Taking Derivative of Av-2exp(s/R): Solve Tds-Pdv

[avocadogirl]

Homework Statement

I have a system that is defined by the equation:

u = Av^-2exp(s/R)

I'm looking for the final temperature of the system knowing that as the system changes, pressure will be halved and (s/R) will be a constant.

The equation which relates energy, temperature, and pressure is du = Tds - Pdv

How do I take the derivative of Av^-2exp(s/R) to get it in the form Tds - Pdv?

Homework Equations

$$\partial$$u$$\partial$$s=T
$$\partial$$u$$\partial$$v=-P
du = $$\partial$$u$$\partial$$s ds + $$\partial$$u$$\partial$$v dv

The Attempt at a Solution

Do I take the partial derivative of the whole thing for v, then add another term that is the partial derivative of the whole thing for s?

-2Av^-3exp(s/R) + partial derivative for exp(s/R) with respect to s = du?

And, when you're taking the derivative of exp(s/R), does it come out something like: 1/Rexp(s/R)?

 

[Mark44]

I think what you're looking for is what's called the total differential, du.

In your problem,
$$du = \frac{\partial u}{\partial s}~ds~+~\frac{\partial u}{\partial v}~dv$$

Starting with u = Av^-2e^s/R, take each partial derivative and substitute it into the preceding formula.

And, when you're taking the derivative of exp(s/R), does it come out something like: 1/Rexp(s/R)?

It depends on which partial you're taking. If you take the partial of u w.r.t s, yes, that's right. If you're taking the partial of u wrt v, though, it doesn't. That partial is 0, since e^s/R is a constant as far as v is concerned.

 

[avocadogirl]

Thank you.