partial derivitives

[what23]

um, i have a test tomorrow in thermo and was wondering if someone could solve this

V=RT/P + bT^3 + cPT^5

i need to take (partial V/ partial T)P meaning P is constant.

thanks

[Homer]

[ QUOTE ]
um, i have a test tomorrow in thermo and was wondering if someone could solve this

V=RT/P + bT^3 + cPT^5

i need to take (partial V/ partial T)P meaning P is constant.

thanks

[/ QUOTE ]

Can you take the derivative of something like V = 4T + 8T^3 + 10T^5?

This is pretty much the same thing unless I have forgotten something (highly possible).

[almostbusto]

R+3bpT^2+5c(p^2)T^4

i don't know anything about physics(but i have taken 5 semesters of calculus) but thats the answer if i am interpreting you correctly. basically treat everything on the right side of the equation as a constant except T and take the derivative, then multiply by p.

[what23]

sweet, thanks guys, thats what i got when i looked mine over.
sadly i haven't taken calc in almost 2 years and partials didn't seem that relevant. can't figure out what to derive or integrate or whatever. anyways thanks again

(mult by P just makes this look nicer and isn't necessary right?)

[Neuge]

[ QUOTE ]
R+3bpT^2+5c(p^2)T^4

i don't know anything about physics(but i have taken 5 semesters of calculus) but thats the answer if i am interpreting you correctly. basically treat everything on the right side of the equation as a constant except T and take the derivative, then multiply by p.

[/ QUOTE ]
You don't multiply by P after taking the derivative with respect to T. It's notation used almost exclusively in thermodynamics to denote which state variable(s) is(are) held constant. It's necessary because unlike true partial derivatives in which all other variables are held constant, other thermodynamic variables are not held constant a priori. The formal representation of it looks like this.

http://img.photobucket.com/albums/v366/neuge/equ.jpg

The correct answer is:

V = R/P + 3bT^2 + 5cPT^4

[almostbusto]

yea listen to neuge, he knows what he is talking about.

i have never seen that notation and definitely didn't consider it was a subscript intended in the OP. so i was working with that i know. i did the derivative part right though

[HajiShirazu]

Thermodynamics is an odd subject. Most of the stuff is actually really simple to understand if you are thinking about it correctly, but the meaning always gets lost in the sea of partial derivatives and notation.

[Duke]

Don't make it out to be so complicated. It's just a derivative, if you think about it.

[what23]

all is well i got the problem right on the test this morning and i believe i got a very good score. thanks guys for the help.