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diff --git a/926/CH3/EX3.5/Chapter3_Example5.sce b/926/CH3/EX3.5/Chapter3_Example5.sce new file mode 100644 index 000000000..cda2af729 --- /dev/null +++ b/926/CH3/EX3.5/Chapter3_Example5.sce @@ -0,0 +1,29 @@ +//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-3, Illustration 5, Page 55
+//Title: Calculation of percentage dissociation of gas
+//=============================================================================
+clear
+clc
+
+//INPUT
+W = 17.2; //Weight of N2O4 in grams
+V1 = 11450; //Volume occupied by N2O4 in cc
+P1 = 720; //Given pressure in mm hg
+T = 100; //Given temperature in degree C
+MW = 92; //Molecular weight of N2O4 in g/g mole
+V2 = 22400; //Volume at standard conditions in cc
+P2 = 760; //Pressure at standard conditions in mm Hg
+T2 = 273; //Temperature at standard conditions in K
+
+//CALCULATIONS
+T1 = T+273; //Given temperature in K
+n1 = W/MW; //No of moles of N2O4 initially present in g mole
+n2 = (V1/V2)*(T2/T1)*(P1/P2); //Total g moles present after dissociation
+x = n2-n1; //g moles of N2O4 dissociated
+X = x*100/n1; //Percentage dissociation
+
+//OUTPUT
+mprintf('\n Percentage dissociation of N2O4 to NO2 is %2.0f percent',X);
+
+//=========================END OF PROGRAM======================================
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