# Can't we draw an indicator diagram for a fast adiabatic process?

We know that an indicator diagram can be drawn for an quasi static process only. An fast process in which there is no sufficient time for the exchange of heat can consider to be adiabatic. Can i draw an indicator diagram for it?

The indicator diagram essentially consists of quantities $$(T,V,P)$$ and these are state functions. "State" refers to temperature, pressure, and the amount and type of substance present. Once the state has been established, state functions can be defined.

Another way to think of state functions is as integrals. Integrals depend on only three things: the function, the lower limit and the upper limit. Similarly, state functions depend on three things: the property, the initial value, and the final value.

Take an example of $$\triangle H$$ as a function of time. This is a state function.

$$\int^{t_{f}}_{t_{0}} H(t)dt = H(t_f)-H(t_i)$$

which is equivalent to: $$\triangle H=H_{\text{final}}-H_{\text{initial}}$$

As represented by the solution to the integral, enthalpy is a state function because it only depends on the initial and final conditions, and not on the path taken to establish these conditions.

So in this case, when we would be drawing an indicator diagram, we'd have only 2 points on the graph. The first point representing the initial state, and the second point representing the final state of the system.

Additionally, there are path functions, which can not be solely defined at a point. These quantities depend on the path of the process.

For example work done by a system and heat within a system are represented as follows:

$$\oint {\text{dw}} \ \$$, $$\oint {\text{dq}} \ \$$ i.e the cyclic integral

A path function is an inexact or imperfect differential. The cyclic integral of all state functions is 0, however this may or may not be the case for path functions.

You cannot.

Because, in order for an indicator diagram to be drawn, the system has to have well-defined P, T, etc. If the process is reversible, this is obviously doable. That means, at least quasistatic.

Once you try to do something fast, there will be variance within the system. If you changed V very quickly, then there will be pressure P waves inside the system. There would not be one single P value to define for the entire system any more. Then you cannot draw the indicator diagram.