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In this problem you will investigate the behavior of a van der Waals fluid near the critical point. It is easiest to work in terms of reduced variables throughout.

(a) Expand the van der Waals equation in a Taylor series in , keeping terms through order . Argue that, for T sufficiently close to Tc, the term quadratic in (V-VC)becomes negligible compared to the others and may be dropped.

(b) The resulting expression for P(V) is antisymmetric about the point V = Ve. Use this fact to find an approximate formula for the vapor pressure as a function of temperature. (You may find it helpful to plot the isotherm.) Evaluate the slope of the phase boundary,dP/dT

( c) Still working in the same limit, find an expression for the difference in volume between the gas and liquid phases at the vapor pressure. You should find Vg-VlTc-T.8, where (3 is known as a critical exponent. Experiments show that (3 has a universal value of about 1/3, but the van der Waals model predicts a larger value.

(d) Use the previous result to calculate the predicted latent heat of the transformation as a function of temperature, and sketch this function.

The shape of the T = Tc isotherm defines another critical exponent, called P-PcV-VcCalculate 5 in the van der Waals model. (Experimental values of 5 are typically around 4 or 5.)

A third critical exponent describes the temperature dependence of the isothermal compressibility, K=-t This quantity diverges at the critical point, in proportion to a power of (T-Tc) that in principle could differ depending on whether one approaches the critical point from above or below. Therefore the critical exponents 'Y and -y' are defined by the relations

T-Tc-Tc-T-'

Calculate K on both sides of the critical point in the van der Waals model, and show that 'Y = -y' in this model.

Short Answer

Expert verified

(a) The van der wall force in Taylor seriesPcVcNkTe=127ab23NbN278ba=38=0.375

(b) The two curves becoming indistinguishable over the range.

(c) The difference between volume between the gas and liquid phases at the vapor pressure.

vg-vl=(1+21-t)-(1-21-t)=41-t

(d) the predicted latent heat of the transformation as a function of temperature

LVkTc=38

Step by step solution

01

Part(a) Step 1: Given information

We have been givenp=8t(3v-1)-1-3v-2

02

Part(a) Step 2: Simplify

The terms used in:

3pv3=-1296t(3v-1)-4+72v-5

Therefore, to plot isotherm and perform the Maxwell construction

Series8t/(3v-1)-3/v~2,{v,1,3}

03

Part(b) Step 1: Given information

We have been given, As V-1even gets closer to. the two curves becoming indistinguishable over the range.

04

Part(b) Step 2; Simplify

Constant-pressure line that results in equal area enclosed by two loops

05

Part(c) Step 1: Given information

We have been givenvg-vl=(1+21-t)-(1-21-t)=41-t

The volume of the liquid and gas at transition pressure are just the values of vat transition pressure found in the above parts.

06

Part(c) Step 2: Simplify

The term we will get:

4t-3=4t-3-6(t-1)(v-1)-32(9t-8)(v-1)3
07

Part(d) Step 1;Given information

We have been givenL=TVg-VldPdT=PcVcvg-vldpdt=38NkTcvg-vldpdt

This equation describes a parabola opening to the left descresing to left as t1

08

Part(d) Step 2: Simplify

The term we get:

LNkTc=3841-t4=61-t

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Most popular questions from this chapter

A formula analogous to that for CP-CVrelates the isothermal and isentropic compressibilities of a material:

T=S+TV2CP.

(Here S=-(1/V)(V/P)Sis the reciprocal of the adiabatic bulk modulus considered in Problem 1.39.) Derive this formula. Also check that it is true for an ideal gas.

Sulfuric acid, H2SO4,readily dissociates intoH+andHSO4-H+andHSO4-ions

H2SO4H++HSO4-

The hydrogen sulfate ion, in turn, can dissociate again:

HSO4-H++SO42-

The equilibrium constants for these reactions, in aqueous solutions at 298 K, are approximately 10 and 10*, respectively. (For dissociation of acids it is usually more convenient to look up K than G. By the way, the negative base-10 logarithm of K for such a reaction is called pK, in analogy to pH. So for the first reaction pK = -2, while for the second reaction pK = 1.9.)

(a) Argue that the first reaction tends so strongly to the right that we might as well consider it to have gone to completion, in any solution that could possibly be considered dilute. At what pH values would a significant fraction of the sulfuric acid not be dissociated?

(b) In industrialized regions where lots of coal is burned, the concentration of sulfate in rainwater is typically 5 x 10 mol/kg. The sulfate can take any of the chemical forms mentioned above. Show that, at this concentration, the second reaction will also have gone essentially to completion, so all the sulfate is in the form of SOg. What is the pH of this rainwater?

(c) Explain why you can neglect dissociation of water into H* and OH in answering the previous question. (d) At what pH would dissolved sulfate be equally distributed between HSO and SO2-?

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 dT=27TPdP-27LnRdnw

where nw is the number of moles of water vapor present, L is the latent heat of vaporization per mole, and I've assumed f=7/5for 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 dP/dzanddT/dz, and the vapor pressure PvT. Use the Clausius-Clapeyron relation to eliminate dP/dT.

(c) Combine the results of parts (a) and (b) to obtain a formula relating the temperature gradient, dT/dz, to the pressure gradient, dP/dz. 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 dTdz=-27MgR1+PvPLRT1+27PvPLRT2

where " width="9">

(d) Calculate the wet adiabatic lapse rate at atmospheric pressure (I bar) and 25C, then at atmospheric pressure and 0C. Explain why the results are different, and discuss their implications. What happens at higher altitudes, where the pressure is lower?

Consider the production of ammonia from nitrogen and hydrogen,

N2+3H22NH3

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

Assume that the air you exhale is at 35掳C, with a relative humidity of 90%. This air immediately mixes with environmental air at 5掳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.)

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