# Calc 3/Thermo differentiation

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1. Nov 4, 2014

### kamu

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

2. Relevant equations
au/as=T
au/av=p
S/R=ln[(v-b)(u+a/c)^2]

3. The attempt at a solution
1/T=1/au/as=as/au
S=ln[(v-b)(U+a/v)^2]R
as/au=[(v-b)2(U+a/v)(1)]R/[(v-b)(U+a/v)^2]=2R/(U+a/v)=1/T
T=(U+a/v)/2R
U=2RT-a/v
au/av=-P
au=-Pav
integrate au to get u=-pv+c

u=-pv+c=2RT-a/v
-p=2RT-a/v-C
p=a/v-2RT+c

uploaded my work in image format.

One of the main problem I have is that we haven't learned multivariable integration(or partial integration I don't know what it's called.)

edit:
more attempts

Last edited: Nov 4, 2014
2. Nov 4, 2014

### vela

Staff Emeritus
You found an expression for U as a function of V, and you're told $P = -\frac{\partial U}{\partial V}$. Why are you integrating? Just differentiate U with respect to V to find the pressure.

3. Nov 4, 2014

### kamu

Hi vela, I need to express pressure as a function of temperature and volume. I need to get both T and V in the expression. Which is the problem I'm having.

taking the au/av would give me a/v^2

4. Nov 4, 2014

### kamu

Another attempt.

5. Nov 4, 2014

### Staff: Mentor

You already correctly obtained $\left(\frac{\partial U}{\partial S}\right)_V=T=\frac{(U+\frac{a}{V})}{2R}$. Now please show us what you get for $\left(\frac{\partial U}{\partial V}\right)_S=-P$. You will need to combine the two relationships.

Chet

6. Nov 4, 2014

### kamu

Hey Chet. In my previous posts I listed how I derived au/av from the chain rule in implicit differentiation. They might not be correct but
au/av=e^(S/R) (-2/(u+a/v))^3 is what I got for fourth attempt

au/av= 2R[((u+a/v)(-a/v^2)]/[(v-b)(u+a/v)^2]/[2R((v-b)(u+a/v))/((v-b)(u+a/v)^2)] is what I got for my second attempt

au/av=-2R/(v-b)(u+a/v)(v^2)/[(2R/(u+a/v))+1/T] is what I got on the third attempt

my derivation are all in the photos.(Not that I think they are right :(. So far unfortunately only my last attempt actually gave me a expression with T in it, T always somehow cancels out in the end in other attempts.

7. Nov 5, 2014

### kamu

I took a simpler/different approach

the answer looks appropriate but I'm not sure if I messed up somewhere.

8. Nov 5, 2014

### Staff: Mentor

I get:
$$\frac{S}{R}=ln(V-b)+2ln(U+\frac{a}{V})$$
So,
$$\frac{dS}{R}=\frac{dV}{V-b}+\frac{2}{(U+\frac{a}{V})}(dU-\frac{a}{V^2}dV)$$
At constant S, this becomes:
$$0=(\frac{1}{V-b}-\frac{2a/V^2}{(U+\frac{a}{V})})dV +\frac{2}{(U+\frac{a}{V})}dU$$
Solving for $\partial U/\partial V$ gives:
$$\left(\frac{\partial U}{\partial V}\right)_S=-\frac{(U+a/V)}{2(V-b)}+a/V^2=-P$$
Does this match any of the results from your attempts?

Chet

9. Nov 5, 2014

### kamu

Hi Chet, yes it matches the results from my last attempt. If I sub in U=2RT-a/v for U in your answer I would get p=RT/(v-b)-a/v^2

$$\left(\frac{\partial U}{\partial V}\right)_S=-\frac{(2RT-a/v+a/V)}{2(V-b)}+a/V^2=-P$$

$$\left(\frac{\partial U}{\partial V}\right)_S=\frac{(RT)}{(V-b)}-a/V^2=P$$

10. Nov 5, 2014

### kamu

I forgot to mention. Thank you for the help Chet! I really appreciate it.