Lead is compressed reversibly with T const, find dS

In summary, the conversation discusses the change in entropy when lead is compressed from 1 bar to 1000 bars at a constant temperature of 300K. The Maxwell relation and other equations are used to determine the change in entropy, with the final conclusion being that the volume change is small enough to consider the integrand as constant.
  • #1
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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!
 
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  • #2
Looks correct to me.
 
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Likes gcuf and Kara386
  • #3
Chestermiller said:
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?
 
  • #4
Kara386 said:
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.
 
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1. What is the definition of "Lead is compressed reversibly"?

Lead is compressed reversibly means that the compression process is carried out slowly and without any friction or energy loss, allowing the system to reach equilibrium at each step.

2. How is the compression of lead reversible with constant temperature (T)?

The compression of lead is reversible with constant temperature because the system is compressed slowly and smoothly, allowing the temperature to remain constant throughout the process.

3. How is the entropy (S) affected by the reversible compression of lead?

The reversible compression of lead does not affect the entropy of the system, as the process is carried out at a constant temperature and there is no change in the number of microstates.

4. What is the mathematical equation for finding the change in entropy (dS) during the reversible compression of lead?

The equation for finding the change in entropy during reversible compression of lead is dS = nRln(V2/V1), where n is the number of moles, R is the gas constant, and V2 and V1 are the final and initial volumes, respectively.

5. What is the significance of finding dS during the reversible compression of lead?

Finding dS during the reversible compression of lead allows us to calculate the change in entropy of the system, which is a measure of the disorder or randomness of the particles. This information can be useful in understanding the thermodynamic properties and behavior of lead under different conditions.

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