Why U is a State Function if W(adiabatic) Does Not Depend on State?

[asdf1]

if work is not a state function and internal energy is a state function,
then why does w(adiabatic)=U is a state function?

 

[notawretcheddrunk]

well i think that in \Delta U = q+w q and w are not state functions but processes, so you can ask the same question here i guess. although I am not sure how this can be, somebody please give a mathematical explanation?

 

[FunkHaus]

The answer here might lie in the difference between exact and inexact differentials. An exact differential is the differential of a scalar function, call it F. For a function of two variables x and y, we have:
dF = P(x,y)dx + Q(x,y)dy
By the commutation of second order partial derivatives, we find that if
\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}
then the quantity
Pdx + Qdy
is an exact differential.
Now, say that x and y are also functions of some variables s and t. It may be true that dx = Ads + Bdt is not an exact differential. Likewise, dy = Cds + Ddt may also be inexact. However, if we add them together we can get an expression
Pdx + Qdy
that is exact (that is, Pdx + Qdy = dF).

This is what's going on when one discusses the first law of thermodynamics:
dE = dQ + dW
dE is an exact differential. dQ and dW are inexact, but their sum is exact. By the way, exact differentials are related to path independence in terms of integration. When you have more than two variables you are integrating over, there is a notion of "path". But because an exact differential can be reduced to the differential of a scalar function (ie just one variable now) there are no longer multiple paths--that is, all paths are equivalent.

Hope this helps!

 

[asdf1]

thanks~
but during an adiabatic process, w=U, so how are you supposed to know whether w and U are or aren't state functions?

 

[LeonhardEuler]

Because the change in a state function does not depend on the path taken. Suppose the adiabatic process was running an electric current through a resistor in a gas. Taking as the system the gas with the resistor in it, the only energy transfer is the electric work across the boundary, so the process is adiabatic. The temperature and pressure of the gas increase, but the volume remains the same. In this case \Delta U = w. Now suppose that instead of having a current flow through it, the system is just heated the same final temperature and no work is done. In this case \Delta U = q. In both cases \Delta U is the same, but q and w are different depending on which path is taken. This is the difference: state functions do not depend on the path taken.

 

[asdf1]

but is kinda seems like in an adiabatic function, work will then become a state function too...

 

[Quasi Particle]

No, work is not a state function, only U is. As said above:

U = Q + W

You can get the same value for U for many different combinations of Q and W. In this special case, Q = 0 so U = W.

 

[Yegor]

Are state functions defined for entire thermodynamics or for certain processes? If last then it's possible to say that in adiabatic process work is state function.

 

[LeonhardEuler]

This is what a state function means: I can walk into a room and look at a system and take measurements of things like temperature, pressure, volume, etc. I can then leave and come back later, having no idea what happened while I was gone, but just by taking measurements of the current state of the system I can tell how much any state function changed between the time I left and the time I came back. I can't do this with work because I don't know whether the process was adiabtic or not. Doing work can raise the temperature of a system. So can heating it. It is impossible for me to tell which it was by measuring the T, P, V... at the beginning and end of a process.

 

[asdf1]

so in a adiabatic process, work isn't a state function?

 

[ddesai]

Yes, if you are restrict yourself to adiabatic paths, then the work form can be written as the dw, where w is a function of thermodynamic variables.

This means the work done will only depend on the endpoints of the path -- as long as we are only considering paths along which dq = 0.

 

[asdf1]

i see... thanks! :)