Skip to 0 minutes and 0 secondsHere, we are going to derive the isoentropic pressure - temperature relationship using a property relation. Previously, we have obtained equation of states for adiabatic, reversible changes in a closed system of ideal gas in week 2. The equation of state was P1 times (V1 to gamma) is the same with P2 times (V2 to gamma). The adiabatic, reversible process is a isoentropic process. So this equation of state is kind of isoentropic pressure - volume relation, and can be converted to pressure - temperature relation. The same equation can be obtained using property relations. The isoentropic P - T relation is dT over dP at constant S. The mathematical manipulation changes the equation like this.
Skip to 1 minute and 3 seconds(dT over dP) at constant S is - (dS over dP at constant T) over (dS over dT at constant P). The bottom (dS over dT at constant P) is Cp over T. The numerator (dS over dP at constant T) looks familiar. It can be obtained from dG as before and it is - (dV over dT) at constant P. Thus, the equation become like this. - T over Cp times (- dV over dT at constant P). So, the isoentropic P - T relation (dT over dP) at constant S is (T over Cp) times (dV over dT at constant P). Let's apply this equation to ideal gas.
Skip to 1 minute and 54 secondsFor ideal gas, PV equals to RT so (dV over dT) at constant P is (R over P). Then, isoentropic (dT over dP) becomes (T over Cp) times (R over P). Separation of variables makes the equation (dT over T) equals (R over Cp) times (dP over P). Integrate both sides gives natural log (T2 over T1) equals to (R over Cp) times log (P2 over P1). Change the logarithmic equation into index equation. Rearrange this equation with PV equals to RT relation. Then it becomes P times (V to gamma) is constant. This relation is the same result for the previous adiabatic, reversible expansion of an ideal gas since adiabatic reversible process is an isoentropic process.
Skip to 3 minutes and 3 secondsNow, let's apply the isoentropic temperature - pressure relations to solids. The isoentropic (dT over dP) is (T over Cp) times (dV over dT) at constant P as before. Here, (dV over dT) can be obtained from thermal expansion coefficient. (1 over V) times (dV over dT) is alpha V and it is 3 times alpha L. So the equation becomes (T over Cp) times V times (3 alpha L). Separation of variables results in this equation. (dT over T) is 3 times (V alpha L) over Cp times dP. Integrate both sides, assuming that V, alpha V and Cp are independent of pressure, thus can be regarded as constant. This assumption is usually acceptable for solids.
Skip to 4 minutes and 6 secondsThe equation is now log (T2 over T1) equals 3 times (V alpha L) over Cp times (P2-P1). Let's get some real values. The pressure of a solid copper increases from 1 to 10 atm at 300 K. The calculated temperature increase is only 0.003 K.
Isoentropic P-T relation
We have derived the equation of state of an ideal gas under adiabatic reversible changes in week 2(‘Equation of state for adiabatic, reversible changes of ideal gases’). The equation of state in this case shows relations between temperature and pressure or pressure and volume.
We can derive this relation using a property relation. Since the adiabatic reversible change is an isoentropic change, our task is to derive an expression for pressure dependence of temperature at constant entropy. The obtained temperature-pressure relation or pressure-volume relation is the same as before for the ideal gas under adiabatic, reversible changes. The temperature change upon pressure change at constant entropy for solids is also estimated with some numerical values.