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=10-5m31kJk·bar=10-5m3

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Most popular questions from this chapter

The formula for Cp-Cv derived in the previous problem can also be derived starting with the definitions of these quantities in terms of U and H. Do so. Most of the derivation is very similar, but at one point you need to use the relation P=-(F/V)T.

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.

Use the result of the previous problem to estimate the equilibrium constant of the reactionN2+3H22NH3at 500° C, using only the room- temperature data at the back of this book. Compare your result to the actual value of K at 500° C quoted in the text.

Let the system be one mole of argon gas at room temperature and atmospheric pressure. Compute the total energy (kinetic only, neglecting atomic rest energies), entropy, enthalpy, Helmholtz free energy, and Gibbs free energy. Express all answers in SI units.

A muscle can be thought of as a fuel cell, producing work from the metabolism of glucose:

C6H12O6+6O26CO2+6H2O

(a) Use the data at the back of this book to determine the values of ΔHand ΔGfor this reaction, for one mole of glucose. Assume that the reaction takes place at room temperature and atmospheric pressure.

(b) What is maximum amount of work that a muscle can perform , for each mole of glucose consumed, assuming ideal operation?

(c) Still assuming ideal operation, how much heat is absorbed or expelled by the chemicals during the metabolism of a mole of glucose?

(d) Use the concept of entropy to explain why the heat flows in the direction it does?

(e) How would your answers to parts (a) and (b) change, if the operation of the muscle is not ideal?

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