Skip to 0 minutes and 0 secondsThis is the final example of applying property relations to a real situation. Thermoelastic effect is the change in temperature of solids upon elastic or reversible deformation under adiabatic condition. What we want to know in thermoelastic effect is the temperature change under deformation by uniaxial force f. It is dT over df. Here we have blocks of solids. Uniaxial force f is acting on a solid. The deformation is assumed within elastic limits. When solid is under tension, the solid will be elongated to the direction of the uniaxial force. The force, by convention, is positive when it is tension. When solid is under compression, the solid will shrink to the direction of the uniaxial force.
Skip to 1 minute and 7 secondsThe force, by convention, is negative when it is compression. When we consider energy functions, we have to include "work" other than P - V work in this case. We are looking for dT over df. Mathematically, (dT over df) at constant S can be converted into - ((dS over df) at constant T) over ((dS over dT) at constant f). Here the denominator is obtained from heat capacity. Since heat capacity at constant f is T times (dS over dT) at constant f, (dS over dT) at constant f can be written as (Cf over T). The equation becomes like this. - T times (dS over df) at constant T over (nCf).
Skip to 2 minutes and 8 secondsHere n is the number of moles of material and Cf is the heat capacity at constant force. To get (dS over df) at constant T, let's start from the thermodyanmic potential dU. dU is TdS - PdV, but in this case under the uniaxial force, we need additional work f dl. Here the uniaxial work has different sign from the P - V work. By sign convention, work is positive when the work is done on the system. For P - V work, if the volume increases with a positive dV, the work is done by the system, so P - V work is negative. Therefore P - V work on the system is - PdV.
Skip to 3 minutes and 7 secondsHowever, f is positive when it is tensile, so when it elongates with a positive dl, the work fdl is done on the system. Thus the uniaxial work is + fdL. Let's examine G now. G is H - TS and U + PV - TS. The differential form dG is thus dU + PdV + VdP - TdS - SdT. Insert the value for dU then dG becomes - SdT + VdP + fdl. Starting from this equation, dG is - SdT + fdl at constant pressure. Exactness theorem applies to get a property relation here. The property relation is (dS over dL) at constant T is - (df over dT) at constant l. Let's look at the left side.
Skip to 4 minutes and 22 seconds(dS over dL) at constant T is mathematically the same with (dS over df) times (df over dl), both at constant T. The right side (df over dT) at constant l is also mathematically the same with this. - (dl over dT) at constant f over (dl over df) at constant T. Then equate those two on the left and right. The right side can be written as ((dl over dT) at constant f) times ((df over dl) at constant T). Now we see the same things on both sides. Cancel out the same things. Then the equation becomes like this. (dS over df) at constant T is (dl over dT) at constant f.
Skip to 5 minutes and 23 secondsNow we have a equation for (dS over df) at constant T. Let's turn back to dT over df. Insert the equation, just derived here. The equation is now - T times (dl over dT) at constant f over nCf. (dl over dT) can be obtained from the linear thermal expansion coefficient, alpha l. The equation looks now simple. It is - (T l alpha l) over (nCf). Usually, the driving force for elastic deformation is given as a stress not a force. The stress sigma is force per area. Force f is sigma times area. So (df over d sigma) is A. Since the driving force is given as a stress, we need (dT over d sigma) instead of (dT over df).
Skip to 6 minutes and 37 secondsMathematical manipulation will give us the result. (dT over d sigma) equals to (dT over df) times (df over d sigma). Here, the relations obtained previously, are applied. (dT over df) is - (T l alpha l) over (nCf). (df over d sigma) is A. It is now - (T l alpha l) over (nCf) times A. L times A is the volume V. It is now - (T V alpha l) over (nCf). V over n is the molar volume V underbar. So finally it is - (T V underbar alpha l) over Cf. (dT over d sigma) is now - (T V underbar alpha l) over Cf.
Skip to 7 minutes and 47 secondsHeat capacities at constant force, stress, and pressure are essentially the same, so we can change Cf with Cp. So the equation becomes - (T V underbar alpha l) over Cp. Integrate by assuming delta T is small, then everything on the right side can be regarded as constant. The integration is approximated to delta T is - (T V under alpha l) over Cp times delta sigma.
Thermoelastic effect is the temperature change resulting from stretching or contracting of an elastic material.
The temperature change in this case can be calculated mathematically from the stress dependence of the temperature at constant entropy. Through multiple mathematical manipulation as well as a property relation, this dependence of temperature on applied pressure is evaluated. It depends on temperature, thermal expansion coefficient and heat capacity.