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Chapter 5: Q 5.26 (page 171)

How can diamond ever be more stable than graphite, when it has

less entropy? Explain how at high pressures the conversion of graphite to diamond

can increase the total entropy of the carbon plus its environment.

Short Answer

Expert verified

As the pressure increases entropy increases.

Step by step solution

01

Given information 

Consider a system; the equilibrium state occurs when the system's entropy maximises, but the entropy of the diamond is less than the entropy of the graphite, despite the fact that the diamond is more stable. This is because the total entropy of the system plus the environment tends to increase, and in the process of Graphite to Diamond conversion under high pressure, the graphite takes up less space, leaving more space for the surrounding material, the entropy is proportional to volume.

02

Conclusion

As a result, this space permits the entropy of the surrounding material to rise. This can also be illustrated using the pressure-entropy relationship:

P=∂S∂VU

so as the pressure increases entropy increases.

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Most popular questions from this chapter

Assume that the air you exhale is at 35°C, with a relative humidity of 90%. This air immediately mixes with environmental air at 5°C and unknown relative humidity; during the mixing, a variety of intermediate temperatures and water vapour percentages temporarily occur. If you are able to "see your breath" due to the formation of cloud droplets during this mixing, what can you conclude about the relative humidity of your environment? (Refer to the vapour pressure graph drawn in Problem 5.42.)

Ordinarily, the partial pressure of water vapour in the air is less than the equilibrium vapour pressure at the ambient temperature; this is why a cup of water will spontaneously evaporate. The ratio of the partial pressure of water vapour to the equilibrium vapour pressure is called the relative humidity. When the relative humidity is 100%, so that water vapour in the atmosphere would be in diffusive equilibrium with a cup of liquid water, we say that the air is saturated. The dew point is the temperature at which the relative humidity would be 100%, for a given partial pressure of water vapour.

(a) Use the vapour pressure equation (Problem 5.35) and the data in Figure 5.11 to plot a graph of the vapour pressure of water from 0°C to 40°C. Notice that the vapour pressure approximately doubles for every 10° increase in temperature.

(b) Suppose that the temperature on a certain summer day is 30° C. What is the dew point if the relative humidity is 90%? What if the relative humidity is 40%?

The enthalpy and Gibbs free energy, as defined in this section, give special treatment to mechanical (compression-expansion) work, -PdV. Analogous quantities can be defined for other kinds of work, for instance, magnetic work." Consider the situation shown in Figure 5.7, where a long solenoid ( Nturns, total length N) surrounds a magnetic specimen (perhaps a paramagnetic solid). If the magnetic field inside the specimen is B→and its total magnetic moment is M→, then we define an auxilliary field H→(often called simply the magnetic field) by the relation

H→≡1μ0B→-M→V,

where μ0is the "permeability of free space," 4π×10-7N/A2. Assuming cylindrical symmetry, all vectors must point either left or right, so we can drop the -→symbols and agree that rightward is positive, leftward negative. From Ampere's law, one can also show that when the current in the wire is I, the Hfield inside the solenoid is NI/L, whether or not the specimen is present.

(a) Imagine making an infinitesimal change in the current in the wire, resulting in infinitesimal changes in B, M, and H. Use Faraday's law to show that the work required (from the power supply) to accomplish this change is Wtotal=VHdB. (Neglect the resistance of the wire.)

(b) Rewrite the result of part (a) in terms of Hand M, then subtract off the work that would be required even if the specimen were not present. If we define W, the work done on the system, †to be what's left, show that W=μ0HdM.

(c) What is the thermodynamic identity for this system? (Include magnetic work but not mechanical work or particle flow.)

(d) How would you define analogues of the enthalpy and Gibbs free energy for a magnetic system? (The Helmholtz free energy is defined in the same way as for a mechanical system.) Derive the thermodynamic identities for each of these quantities, and discuss their interpretations.

Derive a formula, similar to equation 5.90, for the shift in the freezing temperature of a dilute solution. Assume that the solid phase is pure solvent, no solute. You should find that the shift is negative: The freezing temperature of a solution is less than that of the pure solvent. Explain in general terms why the shift should be negative.

In a hydrogen fuel cell, the steps of the chemical reaction are

at-electrode:H2+2OH-⟶2H2O+2e-at+electrode:12O2+H2O+2e-⟶2OH-

Calculate the voltage of the cell. What is the minimum voltage required for electrolysis of water? Explain briefly.
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