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Another State Function

Recall that \bgroup\color{blue}$ dq$\egroup is not a perfect differential.

Let's consider \bgroup\color{blue}$ dq$\egroup for an ideal gas undergoing a reversible process.

$\displaystyle \input{equations/81A}$ (11-17)

for an ideal gas

$\displaystyle \input{equations/81B}$ (11-18)

$\displaystyle \input{equations/81C}$ (11-19)

Now divide through by \bgroup\color{blue}$ T$\egroup

$\displaystyle \input{equations/81D}$ (11-20)

Notice that we have separated the equation into something that is integrable over segments of \bgroup\color{blue}$ dT$\egroup and \bgroup\color{blue}$ dV$\egroup and thus over any curve.

Therefore, \bgroup\color{blue}$ dq_{rev}/{T}$\egroup is a ``perfect differential'' and it must then be a state function for an ideal gas.



next up previous
Next: About this document ... Up: Lecture_11_web Previous: A New Thermodynamic State
W. Craig Carter 2002-09-28