## Description

1 van der Waals CO2 redux

Last homework you worked with the fundamental van der Waals equation:

svdW(u,v) = s0 + Rln(v โ b) + cRln(u + a/v) (1) 1. Construct the Helmholtz potential of a van der Waals fluid, avdW(T,v). Hints: First differentiate to find the relationship between u and T.

2. Recall the example done in class of a piston separating two chambers of volumes 1 and 10L, respectively, each containing 1 mole of ideal gas, the whole system held isothermal at 273K. We showed in class that โ2.5kJ of work could be extracted by allowing the piston to adjust the volumes to 5L and 6L. Compare the amount of work available from the same process and conditions if the fluid was van der Waals CO2.

3. Compare the ideal and van der Waals CO2 cases if the volumes started at 0.1 and 1.0L and ended at 0.5 and 0.6L.

4. Take the same case of a system of total volume of 1.1L, but now put 1 mole of van der Waals CO2 fluid on one side of the piston and 1 mole of ideal gas on the other. If the piston was allowed to freely move, where would it end up?

2 Manipulating thermodynamic derivatives

As Prateek might say, the susceptibilities, cP, ฮฑ, and ฮบT, are โawesome.โ Everything about a one component fluid can be expressed in terms of them, and for good reason! Hereโs some practice with them.

1. Express the derivative in terms of the three susceptibilities.

h

2. Determine the susceptibilities of a van der Waals fluid.

3. Plot vs. v for T from โ50 to 50โฆC in 20โฆC increments for van der Waals CO2. Under what conditions of temperature and volume is the derivative negative?h

3 Practice with potentials

While a Carnot refrigerator sure would be efficient, in practice we donโt like to wait an eternity for our โfridges to get cold. Practical refrigerators use the Joule-Thomson effect and irreversible, isenthalpic expansion to create the cold we love. Letโs see how cold we can make things with van der Waals CO2.

1. Construct the enthalpy hvdW(T,v) of a van der Waals fluid.

2. Determine the change in temperature when van der Waals CO2 is isenthalpically expanded from 10โ4 to 10โ3 m3 molโ1, starting from 10โฆC.

3. A more reliable estimate of the change in temperature can be gotten from thermodynamic properties charts. Thermodynamic charts for CO2 refrigerant are available at link. Use the pressure-enthalpy chart at the end of this document to determine the change in temperature for the same isenthalpic expansion as in the previous question.

4 A departure from ideality

The examples above assume the heat capacity of CO2 to be a constant. More generally heat capacities are functions of temperature and pressure. (Weโll learn why when we study statistical mechanics.) Itโs common to report heat capacities in a hypothetical ideal gas, pressure-independent limit, cigp (T). For instance, the molar heat capacity of ideal CO2 is reported to be:

cigp (t) = โ11.401074 โ 55.231532t + 5.149108t2 โ 0.29158t3 + 0.110128tโ2 + 115.93493t1/2

where t = T(K)/1000.

1. Find the change in molar entropy of โideal gasโ CO2 when heated from 290 to 350K.

2. To find the real change in entropy, we would need to know the difference in entropy between the ideal gas and real states of CO2 at identical T and v conditions. This difference is called a โdeparture function.โ To create this, we construct a hypothetical path between the real and ideal gas states, first expanding the van der Waals gas from v to v โโ, where it behaves ideally, and then compressing from โ to v as an ideal gas. Express in terms

T of susceptibilities.

1. Construct for ideal CO2.

T

2. Construct for van der Waals CO2.

T

3. Construct the departure function SvdW(T,v) โ Sig(T,v) by appropriate integration. Hint: This difference is equivalent to adding SvdW integrated along the path v โโ and Sig integrated along the path โโ v.

4. Use the departure function (twice) to compute the molar entropy change when van der Waals CO2 is heated from 290K and 1.0L/mol to 350K and 0.15L/mol.

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