Lead is compressed reversibly with T const, find dS

[Kara386]

Homework Statement

Lead is compressed reversibly and with temperature kept constant at 300K, and from pressure of 1bar to 1000 bar. Assuming the change in volume is small, what is the change in entropy S? Use the Maxwell relation derived from Gibbs free energy function. V = $10^{-3}m^3$.

Homework Equations

$(\frac{\partial{V}}{\partial{T}})_p = -(\frac{\partial{S}}{\partial{p}})_T$ Maxwell Relation
$\kappa = \frac{1}{V}(\frac{\partial{V}}{\partial{T}})_p = 8\times 10^{-5}K^{-1}$
$\beta = -\frac{1}{V}(\frac{\partial{V}}{\partial{p}})_T = 2.2\times 10^{-6}bar^{-1}$ Where the T is meant to be a subscript denoting T constant for that derivative, but I can't make it look much like a subscript.

3. The Attempt at a Solution
From the Maxwell Relation,
$\kappa V = -(\frac{\partial{S}}{\partial{p}})_T$
And since the process is isothermal and I know dP, I'm wondering whether I can't just rearrange so that
$\kappa V \partial p = \partial S$
But that seems like wrong maths in terms of how you're allowed to treat derivatives. I suppose I'm essentially asking if that manipulation is allowed, and if not, I've tried using cyclic relationships, the central equation and other Maxwell relations but I can't seem to get anywhere. So what would I do instead?!

Thanks for any help!

 

[Chestermiller]

Looks correct to me.

 

[Kara386]

Looks correct to me.

OK, thanks! But why can I do that without integrating? Because doesn't $dp$ represent a small change in $p$, while we're looking at a change of $999bar$ in this case?

 

[Chestermiller]

OK, thanks! But why can I do that without integrating? Because doesn't $dp$ represent a small change in $p$, while we're looking at a change of $999bar$ in this case?

Between 1 bar and 1000 bars, the volume only changes of 0.2%. So, it is essentially constant. So, the integrand is essentially constant.