- #1

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show that (δT/δV)s = - (δP/δS)v

solution (δ/δV (δU(S,V)/δS)v)s = (δ/δS(δU(S,V)/δV)s)v

(δ/δV (δ(TdS - PdV)/δS)v)s = (δ/δS(δ(TdS-PdV)/δV)s)v

(δT/δV)s = -(δP/δS)v

I don't understand the substitution or the last step

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- Thread starter AndrewBworth
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- #1

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show that (δT/δV)s = - (δP/δS)v

solution (δ/δV (δU(S,V)/δS)v)s = (δ/δS(δU(S,V)/δV)s)v

(δ/δV (δ(TdS - PdV)/δS)v)s = (δ/δS(δ(TdS-PdV)/δV)s)v

(δT/δV)s = -(δP/δS)v

I don't understand the substitution or the last step

- #2

Chestermiller

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##dU=TdS-PdV##

We must also have that:

$$dU = \left(\frac{\partial U}{\partial S}\right)_VdS+\left(\frac{\partial U}{\partial T}\right)_SdV$$

Therefore, comparing both equations, we have:

$$T=\left(\frac{\partial U}{\partial S}\right)_V$$

$$-P=\left(\frac{\partial U}{\partial T}\right)_S$$

- #3

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- #4

Chestermiller

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So their notation in hinky. Can you get to the final result from my last two equations?

Chet

- #5

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Sure no sweat.

- #6

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Ah I figured it out

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