Find Isobaric Expansion & Pressure-Volume Coefficient for Solid

AI Thread Summary
The discussion centers on finding the isobaric expansion and pressure-volume coefficients for a solid, with participants analyzing the derivatives of volume with respect to temperature and pressure. The initial calculations presented involve deriving the coefficients, but confusion arises regarding the relationship between the variables and the correctness of the textbook solution. It is suggested that the textbook may have provided a polynomial in temperature rather than the expected form of the derivatives. Participants clarify that the answers should be expressed as partial derivatives, indicating a potential mix-up in the original question and its corresponding answer. The conversation concludes with a participant expressing gratitude for the clarification received.
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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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