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In working high-pressure geochemistry problems it is usually more

convenient to express volumes in units of kJ/kbar. Work out the conversion factor

between this unit and m3

Short Answer

Expert verified

The conversion factor is1kJk·bar=10-5m3

Step by step solution

01

Step 1. Given information 

We used kJ/kbar as the unit of pressure in the high-pressure geochemical problem.

1kJkbar=1·103·J103·bar=1·Jbar

02

Calculation

Using the conversion between the units of pressure 1 bar = 105 Pa, we get

=1Jbar×lbar105N/m2=10-5JN/m2=10-5×N·MN/m2=10-5m3∴1kJk·bar=10-5m3

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Most popular questions from this chapter

By subtracting μNfrom localid="1648229964064" U,H,F,orG,one can obtain four new thermodynamic potentials. Of the four, the most useful is the grand free energy (or grand potential),

Φ≡U-TS-μN.

(a) Derive the thermodynamic identity for Φ, and the related formulas for the partial derivatives ofΦwith respect toT,V, and μN

(b) Prove that, for a system in thermal and diffusive equilibrium (with a reservoir that can supply both energy and particles), Φtends to decrease.

(c) Prove thatϕ=-PV.

(d) As a simple application, let the system be a single proton, which can be "occupied" either by a single electron (making a hydrogen atom, with energy -13.6eV) or by none (with energy zero). Neglect the excited states of the atom and the two spin states of the electron, so that both the occupied and unoccupied states of the proton have zero entropy. Suppose that this proton is in the atmosphere of the sun, a reservoir with a temperature of 5800Kand an electron concentration of about 2×1019per cubic meter. Calculate Φfor both the occupied and unoccupied states, to determine which is more stable under these conditions. To compute the chemical potential of the electrons, treat them as an ideal gas. At about what temperature would the occupied and unoccupied states be equally stable, for this value of the electron concentration? (As in Problem 5.20, the prediction for such a small system is only a probabilistic one.)

The partial-derivative relations derived in Problems 1.46,3.33, and 5.12, plus a bit more partial-derivative trickery, can be used to derive a completely general relation between CPandCV.

(a) With the heat capacity expressions from Problem 3.33 in mind, first considerSto be a function of TandV.Expand dSin terms of the partial derivatives (∂S/∂T)Vand (∂S/∂V)T. Note that one of these derivatives is related toCV

(b) To bring in CP, considerlocalid="1648430264419" Vto be a function ofTand P and expand dV in terms of partial derivatives in a similar way. Plug this expression for dV into the result of part (a), then set dP=0and note that you have derived a nontrivial expression for (∂S/∂T)P. This derivative is related to CP, so you now have a formula for the difference CP-CV

(c) Write the remaining partial derivatives in terms of measurable quantities using a Maxwell relation and the result of Problem 1.46. Your final result should be

CP=CV+TVβ2κT

(d) Check that this formula gives the correct value of CP-CVfor an ideal gas.

(e) Use this formula to argue that CPcannot be less than CV.

(f) Use the data in Problem 1.46 to evaluateCP-CVfor water and for mercury at room temperature. By what percentage do the two heat capacities differ?

(g) Figure 1.14 shows measured values of CPfor three elemental solids, compared to predicted values of CV. It turns out that a graph of βvs.T for a solid has same general appearance as a graph of heat capacity. Use this fact to explain why CPand CVagree at low temperatures but diverge in the way they do at higher temperatures.

In constructing the phase diagram from the free energy graphs in Figure 5.30, I assumed that both the liquid and the gas are ideal mixtures. Suppose instead that the liquid has a substantial positive mixing energy, so that its free energy curve, while still concave-up, is much flatter. In this case a portion of the curve may still lie above the gas's free energy curve at TA. Draw a qualitatively accurate phase diagram for such a system, showing how you obtained the phase diagram from the free energy graphs. Show that there is a particular composition at which this gas mixture will condense with no change in composition. This special composition is called an azeotrope.

Prove that the entropy of mixing of an ideal mixture has an infinite slope, when plotted vs. x, at x = 0 and x= 1.

The first excited energy level of a hydrogen atom has an energy of 10.2 eV, if we take the ground-state energy to be zero. However, the first excited level is really four independent states, all with the same energy. We can therefore assign it an entropy of S =kln(4) , since for this given value of the energy, the multiplicity is 4. Question: For what temperatures is the Helmholtz free energy of a hydrogen atom in the first excited level positive, and for what temperatures is it negative? (Comment: When F for the level is negative, the atom will spontaneously go from the ground state into that level, since F=0 for the ground state and F always tends to decrease. However, for a system this small, the conclusion is only a probabilistic statement; random fluctuations will be very

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