Find Isobaric Expansion & Pressure-Volume Coefficient for Solid

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SUMMARY

The discussion focuses on the calculation of isobaric expansion and pressure-volume coefficients for solids, specifically using the equation V = cT^2 - bpT. The user correctly derives the isobaric expansion coefficient as ∂V/∂T = 2cT - bp and questions the textbook's solution, which presents a polynomial in temperature without addressing the required form of the derivatives. The conversation highlights the importance of clarity in mathematical expressions and the potential confusion arising from misinterpretation of variables.

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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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