Path functions in thermodynamics: A short note

A brief introduction to state and path function

This post is about path function

Whenever we define a substance one of the first things mentioned is the state of the compound. "State" refers to temperature, pressure, and the amount and type of substance present. Once the state has been established, state functions can be defined. Thermodynamic variables such as the internal energy (U), volume, pressure, temperature, entropy (S ), and free energy (G), all these quantities can be used to specify the state of a system. They are properties of the current state of the system, and they do not depend on the way the system got to that state. They are all state functions.

?Path function in Thermodynamics

A thermodynamic property that depends on the path between the initial and final state is known as the path function. The path functions depend on the path taken or covered between two (initial and final) states. For example, work and heat. If different paths are chosen to reach from one point to another point, the work done will be different however you reach the same point in each case. So, work is not a state function as we cannot say that a system will have a specific amount of work at a specific state. There is no such thing as an amount of work or heat in a system. The amounts of heat and work that "flow" during a process connecting specified initial and final states depend on how the process is carried out.

Differential and inexact differential

Quantities whose values are independent of path are called state functions, and their differentials are exact ( dP ,?dV ,?dG , dT ...). Quantities that depend on the path followed between states are called path functions, and their differentials are inexact ( dw ,?dq ). When we integrate an exact differential, the result depends only on the final and initial points, but not on the path chosen. However, when we integrate an inexact differential, the path will have a huge influence on the result, even if we start and end at the same points.

?We will discuss two typical cases for path functions one for heat and the other for work

Heat transfer: Path function

?A typical case of the path function of heat transfer is how does an ice cube choose how quickly to melt in an equal mass of water in two containers one in a vacuum other under pressure. The only difference is one container is under vacuum and the other under pressure.

Explanation

?Ice melting is a phase transition. It takes place isothermally. Melting ice is endothermic (surrounding temperature decreases) taking heat from the environment, feeling cold, because it requires energy to break ice bonds.

The source of heat in both cases is water. Under vacuum, because of lower saturation temperature, the water will be at a lower temperature with smaller sensible heat to supply to the ice to melt compared to the other case where water is under pressure.

Therefore, the rate of heat transfer from water to ice will be slower in a vacuum.

?Water phase diagram

Please look at the phase diagram of the water above. You would find no phase transition of ice below atmospheric pressure because there is no supplier of heat.

Below certain atmospheric pressure, ice will sublime to vapor at a lower energy level.

Specific heat of ice = 2.2 KJ/KG/K

Specific heat of vapor = 1.996 KJ/KG/K

?Work transfer: Path function

Question: Isothermal and adiabatic processes are two typical thermodynamic processes. Which process involves more work.

?Isothermal process

An isothermal process is a type of thermodynamic process in which the temperature T of a system remains constant: ΔT = 0. This typically occurs when a system is in contact with an outside thermal reservoir, and a change in the system occurs slowly enough to allow the system to be continuously adjusted to the temperature of the reservoir through heat exchange. For an ideal gas the change in internal energy for an isothermal process, ?U = 0.

Isothermal compression and expansion

In the isothermal compression of a gas, there is work done on the system to decrease the volume and increase the pressure. Doing work on the gas increases the internal energy and will tend to increase the temperature. To maintain the constant temperature energy must leave the system as heat and enter the environment. If the gas is ideal, the amount of energy entering the environment is equal to the work done on the gas, because internal energy does not change. For isothermal expansion, the energy supplied to the system does work on the surroundings.

Adiabatic process

an adiabatic process is a type of thermodynamic process that occurs without transferring heat or mass between the thermodynamic system and its environment, ?Q = 0. Unlike an isothermal process, an adiabatic process transfers energy to the surroundings only as work.

Adiabatic compression and expansion

Since in an adiabatic process the ?Q = 0 the gas uses its own internal energy to expand and that causes a drop in temperature. The adiabatic compression of a gas causes a rise in the temperature of the gas since the compression of a gas increases the internal energy of the gas.

?Work transfer in Isothermal and Adiabatic expansion?

?Explanation

The work involved in an isothermal process is more than an adiabatic process. The area under the blue isothermal line shows the work generated by an isothermal process which is bigger than the area under the red adiabatic line.

The reason is in an isothermal process, ?T=0. The change in internal energy ?U=0, therefore Q = W. That is the entire heat Q goes for doing work, W.

In an adiabatic process since ?Q=0, the heat for expansion comes from its internal energy. The gas cools losing kinetic energy and pressure. That gives a steeper fall in pressure as the gas expands. You can see a comparatively sharper slope in the above PV graph compared to a flatter line for isothermal expansion. This makes the area under the isothermal line more and therefore work compared to an adiabatic process.?

?Calculation

Isothermal work vs Adiabatic work

Let us assume 1 mole of an ideal gas expands from 0.1 m3 to 1 m3 at 300 k.

What is the work done by each process?

?Isothermal work:

?

?W = 0.5743 X 10^4 J

Adiabatic work

The assumption remains the same. 1 mole ideal gas is expanding from 0.1 M3 to 1 M3 at 300 k adiabatically. Y = Cp/Cv assumed to be 1.685

W = 2.9392 x 10^3 J

Thus, roughly isothermal process under identical conditions could produce 1.95 times more work than the adiabatic process.

Work transfer in Isothermal and Adiabatic compression of gases?

This is opposite to isothermal expansion. Here the adiabatic compression does more work than isothermal compression

Explanation

In isothermal compression since ?T =0, the heat generated by the piston goes to the surroundings. While in the adiabatic process since ?Q = 0, the heat increases the gas temperature, and the gas expands within the cylinder. The piston requires more work to compress the gas.

Credit: Google