initial commit / add all books

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+clear;

+//clc();

+

+// Example 6.1

+// Page: 108

+printf("Example-6.1  Page no.-108\n\n");

+

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

+T = 20;//[C]

+m_1 = 0;//[molal]

+m_2 = 1;//[molal]

+// The data given in the figure 6.2 , as reported in book, can be repersented with excellent accuracy by a simple data fitting equation

+//V = 1.0019+0.054668*m-0.000418*m^(2);

+// Where 'V' is( solution volume, liters per 1000g of water ) and 'm' is the molality of ethanol in water

+//The partial molal volume is obtained by differentiating the expression of the 'V' with respect to 'm'

+// v_ethanol = dV/dm = 0.054668-2*0.000418*m

+// So that at zero molality 

+m = 0;//[molal]

+// the partial molal volume is 

+v_1 = 0.054668-2*0.000418*m;//[L/mol]

+// and at

+m = 1;//[molal]

+v_2 = 0.054668-2*0.000418*m;//[L/mol]

+v_1 = v_1*1000;//[cm^(3)/mol]

+v_2 = v_2*1000;//[cm^(3)/mol]

+printf("Partial molal volume of ethanol in water at zero molality is  %f cm^(3)/mol\n",v_1);

+printf(" Partial molal volume of ethanol in water at unity molality is %f cm^(3)/mol",v_2);