Find Isobaric Expansion & Pressure-Volume Coefficient for Solid

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Homework Help Overview

The discussion revolves around the isobaric expansion and pressure-volume coefficient for a solid, with participants analyzing the relationships between volume, temperature, and pressure as described by a given equation.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the derivatives of volume with respect to temperature and pressure, questioning the correctness of their calculations and the relationship between variables. There is also a discussion about potential confusion regarding the notation of pressure.

Discussion Status

Some participants have provided insights into the derivatives and questioned the accuracy of the textbook solution, suggesting that the original question may have been altered. There is a recognition of the need for clarity in the formulation of the problem and its answers.

Contextual Notes

There is uncertainty regarding the notation used for pressure (p vs. P) and the potential misalignment between the question and the provided answers. Participants express confusion about the expectations for the form of the answers.

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Homework Statement
The problem I am trying to solve is,
Find the isobaric expansion coefficient ##\frac{dV}{dT}## and the isothermal pressure-volume coefficient ##\frac{dV}{dP}## of a solid that has equation of state ##V+bpT–cT^2=0##
Relevant Equations
Equation of state of solid: ##V+bpT–cT^2=0##
The answer to this problem is
1686990202428.png

However, I am confused how this relates to the question.

My working is,
##V = cT^2 - bpT##
##\frac{dV}{dT} = 2cT - bp## (I take the partial derivative of volume with respect to temperature to get the isobaric expansion coefficient)

##\frac{dV}{dP} = 0## (I take the partial derivative of volume with respect to pressure to get the isothermal pressure-volume coefficient)

If someone please knows whether I am correct or not then that would be greatly appreciated!

Many thanks!
 
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I suspect that p and P are supposed to be the same.
 
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haruspex said:
I suspect that p and P are supposed to be the same.
Thank you for your reply @haruspex!

That gives ##\frac{dV}{dP} = -bT##. Do you please know whether the textbook solution is wrong? It appears they just give a polynomial in temperature.

Many thanks!
 
ChiralSuperfields said:
Thank you for your reply @haruspex!

That gives ##\frac{dV}{dP} = -bT##. Do you please know whether the textbook solution is wrong? It appears they just give a polynomial in temperature.

Many thanks!
It is obviously not an answer to the question. It merely restates the given equation. The answers must take the form ##\frac{dV}{dT}=## etc. Or better, ##\frac{\partial V}{\partial T}=## etc.
Perhaps the original question was the reverse: it provided the partial derivatives and asked for the state equation. Someone changed the question but forgot to change the answer.
 
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haruspex said:
It is obviously not an answer to the question. It merely restates the given equation. The answers must take the form ##\frac{dV}{dT}=## etc. Or better, ##\frac{\partial V}{\partial T}=## etc.
Perhaps the original question was the reverse: it provided the partial derivatives and asked for the state equation. Someone changed the question but forgot to change the answer.
Thank you for your help @haruspex ! I understand now :)
 

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