[Thermo.] thermal expansion and isothermal compressibility coefficients.

AI Thread Summary
The discussion focuses on calculating the final pressure of a metal when its temperature is raised and determining the maximum achievable temperature under pressure constraints. The thermal expansion coefficient is given as 5.0 × 10^-5 °C^-1, and the isothermal compressibility is 1.2 × 10^-6 atm^-1. Initial calculations led to a negative final pressure, indicating a potential error in the application of the equations. A key insight suggests that the volume change (dV) is negligible due to the insulating Invar layer, which alters the expected outcomes. The isothermal compressibility cannot be negative, reinforcing the need for accurate application of the relevant thermodynamic equations.
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Homework Statement



A certain metal whose thermal expansion coefficient \beta is 5,0 × 10^-5 °C^-1 and whose isothermal compressibility \kappa_T is 1,2 × 10^-6 atm^-1 is at an initial pressure of 1 atm and an initial temperature of 20°C. A thick layer of Invar is thermally insulating the system. The Invar's coefficients are not to be taken into consideration.

a) What would be the final pressure if the temperature raised to 32°C?

b) If the layer could resist to a pressure up to 1200 atm, what would be the highest possible temperature achievable?

Homework Equations



\kappa_T = \frac{-1}{V} (\frac{\partial V}{\partial P})_T\beta = \frac{1}{V} (\frac{\partial V}{\partial T})_P

The Attempt at a Solution



We know from the statement that:

\kappa_T = \frac{-1}{V} (\frac{\partial V}{\partial P})_T = 1,2 \times 10^{-6}

\beta = \frac{1}{V} (\frac{\partial V}{\partial T})_P = 5,0 \times 10^{-5}

a)

Working out both expressions to a common equation gives us

- 1,2 \times 10^{-6} dP = \frac{dV}{V}

5,0 \times 10^{-5} dT = \frac{dV}{V}

Therefore,

- 1,2 \times 10^{-6} dP = 5,0 \times 10^{-5} dT

Integrating,

- 1,2 \times 10^{-6} \int_{P_i}^{P_f}\ dP = 5,0 \times 10^{-5} \int_{T_i}^{T_f}\ dT

- 1,2 \times 10^{-6} (P_f - 1) = 5,0 \times 10^{-5} (32 - 20)

P_f = 1 - 500 = -499\ atm

Is this correct? A negative pressure?

b)

- 1,2 \times 10^{-6} (1200 - 1) = 5,0 \times 10^{-5} (T_f - 20)

T_f = -8.776\ degrees\ Celsius

So what am I doing wrong here?______________________________

EDIT: Is it possible that \kappa_T = - 1,2 \times 10^{-6} instead of + 1,2 \times 10^{-6} ?

That would work out, it seems.
 
Last edited:
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Because the invar has remarkably low thermal expansion coefficient and it covers the metal, the metal barely expands: dV = 0 (this is the important thing you missed!).
One thing to notice: dV = \frac{\partial V}{\partial P}dP + \frac{\partial V}{\partial T}dT (this is where you made the mistake!).
You can work things out from here, can't you? :smile:

P.S: K_T cannot be negative, because the larger the pressure is, the more the metal is compressed, the smaller the volume becomes.
 
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