When plotting graphs and performing numerical calculations, it is convenient to work in terms of reduced variables, Rewrite the van der Waals equation in terms of these variables, and notice that the constants a and b disappear.

Short Answer

Expert verified

p=8t3v-1-3v2

Step by step solution

01

Given information 

P=pPcT=tTcV=vVc

and

role="math" localid="1646979368068" P=NkT(V-Nb)-aN2V2 (van der Waal's equation)

02

 Substituting the values of P, V and T in the van der Waal equation.

P=NkT(V-Nb)-aN2V2pPc=NktTc(vVc-Nb)-aN2(vVc)2

03

 Substituting the values of Pc, Vc and Tc in equation. 

p127ab2=N(3Nbv-Nb)827abt-aN2(3Nbv)2

further solving the equation we get

p=8t3v-1-3v2

The van der Waal's equation is independent of constants a and b when represented in reduced variables

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Most popular questions from this chapter

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.

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

Vc=3Nb, Pc =127ab2 and kTc=827ab

Derive the thermodynamic identity for G (equation 5.23), and from it the three partial derivative relations 5.24.

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

κT=κS+TVβ2CP.

(Here κS=-(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.

Osmotic pressure measurements can be used to determine the molecular weights of large molecules such as proteins. For a solution of large molecules to qualify as "dilute," its molar concentration must be very low and hence the osmotic pressure can be too small to measure accurately. For this reason, the usual procedure is to measure the osmotic pressure at a variety of concentrations, then extrapolate the results to the limit of zero concentration. Here are some data for the protein hemoglobin dissolved in water at 3oC:

Concentration (grams/liter)h (cm)
5.62.0
16.66.5
32.512.8
43.417.6
54.022.6

The quantity his the equilibrium difference in fluid level between the solution and the pure solvent,. From these measurements, determine the approximate molecular weight of hemoglobin (in grams per mole).

An experimental arrangement for measuring osmotic pressure. Solvent flows across the membrane from left to right until the difference in fluid level,h, is just enough to supply the osmotic pressure.

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