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Dependence of entropy on pressure at constant temperature

(dS,dP) at constant T
Welcome to thermodynamics in energy engineering week 6. We are going to demonstrate the usefulness of the property relations by applying them to various situations. Here is the first example to demonstrate the usefulness of the property relations. We are going to calculate the entropy change of isothermal pressure change from 1 to 10 atm. Pressure changes at constant T, so we need to know dS over dP at constant T. dS over dP at const T can be obtained from dG. Look at the two independent variables. They are T and P. Which thermodynamic potential has T, P as variables? It’s G. Applying the exactness property to G gives this property relation.
-(dS over dP) at constant T is (dV over dT) at constant P. Let’s first assume that our system is the ideal gas. For ideal gas, (dV over dT) at constant P is (R over V) from the equation of state. So the entropy change is now integration of - (R over P) dP from 1 to 10 atm. The result is -19.14 J/mole.k. Second, let’s assume our system as a solid under isothermal compression. The property relation also holds in this case. Here, the equation of state for ideal gas is not valid. Instead, we can use the alpha v, the volume expansion coefficient defined like this. Alpha v is (1 over V) times (dV over dT) at constant p.
So (dV over dT) at constant P is alpha v times V. The entropy change is then integration of V times (alpha v) dP from 1 to 10 atm. If V and alpha v do not depend on P or if the pressure range of interest is small, we can regard V and alpha v as constant. So the entropy change is - V times alpha V times pressure change. Let’s insert the real value of copper to get actual entropy change. For copper, specific volume is 7.09 times 10 to -6 cubic meter per mole, and the linear expansion coefficient is 1.67 times 10 to -6 meter per meter K.
The relation between volume expansion coefficient and the linear expansion coefficient is alpha v is three times alpha L. The entropy change with this value is - three times (7.09 times 10 to -6) times (1.67 times 10 to -6) times (10-1) atm times unit conversion factor (1.013 times 10 to 5 ) newton (meter squared) per atm. The entropy change is 3.24 times 10 to -4 J/mole.K. Which is much much smaller than the entropy change of ideal gas under the same compression.

The dependence of entropy temperature is simple known. It is heat capaity over temperautre. How about the dependence of entropy on pressure at constant temperature?

At a first glace, it cannnot be simply obtained. We can derive this dependence using a property relation. With the help of the property relation, the dependence of entropy on pressure can be expressed with more familar form: the temperature dependence of volume. Numerical excercise is provided for an ideal gas and a solid.

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Thermodynamics in Energy Engineering

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