Recent content by kamu

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

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

I took a simpler/different approach the answer looks appropriate but I'm not sure if I messed up somewhere.

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...

Another attempt.

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

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

Hey BvU, thanks for the help. I appreciate it. Anyway the project was due yesterday so I didn't have a lot of time retype everything. Mr.Vickson I will take the advice and be sure to type it in a better format in the future.

for (i) it says to compare |r|=|r x v|^2/(GM+|c| cos(theta) to the formula in the textbook a(1-e^2)/(1+ecos(theta)), not sure what I'm suppose to do here. I'm also insecure about the correctness of my work because I can't remember all the cross product and dot product properties to see if I'm...

Uploading photos of more attempts

Homework Statement The Attempt at a Solution Honestly I'm completely lost, this is an assignment for my calc 3 class. I tried to do (a) but I think I'm completely off track so any helps appreciated. r x a=0 |r x a|=0=|r||a|sin(theta) r cannot be 0 since it's in the denominator Assuming...