Pour tea into three containers at breakfast. One is an open steel tumbler, one is a small pressure cooker with its lid locked, and one is a vacuum flask. By lunch the tumbler is cold and has lost some tea as vapour. The cooker is cold, but not a drop is missing. The flask still pours hot tea.
Drawing a boundary
Thermodynamics studies how energy moves as heat and as work. Before counting any energy, we must say what we are looking at. The part of the world we study is the thermodynamic system: a gas in a cylinder, the water in a cooker, a cup of tea. Everything else is its surroundings. The two are separated by a boundary, real or imagined. Heat, work and matter are counted when they cross this boundary. This chapter uses \(R = 8.31\,\text{J/(mol K)}\) and \(1\,\text{atm} = 1.013 \times 10^5\,\text{Pa}\).
A boundary can let different things through. If both matter and energy can cross it, the system is an open system. If energy can cross it but matter cannot, the system is a closed system. If neither can cross, the system is an isolated system. Most of this chapter studies a closed system: a fixed amount of gas in a cylinder with a piston.
An open system can exchange both matter and energy with its surroundings. A closed system can exchange energy, as heat or work, but not matter. An isolated system can exchange neither matter nor energy.
Fig 1.1 settles the tea. The open tumbler lost heat to the air and lost vapour too, so it was an open system. The locked cooker lost heat through its steel walls, but no tea escaped, so it was closed. The vacuum flask has a vacuum between two silvered walls, which blocks conduction and convection and reflects radiation. Almost no heat crosses it, so the tea inside is nearly an isolated system and stays hot.
A pressure cooker on a stove has its whistle held firmly down, so no steam escapes. What kind of system is the water and steam inside?
The state of a gas
How do we describe a gas in a cylinder? We do not follow its \(10^{23}\) molecules. We measure a few large-scale quantities: the pressure \(p\), the volume \(V\), the temperature \(T\) and the amount \(n\) in moles. These are state variables. When they are known, the state of the gas is known.
These numbers have a meaning only in equilibrium. The gas must be in mechanical equilibrium, with one pressure throughout and no flow. It must also be in thermal equilibrium, with one temperature throughout. A gas that is still swirling after a bang has no single pressure. A gas warmer on one side than the other has no single temperature.
The state variables are not independent. An equation of state links them. For an ideal gas it is \(pV = nRT\), from the last chapter. So for a fixed amount of gas, any two of \(p\), \(V\) and \(T\) fix the third. One mole at \(300\,\text{K}\) and \(1\,\text{atm}\) must have \(V\) \(= RT/p\) \(= 8.31 \times 300/(1.013 \times 10^5)\) \(= 0.0246\,\text{m}^3\), which is \(24.6\,\text{L}\).
So each equilibrium state of a fixed amount of gas can be drawn as one point on a graph of \(p\) against \(V\). The temperature is not lost; it is \(pV/nR\) at that point. This graph is the indicator diagram, or \(p\)-\(V\) diagram. It is to thermodynamics what the position-time graph was to kinematics.
Two moles of an ideal gas fill \(20\,\text{L}\) at \(400\,\text{K}\). What is its pressure, in kPa? Take \(R = 8.31\,\text{J/(mol K)}\).
The law that comes first
The chapter on temperature met this idea in words. Put two bodies in contact, and heat flows until they reach thermal equilibrium. Now it can be stated as a law, the first of four laws of thermodynamics. It was named last, but it comes before the others, so it is called the zeroth law.
Let two bodies A and B each be in thermal equilibrium with a third body C. Then A and B are in thermal equilibrium with each other.
The law sounds obvious, but it carries weight. It says that thermal equilibrium can be labelled by one number that the bodies share. That number is the temperature. Two bodies are in equilibrium exactly when their temperatures are equal. So the zeroth law of thermodynamics is what makes temperature a property of a body.
A thermometer uses the law every time. The thermometer is the third body C. Touch it to a patient and wait until it reaches equilibrium with the patient; read it. Touch it to a second person and read it again. If the two readings agree, the zeroth law says the two people would be in equilibrium with each other. We never need to bring them together.
Which physical quantity does the zeroth law allow us to define?
By 1850 the first and second laws of thermodynamics were famous. Nobody had thought to state the law that temperature itself rests on. In the 1930s the British physicist Ralph Fowler noticed the gap. The rule about a third body was more basic than the first law, yet it had no name. In their 1939 book, Fowler and Edward Guggenheim called it the zeroth law, so that the famous numbering could stay.
A process slow enough to draw
A process takes a system from one state to another. If the system is in equilibrium at every stage, each stage is a point on the \(p\)-\(V\) diagram. The points join into a line, the path of the process. Such a process is a quasi-static process. It must be slow, so that the gas settles at every step before the next one.
Play both chips in Fig 1.2. The gas holds \(pV/RT\) \(= 300\,000 \times 0.010/(8.31 \times 300)\) \(= 1.20\,\text{mol}\). In the slow chip, the load falls gradually and the state point slides smoothly from state 1 to state 2 along the curve \(pV = 3000\,\text{kPa L}\). In the sudden chip, the piston shoots up and bounces. For a moment the gas near the piston is thinner than the gas below. There is no one pressure to plot, so the point vanishes and returns only at state 2.
How slow is slow? The pressure inside a gas evens out at about the speed of sound. Sound crosses a \(10\,\text{cm}\) cylinder in \(0.1/340 = 0.3\,\text{ms}\). A piston that takes a second, or even a tenth of a second, to move is slow enough. A balloon that bursts in a millisecond is not.
The two ends of a sudden process can be equilibrium states, so both are points on the diagram. The line between them is not a path. Never join the ends of a sudden process with a curve and find an area under it. The next lesson finds work from exactly that area, and it needs a real path.
Which of these can be drawn as a line on a \(p\)-\(V\) diagram?
A student plots the start and the end of a gas that was released suddenly, and joins them with a straight line. What is wrong with the line?
The start and end are equilibrium states, so both points are correct. Between them the gas was not in equilibrium: it had no single pressure or temperature. So the states in between cannot be plotted, and the line is not a path the gas followed. Any area under it has no meaning as work.
- Calling a pressure cooker with its whistle shut an isolated system.Heat still enters through the base. With matter kept in and energy let in, it is a closed system.
- Using p or T for a gas that is not in equilibrium.A swirling or unevenly heated gas has no single pressure or temperature. State variables need equilibrium.
- Drawing a path for a sudden process.Only a quasi-static process has a path. For a sudden one, mark the two end states and nothing between.