Please help show that ${(\frac{\partial T}{\partial P})}_{S} = \frac{V α T}{C_{P}}$ $((delT)/(delP))_S = (ValphaT)/C_P$ ?

Question

Please help show that ${(\frac{\partial T}{\partial P})}_{S} = \frac{V α T}{C_{P}}$ $((delT)/(delP))_S = (ValphaT)/C_P$ ?

1 Answer

Impact of this question

3043 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-25T06:46:09

Okay, considering that ${(\frac{\partial T}{\partial P})}_{S}$ $((delT)/(delP))_S$ contains $S$ $S$ , $T$ $T$ , and $P$ $P$ , I would start from $S = S (T, P)$ $S = S(T,P)$ and take the total derivative:

$d S = {(\frac{\partial S}{\partial T})}_{P} d T + {(\frac{\partial S}{\partial P})}_{T} d P$ $dS = ((delS)/(delT))_PdT + ((delS)/(delP))_TdP$ $(1)$ $" "" "bb((1))$

Now, if we divide by $\partial P$ $delP$ at a constant $S$ $S$ , we'd get:

$cancel(((delS)/(delP))_S)^(0) = ((delS)/(delT))_P((delT)/(delP))_S + ((delS)/(delP))_Tcancel(((delP)/(delP))_S)^(1)$

$" "" "bb((2))$

since a constant $S$ forces a derivative of $S$ with respect to any variable to go to $0$ , and $(dP)/(dP) = 1$ no matter what else is held constant.

From this question, we can refer back and recall that

$((delS)/(delT))_P = 1/T ((delH)/(delP))_T = C_P/T$ .

$bb((3))$

Now we need to figure out what $((delS)/(delP))_T$ is.

The bottom of the derivative shows $T$ and $P$ , which are the natural variables of the Gibbs' free energy, so from the Maxwell relation:

$dG = -SdT + VdP$ ,

we can either recall the cyclic relation $((delS)/(delP))_T = -((delV)/(delT))_P$ , or rederive it. Again, from the total derivative of $G(T,P)$ :

$dG = ((delG)/(delT))_PdT + ((delG)/(delP))_TdP$ ,

showing that $-S = ((delG)/(delT))_P$ and $V = ((delG)/(delP))_T$ . Since the order of second partial differentiation doesn't matter,

$((del^2G)/(delPdelT))_(P,T) = ((del^2G)/(delTdelP))_(T,P)$ ,

meaning that

$-((delS)/(delP))_T = ((delV)/(delT))_P$ . $" "" "bb((4))$

Therefore, from $bb((2))$ , subtract over $((delS)/(delP))_T$ , plug in $bb((3))$ and $bb((4))$ , and we have:

$((delV)/(delT))_P = C_P/T((delT)/(delP))_S$

$=> ((delT)/(delP))_S = T/C_P ((delV)/(delT))_P$

Finally, recall that $alpha = 1/V((delV)/(delT))_P$ , so that we have:

$=> color(blue)(((delT)/(delP))_S = (ValphaT)/C_P)$

Science

Math

Humanities

... and beyond

Please help show that ${(\frac{\partial T}{\partial P})}_{S} = \frac{V α T}{C_{P}}$ $((delT)/(delP))_S = (ValphaT)/C_P$ ?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

Please help show that ((delT)/(delP))_S = (ValphaT)/C_P(∂T∂P)S=VαTCP((delT)/(delP))_S = (ValphaT)/C_P?

1 Answer

Related questions

Impact of this question

Please help show that ${(\frac{\partial T}{\partial P})}_{S} = \frac{V α T}{C_{P}}$ $((delT)/(delP))_S = (ValphaT)/C_P$ ?