Topic 4: The First Law of Thermodynamics#
Introduction#
In general, a thermodynamic system is a collection of objects that is convenient to regard as a unit, and that may have the potential to exchange energy with its surroundings (e.g. popcorn kernels heated in a pot expand and does work as it exerts an upward force and moves the lid). Such a process, in which there are changes in the state of a thermodynamic system is a thermodynamic process. The first law of thermodynamics relates heat supplied, work done and internal energy. In fact, it is simply a statement of energy conservation:
Here, an isolated system is one which is not in any sort of contact with the outside world: it is not exerting or subjected to any forces, it is not supplying or being supplied with heat or any other form of energy. In this form, the law seems like a statement of the obvious. The first law is more useful if we permit some contact with the outside world
This can be expressed mathematically as
where \(\Delta U\) is the change in internal energy, \(Q\) is the heat supplied to the system, and \(W\) is the work done on the system. Note the sign conventions here: if the system actually radiates heat, \(Q\) < 0, and if the system does work on its surroundings, \(W\) < 0.
Heat and work#
Both heat and work are forms of energy. Heat is the energy that flows between two systems in thermal contact with each other if they are not in thermal equilibrium, i.e. heat flow results from a difference in temperature. We have seen that the temperature of a body is related to the average kinetic energy of its constituent atoms, so we can conclude that heat is energy transfer associated with random motion of atoms.
Work, on the other hand, is energy transfer associated with directed motion, and in particular with the application of a force. In one dimension, we can write
In the next section, we will see that we need to treat this integral with great care, because it is path dependent. Suppose I need to
move a box from my office (E43) to Clive Tadhunter’s office (E37), which is no the other side of the mid-corridor fire door. If I can simply
move it the 10 m or so down the corridor, I will have done some amount of work \(F \Delta x\) where \(\Delta x\) in this case is 10 m. However,
if the fire door is jammed shut because of some fault, and I have to talk down several flights of stairs to the Hicks SE exit, up Hounsfield Road, into the main entrance on D floor, up
the stairs to E floor, and along the
corridor to E37, I will have done considerably
more work to achieve the same end result.
The same is true of heat: the amount of heat I need to supply to raise the temperature of an object by a certain amount \(\Delta T\) depends on the way that I do it, in other words the path that the temperature takes from \(T\) to \(T + \Delta T\). This path dependence of heat and work turns out to be critical to practical applications of thermodynamics, as we shall see later in the course. The difference between the ordered motion of work and the disordered motion of heat is also very important, as we will see when we consider the Second Law and the concept of entropy.
Some examples of work#
In general, we have, for a force \(F\) applied over a small displacement \(\Delta x\), \(\Delta W = F \Delta x\). (This is assuming that the force and the displacement are parallel; if they are not we need to write \(\Delta W = {\textbf F} \cdot \Delta {\textbf x}\) where \(\textbf{F}\) and \(\Delta \textbf{x}\) are vectors).
For gases, it is usually more useful to work in terms of pressure: since \(f = PA\) where \(A\) is the area over which the pressure acts, the work done by pressure \(P\) is \( PA \Delta x = P \Delta V\), where \(\Delta V\) is the change in the volume of the gas. Note that the work done on the gas, which is what we usually for the First Law is equal to \(- P \Delta V\).
Other forms of work include stretching a spring, where \(\Delta W = f \Delta L\) where \(f\) is the first applied to the spring and \(\Delta L\) is the change in its length, expanding a liquid droplet (\(\Delta W = \gamma \Delta A\) where \(\gamma\) is the surface tension and \(\Delta A\) is the change in area), applying an electric field to a charged particle, and so on. In the rest of this course we shall usually be using \(-P \Delta V\), but it is important to remember that the First Law does not only apply to expanding gases!
Internal energy#
The internal energy of a system is the total energy of its constituent particles. This includes the translational, rotational and vibrational energy that we considered in Section Heat capacity of ideal gases and equipartition of energy, and also the potential energy due to forces between atoms, which we neglected in the ideal gas but which are obviously important in liquids and solids.
Unlike heat and work, internal energy is a function of state: it is determined only be the current properties of the system, not by its history. If I take a system that is in state A (say, a gas with a specified \(P, T\) and \(V\)) and by supplying heat and/or by doing work I take it to state B, the change in its internal energy is not path dependent: whatever combination of heat and work I applied, in whatever order, the change in the internal energy will be the same.
