Sketch a qualitatively accurate graph of G vs. T for a pure substance as it changes from solid to liquid to gas at fixed pressure. Think carefully about the slope of the graph. Mark the points of the phase transformations and discuss the features of the graph briefly.

Short Answer

Expert verified

The slope of the curve is most stepper for gaseous state than liquid state than solid state for a pure substance.

Step by step solution

01

Introduction

Write the expression for Gibbs free energy.

G=U-TS+PV

Here, G is Gibbs free energy, T is the absolute temperature, S is the entropy, P is the pressure and V is the volume.

02

Explanation

A plot between G and T will simply be a line graph where the slope of the graph is the negative entropy -S. The entropy is lowest for the solid and since the slope is negative, the line for the solid has the highest value in the vertical axis. The substance then melts into liquid which results larger value of entropy than solid, making the slope of the graph more stepper and finally it evaporates into a gas, the value of entropy for which is largest. It makes the slope of the graph most stepper.

Draw a graph to show the variation of G and T for a pure substance.

The slope of the curve is most stepper for gaseous state than liquid state than solid state for a pure substance.

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Most popular questions from this chapter

Consider the production of ammonia from nitrogen and hydrogen,

N2+3H22NH3

at 298 K and 1 bar. From the values of ΔH and S tabulated at the back of this book, compute ΔG for this reaction and check that it is consistent with the value given in the table.

Use the result of the previous problem and the approximate values of a and b to find the value of Tc, Pc, Vc/N for N2, H2O and He.

Sketch qualitatively accurate graphs of G vs. P for the three phases of H20 (ice, water, and steam) at 0°C. Put all three graphs on the same set of axes, and label the point corresponding to atmospheric pressure. How would |the graphs differ at slightly higher temperatures?

In the previous section I derived the formula (F/V)T=-P. Explain why this formula makes intuitive sense, by discussing graphs of F vs. V with different slopes.

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

Pv(P)-PvPv=eP-PvV/NkT

where the ratio V/N in the exponent is that of the liquid. (The term Pv(Pv) is just the vapour pressure in the absence of the inert gas.) Thus, the presence of the inert gas leads to a slight increase in the vapour pressure: It causes more of the liquid to evaporate.

(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.

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