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authorpriyanka2015-06-24 15:03:17 +0530
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+clear;
+clc;
+
+// Example: 9.3
+// Page: 339
+
+printf("Example: 9.3 - Page: 339\n\n");
+
+// Solution
+
+//*****Data******//
+Vol = 2000;// [cubic cm]
+y1_1 = 0.96;// [mass fraction of ethanol in laboratory alcohol]
+y2_1 = 0.04;// [mass fraction of water in laboratory alcohol]
+y1_2 = 0.56;// [mass fracion of ethanol in vodka]
+y2_2 = 0.44;// [mass fraction of water in vodka]
+Vbar_water1 = 0.816;// [cubic cm/g]
+Vbar_ethanol1 = 1.273;// [cubic cm/g]
+Vbar_water2 = 0.953;// [cubic cm/g]
+Vbar_ethanol2 = 1.243;// [cubic cm/g]
+Density_water = 0.997;// [cubic cm/g]
+//***************//
+
+// Solution (i)
+// From Eqn 9.9
+Va = y1_1*Vbar_ethanol1 + y2_1*Vbar_water1;// [Volume of laboratory alcohol, cubic cm/g]
+mass = Vol/Va;// [g]
+// Let Mw be the mass of water added in laboratory alcohol.
+// Material balance on ethanol:
+Mw = mass*y1_1/y1_2 - mass;// [g]
+Vw = Mw/Density_water;// [Volume of water added, cubic cm]
+printf("Mass of water added is %d g\n",Mw);
+
+// Solution (ii)
+Mv = mass + Mw;// [Mass of vodka, g]
+Vv = y1_2*Vbar_ethanol2 + y2_2*Vbar_water2;// [Volume of ethanol, cubic cm/g]
+V_vodka = Vv*Mv;// [Volume of vodka obtained after conversion, cubic cm]
+printf("The volume of vodka obtained after conversion is %.d cubic cm\n",V_vodka); \ No newline at end of file