Presentation on theme: "Equilibria Among Three Thermodynamic Systems. Mechanical Equilibrium Consider the apparatus shown below. The internal volumes formed by the pistons and."— Presentation transcript:
Mechanical Equilibrium Consider the apparatus shown below. The internal volumes formed by the pistons and the outer cylinder are filled with three different gases, which thus constitute three thermodynamic systems. All of the bounding surfaces (including the pistons) are well insulated. The movements of pistons A and D are controlled by the experimenter. Pistons B, C, and E can be locked in place or allowed to move freely, without friction.
At the beginning of the experiment, the pressures in the three systems are equal, i.e., P1 = P2 = P3. Pistons B, C, D, and E are locked in place.
Then piston A is moved to the right, decreasing the volume V1 of system 1 and increasing P1. No change is observed in P2 or P3, because these systems are isolated from system 1 and the other surroundings by closed, rigid, and adiabatic boundaries. Each of the three systems reaches a state of statistical equilibrium internally, but there is no equilibrium between systems.
Piston A is locked. If piston B is now unlocked, it will move to the right; P1 will decrease and P2 will increase; work will be done by system 1 on system 2. The motion of piston B will stop eventually, and P1 and P2 will come to new, steady values as systems 1 and 2 attain new states of equilibrium, in which P1=P2. Mechanical equilibrium now exists between systems 1 and 2, because there is no net, unbalanced force on piston B, but not between systems 2 and 3.
Now suppose that the experimenter locks piston B (and leaves pistons C and E locked also) and moves piston D to the left until P3 = P2.
When piston C is unlocked, it does not move, because the forces on it are balanced, i.e., systems 2 and 3 are already in mechanical equilibrium. P2 was not changed in this process, since piston C did not move, and piston B was locked.
Therefore, if piston B is now unlocked, it does not move either, because P1 still equals P2.
We now have mechanical equilibrium between systems 1 and 2 (P1=P2), and mechanical equilibrium between systems 2 and 3 (P2=P3). If piston E is now unlocked, will it move? No, because systems 1 and 3 are already in mechanical equilibrium. We know that P1= P2 and P2 = P3 implies P1= P3.
Thermal Equilibrium Now consider the apparatus shown below. System 1, a gas, is contained in a horizontal, adiabatic cylinder with an adiabatic frictionless piston, the movement of which is controlled by the experimenter. However, the bottom of the cylinder is a rigid, aluminum plate. System 2 is another gas, held in the vertical cylinder, the walls and piston of which are also adiabatic, except where they make contact with the metal bottom of the cylinder containing system 1. In the contact region, the wall of the vertical cylinder is also bare aluminum. System 3 is another gas, in a cylinder just like that of system 1.
There is a weight on piston B, which remains on it throughout the entire experiment. Therefore, the pressure P2 remains constant, although the volume V2 may change because piston B is not locked. If piston A is moved to the right, P1 at first increases,
then decreases. At the same time, volume V2 increases. When the two states reach their new states of equilibrium, it is noted that P1 and V2 are both somewhat larger than their initial values. At this point, systems 1 and 2 are said to be in thermal equilibrium, because the boundary between the systems is not insulated (it is diathermal). We say that heat has flowed between the two systems. It is also noted that P1 is not equal to P2
Now piston A is locked, and systems 1 and 2 are separated. System 2 is brought into contact with system 3,
It is observed that V2 increases as the two systems come to thermal equilibrium.
Piston C is now moved to the right, causing V2 and P3 both to decrease. The position of piston C is adjusted so that V2 is brought back to the value it had when system 2 was in contact with system 1 (i.e., V2=V2A). It is noted that P3 is not equal to P2.
Systems 2 and 3 are now separated and system 3 is brought together with system 1, contacting each other at the bare metal (diathermal) bottoms of the cylinders.
It is observed that the pressures P1 and P3 do not change from the values they took when systems 1 and 3 were separately in thermal equilibrium with system 2. That is, they already are in thermal equilibrium with each other.
Since this experiment can be done using any amounts of any three substances to form the three systems, the general observation can be made that "If two systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each other." This observation has been called the "Zeroth Law of Thermodynamics." By analogy with the case of mechanical equilibrium, we can conclude that there must be a thermodynamic property which serves to indicate when two (or more) systems are in thermal equilibrium. We call this property "temperature." (This word originally had the meaning "neither too hot nor too cold.”) Note that the volume V2 is a measure of the property temperature. In fact, since system 2 is contained in a cylindrical vessel, the height of piston B relative to some datum is also a measure of temperature. System 2 is, in fact, a thermometer. The temperature measured on a scale with an arbitrary zero point is usually called "empirical" or "practical" temperature, and is analogous to gage pressure. The Fahrenheit and Celsius temperature scales are empirical temperature scales. Although we are more accustomed to liquid-filled thermometers, the gas thermometer (system 2) is actually the basis for the absolute temperature scale. Before mercury, brandy was used to fill thermometers (Snapple Real Fact #166).