Example 4.1
One mole of a monatomic ideal gas, say argon, is enclosed in a contained which has one movable wall, so that the gas remains at the same pressure as the outside environment. (Neglect any frictional forces). The gas is heated at a constant pressure by an amount \(\Delta T\).
What is the work done on the gas?
What is the change in the internal energy of the gas?
What is the heat supplied to the gas?
What is the relation between the molar heat capacity at constant pressure, \(c_{P}\), and the molar heat at constant volume, \(c_{V}\)?
If the gas is heated at constant pressure, then since \(PV = RT\) (for one mole), the volume must change. The amount of work done by the gas is \(P \Delta V = R \Delta T\), so the amount of work done on the gas is \(-R \Delta T\). From Equation (16), we know that the change in the internal energy of the gas is \(\Delta U = \frac{3}{2} N_{A} k_{B} \Delta T = \frac{3}{2} R \Delta T\). (As this is a monatomic ideal gas, its internal energy is just the kinetic energy of its atoms: there is no rotational or vibrational degrees of freedom to worry about).
Therefore from the First Law we have \(Q = \Delta U - W = \frac{3}{2} R \Delta T - (- R \Delta T) = \frac{5}{2} R \Delta T\). But by definition we know \(Q = c_{P} \Delta T\) where \(c_{P}\) is the molar heat capacity at constant pressure, and at constant volume we know that \(\Delta U = Q = c_{V} \Delta T\) since no work is done. Therefore as we have previously deduced (Equation (21)) \(c_{V} = \frac{3}{2} R\) and so \(c_{P} = \frac{5}{2} R = c_{V} + R\). Note that although a more complicated gas, such as isobutane from Example 3.2, would have a larger value of \(c_{V}\), it would still be true that \(c_{P} = c_{V} + R\), since the work done by the gas pressure would be the same (provided that the gas could be treated as an ideal gas).
Cyclic process#
A process that eventually returns a system to its initial state is called a cyclic process. For such a process, the final state is the same as the initial state, so the total internal energy change most be zero. If a net quantity of work \(W\) is done on the system, an equal amount of work must have flowed into the system as heat \(Q\).
Every day your body (a thermodynamic system) goes through a cyclic process. Heat is added by metabolizing food, and your body does work in breathing, walking and other activies. If you return to the same state at the end of the day, the net change in your internal energy is zero i.e. \(\Delta U = 0\) so \(Q = -W\).
Example 4.2
One gram of water (1 cm\(^{3}\)) becomes 1671 cm\(^{3}\) of steam when boiled at a constant pressure of 1 atm (101 kPa). Compute (i) the work done vaporizing the water and (ii) its increase in internal energy. (Note the heat of vaporization at this pressure is \(L_{V} = 2.26 \times 10^{6}\) J/kg
The water does work \(P (V_{2} - V_{1}) = 1.69 \times 10^{2}\) J so the work done on the water is \(W = - 1.69 \times 10^{2}\) J. From Eqn (11) \(Q = m L_{V} = 2.26 \times 10^{3}\) J. The first law of thermodynamics holds for thermodynamic processes of all kinds, \(\Delta U = Q + W = 2.09 \times 10^{3}\) J. Over 90% of the heat remains in the system as an increase in internal energy, with the remainder leaving the system as it expands from liquid to vapour.
Example 4.3
Gas in a cylinder expands from 0.110 m\(^{3}\) to 0.320 m\(^{3}\). Heat flows into the gas to keep pressure constant at \(1.65 \times 10^{5}\) Pa. During the expansion, the total heat added is \(1.15 \times 10^{5}\) J.
Find the work done by the gas
Find the change of internal energy of the gas
Does it matter whether the gas is ideal?
Work done on the gas is \(-P dV = -1.65 \times 10^{5} (0.320 - 0.110) = -3.46 \times 10^{4}\) J so work done by the gas is \(3.46 \times 10^{4}\) J.
From the first law, the change of internal energy \(\Delta U = Q + W = 1.15 \times 10^{5} + (-3.46 \times 10^{4}) = 8.03 \times 10^{4}\) J.
No, the ideal gas law is not specified in this solution.