Conservation of energy implies that if we have a thermally insulated system which goes from state 1 to state 2: $$\Delta E_{12}=E(2)-E(1)=W_{12}$$ and the 1st law of thermodynamics requires that this does not depend on the path so the total energy of the system should be a state function, which we call the internal energy $E$.

I do not understand why if it does not depend on the path it should be a state function?

How has the 1st law been applied here, is it just because the system is adiabatic and the state changed only by the performance of work?


So the first law for an system where we don't have mass flows in or out ( a closed system ), is

$Q + W = \Delta U$

Where $Q$ is your net heat added, and $W$ is your net work added, and $U$ is your net internal energy change: internal energy being like the sum of all the different kinetic energies of the molecules. This is why when we add heat, the internal energy goes up, and when we add work the internal energy goes up.

This means, as the system has been defined as thermally isolated, it is therefore adiabatic ($Q = 0$) and the equation becomes

$W = \Delta U$

I think your confusion lies in the definition of a state function:

The definition of a state function is one that it always the same when we are at a particular state (state: a system where we've set the pressure, temperature, volume to specific values etc), no matter how we got there. If you've done $PV$ diagrams (pressure and volume are state variables) a state variable is one defined at a specific point on the $PV$ diagram, no matter how you got there, and no matter how many times you go away and come back, the state at that point must be the same. Therefore when we say $U$ is a state variable, we're saying $U$ is defined by the state, not by the process, and therefore does not depend on the path. That's why the $\Delta s$ are in the equation

If it's still confusing, I'd try KhanAcademy, he's got some awesome videos on this that are much better explained than what I've written



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.