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+clear;

+//clc();

+

+// Example 6.4

+// Page: 113

+printf("Example-6.4  Page no.-113\n\n");

+

+//***Data***//

+m = 1;//[molal] Molality of the solution with respect to ethanol

+M_water = 18;//[g/mol] molecular weight of water

+

+// First we convert molality to mole fraction

+x_ethanol = m/(m + 1000/M_water);

+

+// For the low range of data point on figure 6.5(page 112), we can fit an equation

+// (Specific volume ) = 0.018032 + 0.037002*x_ethanol - 0.039593*x_ethanol^(2) + 0.21787*x_ethanol^(3)

+// This is applicable for (0 < x_ethanol < 0.04 ), which is the case we have

+

+// So

+v_tan = 0.018032 + 0.037002*x_ethanol - 0.039593*x_ethanol^(2) + 0.21787*x_ethanol^(3);//[L/mol]

+

+// Now we will find the derivative of the specific volume with respect to x_ethanol at the known point x_ethanol

+// (dv/dx_ethanol) =  0.037002 - 2*0.039593*x_ethanol + 3*0.21787*x_ethanol^(2)

+// Hence

+v_derv_tan = 0.037002 - 2*0.039593*x_ethanol + 3*0.21787*x_ethanol^(2);//[L/mol]

+

+// By simple geometry from the figure 6.6(page 113) of the book we find

+// a = v_tan + (1-x_tan)*(dv/dx_1)_tan

+// b = v_tan - x_tan*(dv/dx_1)_tan

+

+// We have a = v_ethanol and b = v_water

+x_tan = x_ethanol;

+// So

+v_ethanol = v_tan + (1-x_tan)*(v_derv_tan);//[L/mol]

+v_water = v_tan - x_tan*(v_derv_tan);//[L/mol]

+

+printf(" Partial molar volume of the ethanol in the given solution is %f L/mol\n",v_ethanol);

+printf(" Partial molar volume of the water in the given solution is %f L/mol",v_water);

+

+

+

+