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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=SVU

so as the pressure increases entropy increases.

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Most popular questions from this chapter

Suppose you have a box of atomic hydrogen, initially at room temperature and atmospheric pressure. You then raise the temperature, keeping the volume fixed.

(a) Find an expression for the fraction of the hydrogen that is ionised as a function of temperature. (You'll have to solve a quadratic equation.) Check that your expression has the expected behaviour at very low and very high temperatures.

(b) At what temperature is exactly half of the hydrogen ionised?

(c) Would raising the initial pressure cause the temperature you found in part (b) to increase or decrease? Explain.

(d) Plot the expression you found in part (a) as a function of the dimension- less variable t = kT/I. Choose the range of t values to clearly show the interesting part of the graph.

Suppose you have a liquid (say, water) in equilibrium with its gas phase, inside some closed container. You then pump in an inert gas (say, air), thus raising the pressure exerted on the liquid. What happens?

(a) For the liquid to remain in diffusive equilibrium with its gas phase, the chemical potentials of each must change by the same amount: dμl=dμg Use this fact and equation 5.40 to derive a differential equation for the equilibrium vapour pressure, Pv as a function of the total pressure P. (Treat the gases as ideal, and assume that none of the inert gas dissolves in the liquid.)

(b) Solve the differential equation to obtain

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(c) Calculate the percent increase in vapour pressure when air at atmospheric pressure is added to a system of water and water vapour in equilibrium at 25°C. Argue more generally that the increase in vapour pressure due to the presence of an inert gas will be negligible except under extreme conditions.

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.

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