[Thermo.] thermal expansion and isothermal compressibility coefficients.

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SUMMARY

The discussion centers on the thermal expansion coefficient (\(\beta\)) of a metal at 5.0 × 10-5 °C-1 and its isothermal compressibility (\(\kappa_T\)) at 1.2 × 10-6 atm-1. When the temperature is raised from 20°C to 32°C, the calculated final pressure yields a nonsensical negative value of -499 atm, indicating an error in the approach. The correct interpretation suggests that the volume change (\(dV\)) is negligible due to the insulating properties of Invar, which leads to the conclusion that \(\kappa_T\) should not be negative, as it contradicts the physical behavior of materials under pressure.

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  • Understanding of thermal expansion coefficients
  • Knowledge of isothermal compressibility
  • Familiarity with calculus, particularly integration
  • Basic principles of thermodynamics
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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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