any closed system tends to increase.

Finally, the third law of thermodynamics states that the entropy of a system approaches a constant value as the temperature approaches absolute zero.