Four Special Cases of the First Law of Thermodynamics
The first law of thermodynamics finds application in several special cases:
An adiabatic process is one in which there is no heat transfer into or out of the system. It occurs very rapidly or a system is well insulated that no transfer of energy as heat occurs between the system and its environment. Therefore dQ = 0 in the first law of thermodynamics, which is then:
dQ = 0, dEint = – dW
An isochoric process is one in which there is no change in volume. An isochoric process is a constant-volume process. When the volume of a thermodynamic system is constant, it does no work on its surroundings. Therefore dW = 0 in the first law of thermodynamics, which is then:
dW = 0, dEint = dQ
In an isochoric process, all the energy added as heat (that is, Q is positive) remains in the system as an increase in internal energy (increase in temperature).
A process that eventually returns a system to its initial state is called a cyclic process. At the conclusion of a cycle, all the properties have the same value they had at the beginning.
For such a process, the final state is the same as the initial state, and so the total internal energy change must be zero. Steam (water) that circulates through a closed cooling loop undergoes a cycle. The first law of thermodynamics is then:
dEint = 0, dQ = dW
Thus, the net work done during the process must exactly equal the net amount of energy transferred as heat.
This is an adiabatic process in which no transfer of heat occurs between the system and its environment and no work is done on or by the system. These types of adiabatic processes are called free expansion. It is an irreversible process in which a gas expands into an insulated evacuated chamber. It is also called Joule expansion. For an ideal gas, the temperature doesn’t change (see: Joule’s Second Law), however, real gases experience a temperature change during free expansion. In free expansion Q = W = 0, and the first law requires that:
dEint = 0
A free expansion can not be plotted on a P-V diagram, because the process is rapid, not quasistatic. The intermediate states are not equilibrium states, and hence the pressure is not clearly defined.