How Does the Equation of State for a Solid Change with Temperature and Pressure?

AI Thread Summary
The discussion revolves around finding the equation of state for a solid based on given coefficients of expansion. The integral of the isobaric expansion coefficient leads to a volume equation, but the initial solution was marked incorrect. The correct form of the equation, as per the textbook, is V - bpT + cT^2 = 0. There was confusion regarding the treatment of Vo, which should be considered a function of pressure. Ultimately, the clarification confirmed that the original answer was indeed correct after addressing the problem statement error.
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Homework Statement


[/B]
Find the equation of state of a solid that has an isobaric expansion coefficient
dV/dT = 2cT - bp
and an isothermal pressure-volume coefficient
dV/dp = -bT
(Assume the solid has a volume Vo at zero temperature and pressure. Enter a mathematical equation. Use any variable or symbol stated above as necessary.)

Homework Equations



None, just need to find the integral of the first equation

The Attempt at a Solution


[/B]
So I took the integral of the first equation to find the volume equation:
V = ∫(2cT - bp)dT = cT^2 - bpT + Vo
but Vo = 0, so:
V = cT^2 - bpT
or
V - cT^2 +bpT = 0
However, when I entered this as the answer, it marked it as wrong. Then I found the answer to this question in my textbook and it says that the correct answer is:
V - bpT + cT^2 = 0
So is the textbook wrong? Or did I do something wrong in my solution?
 
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It doesn't look like your answer satisfies the 2nd equation in your problem statement.
 
Chestermiller said:
It doesn't look like your answer satisfies the 2nd equation in your problem statement.
How do I make my answer satisfy the 2nd equation? I'm just a little confused because this isn't something we went over in the lecture.
 
After the first integration, ##V_0## should be considered a function of p, ##V_0(p)##. So, at constant temperature,
$$\left(\frac{\partial V}{\partial p}\right)_T=-bT+\frac{dV_0}{dp}=-bT$$

Wait a minute! You edited the original problem statement. Which is it?
 
Chestermiller said:
After the first integration, ##V_0## should be considered a function of p, ##V_0(p)##. So, at constant temperature,
$$\left(\frac{\partial V}{\partial p}\right)_T=-bT+\frac{dV_0}{dp}=-bT$$

Wait a minute! You edited the original problem statement. Which is it?
The current equations above are the right ones. I originally mistyped the question.
 
hnnhcmmngs said:
The current equations above are the right ones. I originally mistyped the question.
Then your answer is correct.
 
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