# Adiabatic equation in dependence on volume and pressure

1. Jul 9, 2015

### steroidjunkie

1. The problem statement, all variables and given/known data

Find: (a) Equation of state $$f (p, V, T)$$ and (b) Adiabatic equation in dependence on volume and pressure. Internal energy $$U(V, S)=\frac{1}{aV} ln(\frac{S}{\gamma})$$ where a and $\gamma$ are positive constants.

2. Relevant equations

(a) $dU=TdS-pdV \rightarrow$
$T=(\frac{dU}{dS})_V$
$p=-(\frac{dU}{dV})_{S}$

$T=(\frac{d}{dS})_V \cdot \frac{1}{aV} ln(\frac{S}{\gamma})=\frac{1}{aV} \cdot \frac{\gamma}{S} \cdot \frac{1}{\gamma}=\frac{1}{a \gamma V}$
$p=-(\frac{d}{dV})_S \frac{1}{aV} ln(\frac{S}{\gamma})=- \frac{1}{a} \cdot (- \frac{1}{V^2}) \cdot ln(\frac{S}{\gamma})=\frac{1}{aV^2} ln(\frac{S}{\gamma})$

(b) $pV^{\gamma}=NkT$

$\gamma=?$
$\gamma=\frac{C_p}{C_V}$
$C_p=(\frac{dU}{dT})_p$
$C_V=(\frac{dU}{dT})_V$

3. The attempt at a solution

(b) $C_p=(\frac{d}{dT})_p \frac{1}{aV} ln(\frac{S}{\gamma})$

$T=\frac{1}{a \gamma V} \rightarrow T \gamma=\frac{1}{a V}$
$C_p=(\frac{d}{dT})_p T \gamma ln(\frac{S}{\gamma})$
$p=\frac{1}{aV^2} ln(\frac{S}{\gamma}) \rightarrow paV^2=ln(\frac{S}{\gamma})$
$C_p=(\frac{d}{dT})_p T \gamma paV^2=\gamma paV^2$

$C_V=(\frac{d}{dT})_V \frac{1}{aV} ln(\frac{S}{\gamma})$
$C_V=(\frac{d}{dT})_V T \gamma ln(\frac{S}{\gamma})$
$C_V=(\frac{d}{dT})_V T \gamma paV^2=\gamma paV^2$

$\gamma=\frac{C_p}{C_V}= \frac{\gamma paV^2}{\gamma paV^2}=1$

I need help with (b) part of the problem. I know this is not correct and I assume I did something wrong while substituting, but I have no idea what. If you know something please post.
Thanks.

2. Jul 9, 2015

### Staff: Mentor

You need help on part a also. Your equation for T is incorrect. After you get that corrected, you will need to elimate S between the equations for S and T.

In part b, your equation for Cp is incorrect.

Chet

3. Jul 9, 2015

### Staff: Mentor

With regard to part b: how does S vary along an adiabatic reversible path?

4. Jul 10, 2015

### steroidjunkie

Regarding a mistake with (a) $T=\frac{1}{aV} \cdot \frac{\gamma}{S} \cdot \frac{1}{\gamma}=\frac{1}{aSV}$
Thank you for noticing.
This leads to: $C_V=(\frac{d}{dT})_V TSpaV^2=SpaV^2$

(b) The right formula: $C_V=(\frac{dH}{dT})_p$

$H=U+pV=U+U=2U$
$C_p=(\frac{d}{dT})_p (2U)=2 \cdot C_V=2SpaV^2$

$\gamma=\frac{C_p}{C_V}=\frac{2SpaV^2}{SpaV^2}=2$

So, adiabatic equation is: $pV^2=NkT$

5. Jul 10, 2015

### Staff: Mentor

For part a,

$$S=γe^{aV^2p}=\frac{1}{aVT}$$

This is the p-V-T relationship they were looking for.

For part b,
From your final equation for p, you could have immediately seen that pV2 is constant if S is constant along a reversible adiabat. I'm not so sure about the NkT part, however. I don't think that that is correct.

Chet

6. Jul 10, 2015

### steroidjunkie

I see. I could have stated that: $pV^2=const$
and then from the pressure equation: $const=\frac{1}{a}ln(\frac{S}{\gamma})$
Then I substitute variables in constant in order to get a dependence on T:
$pV^2=\frac{1}{a}ln(\frac{S}{\gamma}) \rightarrow apV^2=ln(\frac{S}{\gamma}) \rightarrow e^{apV^{\gamma}}=\frac{S}{\gamma}$
$\rightarrow \gamma e^{ap V^{\gamma}}=S$

and: $T=\frac{1}{aSV} \rightarrow S=\frac{1}{aTV}$

which leads to: $\gamma e^{apV^{\gamma}}= \frac{1}{aTV} \rightarrow e^{apV^{\gamma}}= \frac{1}{a \gamma TV} \rightarrow apV^\gamma$
$= ln(\frac{1}{a \gamma TV}) \rightarrow pV^{\gamma}= \frac{1}{a}ln(\frac{1}{a \gamma TV})$

Solution: $pV^2= \frac{1}{a}ln(\frac{1}{2aTV})$

Thank you one more time.

Last edited: Jul 10, 2015