Need help solving 2.3 and 2.4 (a) (b) molecular weight 500 g/mol, answer the fol

Question

Need help solving 2.3 and 2.4 (a) (b)

molecular weight 500 g/mol, answer the following questions (2.3) A solution is held in a cylinder which is sealed by two pistons, and divided into two chambers by a semi-permeable membrane (see Fig. 2.11) Initially, the concentrations of the two chambers are the same at number density no, and the chambers have the same volume hA (h and A are the height and the cross-section of the chamber). A weight W is placed on top of the cylinder, causing the piston to move down. Assuming that the solution is ideal (a) Calculate the weight concentration c [g/cm3], weight fraction m, and molar fraction xm of sugar in the solution (b) Estimate the osmotic pressure II of the solu tion (2.2) Polystyrene of molecular weight 3x10' g/mol has a radius of gyration Rg of about 100 nm in benzene answer the following questions Answer the following questions (a) Calculate the displacement r of the piston at equilibrium ignoring the density of the solution (b) Calculate the displacement r of the piston tak ing into account of the density of the solution (a) Estimate the overlap concentration at which the polymer molecules start to overlap each other 26 Soft matter solutions (d) Consider the situation that the volume fraction (i 1, 2, ,n) is much smaller than - In this situation, the osmotic pressure is written as n. 72 rn 0 Show that the chemical potentials of the minor Fig. 2.11 componentsi 1,2,.. , n) are given by

Explanation / Answer

Initially:

Height of piston-1 from membrane is:            h
Height of piston-2 from membrane is:            h

Initially Force on piston 2 is: density*gravity*height*area
                                                        = n*g*(h+h)*A

W weight placed on top piston:

Height of piston-1 from membrane is:            h-x
Height of piston-2 from membrane is:            h+x

Force on piston 2 after putting weight on piston 1: density*gravity*height*area
                                                                                        = (n*g*(h – x + h + x) + W)*A
                                                                                        = (n*g*(h + h) + W)*A

On equilibrium - total height between pistons is same as initial condition and there will be no resistance of membrane between two portions so that no density difference and compression of solution occurs:
                                                                                        = h–x +h + x = 2h

Displacement of pistons down after putting weight W is:    x

                                                        W = n*g*displacement*area
                                                        W = n*g*x*A

                                                        x = W/( n*g*A) Ans

2. If Density change for both chambers: n1, n2

Initially Force on piston 2 is: density*gravity*height*area
                                                        = n*g*(h+h)*A = 2*n*g*h*A

Force on piston 2 after putting weight on piston 1: density*gravity*height*area
= (n1*g*(h-x) + n2*g*(h+x))*A + W

On equilibrium:

Displacement of pistons down after putting weight W is:    x

Total Force Balance:

       2*n*g*h*A + W = n1*g*x*A + (n1*g*(h-x) + n2*g*(h+x))*A
2*n*g*h*A + W = x*A (n1*g - n1*g + n2*g) - n1*g*h + n2*g*h

x = (2*n*g*h*A + W + n1*g*h - n2*g*h) / (n1*g - n1*g + n2*g) Ans

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