- #1

asdf1

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then why does w(adiabatic)=U is a state function?

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- Thread starter asdf1
- Start date

- #1

asdf1

- 734

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then why does w(adiabatic)=U is a state function?

- #2

notawretcheddrunk

- 7

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- #3

FunkHaus

- 10

- 1

[tex]dF = P(x,y)dx + Q(x,y)dy[/tex]

By the commutation of second order partial derivatives, we find that if

[tex]\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}[/tex]

then the quantity

[tex] Pdx + Qdy [/tex]

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

[tex] Pdx + Qdy [/tex]

that is exact (that is, Pdx + Qdy = dF).

This is what's going on when one discusses the first law of thermodynamics:

[tex] dE = dQ + dW [/tex]

dE is an exact differential. dQ and dW are inexact, but their sum is exact. By the way, exact differentials are related to

Hope this helps!

- #4

asdf1

- 734

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

- #5

LeonhardEuler

Gold Member

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- #6

asdf1

- 734

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

- #7

Quasi Particle

- 24

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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.

- #8

Yegor

- 147

- 1

- #9

LeonhardEuler

Gold Member

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- #10

asdf1

- 734

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

- #11

ddesai

- 7

- 0

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.

- #12

asdf1

- 734

- 0

i see... thanks! :)

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