Chapter 5: Q 5.24 (page 171)
Go through the arithmetic to verify that diamond becomes more stable than graphite at approximately 15 kbar.
Short Answer
The diamond is more stable than graphite at 15 kbar.
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Chapter 5: Q 5.24 (page 171)
Go through the arithmetic to verify that diamond becomes more stable than graphite at approximately 15 kbar.
The diamond is more stable than graphite at 15 kbar.
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By subtracting from localid="1648229964064" ,or,one can obtain four new thermodynamic potentials. Of the four, the most useful is the grand free energy (or grand potential),
(a) Derive the thermodynamic identity for , and the related formulas for the partial derivatives ofwith respect to, and
(b) Prove that, for a system in thermal and diffusive equilibrium (with a reservoir that can supply both energy and particles), tends to decrease.
(c) Prove that
(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 ) 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 and an electron concentration of about per cubic meter. Calculate for 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.)
In Problem 1.40 you calculated the atmospheric temperature gradient required for unsaturated air to spontaneously undergo convection. When a rising air mass becomes saturated, however, the condensing water droplets will give up energy, thus slowing the adiabatic cooling process.
(a) Use the first law of thermodynamics to show that, as condensation forms during adiabatic expansion, the temperature of an air mass changes by
where nw is the number of moles of water vapor present, L is the latent heat of vaporization per mole, and I've assumed for air.
(b) Assuming that the air is always saturated during this process, the ratio nw/n is a known function of temperature and pressure. Carefully express dnw/dz in terms of , and the vapor pressure . Use the Clausius-Clapeyron relation to eliminate .
(c) Combine the results of parts (a) and (b) to obtain a formula relating the temperature gradient, , to the pressure gradient, . Eliminate Figure 5.18. Cumulus clouds form when rising air expands adiabatically and cools to the dew point (Problem 5.44); the onset of condensation slows the cooling, increasing the tendency of the air to rise further (Problem 5.45). These clouds began to form in late morning, in a sky that was clear only an hour before the photo was taken. By mid-afternoon they had developed into thunderstorms. the latter using the "barometric equation" from Problem 1.16. You should finally obtain
where " width="9">
(d) Calculate the wet adiabatic lapse rate at atmospheric pressure (I bar) and , then at atmospheric pressure and . Explain why the results are different, and discuss their implications. What happens at higher altitudes, where the pressure is lower?
Consider a fuel cell that uses methane ("natural gas") as fuel. The reaction is
(a) Use the data at the back of this book to determine the values of and for this reaction, for one mole of methane. Assume that the reaction takes place at room temperature and atmospheric pressure.
(b) Assuming ideal performance, how much electrical work can you get out of the cell, for each mole of methane fuel?
(c) How much waste heat is produced, for each mole of methane fuel?
(d) The steps of this reaction are
What is the voltage of the cell?
Plot the Van der Waals isotherm for T/Tc = 0.95, working in terms of reduced variables. Perform the Maxwell construction (either graphically or numerically) to obtain the vapor pressure. Then plot the Gibbs free energy (in units of NkTc) as a function of pressure for this same temperature and check that this graph predicts the same value for the vapor pressure.
As you can see from Figure5.20,5.20,the critical point is the unique point on the original van der Walls isotherms (before the Maxwell construction) where both the first and second derivatives ofPPwith respect toVV(at fixedTT) are zero. Use this fact to show that
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