3 files changed, 39 insertions, 39 deletions
diff --git a/821/CH4/EX4.15/4_15.sce b/821/CH4/EX4.15/4_15.sce
index 8f8a03f63..85d31abdb 100755
--- a/821/CH4/EX4.15/4_15.sce
+++ b/821/CH4/EX4.15/4_15.sce
@@ -1,18 +1,18 @@
-T=273;//temperature 0C in kelvin//
-R=8.31*10^7;//Universal gas constant in erg per degree per mole//
-M=28;//molecular weight of N2 in gram per mole//
-printf('Since 22.4Litres of Nitrogen gas at 0C and 1atm pressure will contain 6.023*10^23Molecules.');
-N=2.69*10^19;//no. of molecules in molecules per cm^3//
-Cm=sqrt(8*R*T/(%pi*M));//mean velocity of Nitrogen in cm/sec//
-printf('\nMean velocity of Nitrogen=Cm=%fcm/sec',Cm);
-V=22400;//volume of nitrogen in cm^3//
-p=M/V;//Density of nitrogen in gram per cm^3//
-printf('\nDensity of Nitrogen=p=%f=1.25*10^-3gram per cm^3',p);
-n=10.99*10^-5;//Viscosity of N2 in poise//
-L=(3*n)/(Crms*p);//mean free path of nitrogen in cm//
-printf('\nMean free path of Nitrogen=L=5.81*10^-6cm');
-G=sqrt(1/(1.414*%pi*L*N));//Collision diameter of Nitrogen in cm//
-printf('\nCollission diameter of Nitrogen=G=3.80*10^-8cm');
-K=sqrt(%pi*R*T/M);
-Z11=2*N^2*G^2*K;//number of collisions per second of Nitrogen at 0C and 1atm//
-printf('\nNumber of molecular collisions per second of Nitrogen at NTP=%f=10.52*10^28molecular collisions per sec per cm^3',Z11);
+T=273;//temperature 0C in kelvin//
+R=8.31*10^7;//Universal gas constant in erg per degree per mole//
+M=28;//molecular weight of N2 in gram per mole//
+printf('Since 22.4Litres of Nitrogen gas at 0C and 1atm pressure will contain 6.023*10^23Molecules.');
+N=2.69*10^19;//no. of molecules in molecules per cm^3//
+Cm=sqrt(8*R*T/(%pi*M));//mean velocity of Nitrogen in cm/sec//
+printf('\nMean velocity of Nitrogen=Cm=%fcm/sec',Cm);
+V=22400;//volume of nitrogen in cm^3//
+p=M/V;//Density of nitrogen in gram per cm^3//
+printf('\nDensity of Nitrogen=p=%f=1.25*10^-3gram per cm^3',p);
+n=10.99*10^-5;//Viscosity of N2 in poise//
+L=(3*n)/(Cm*p);//mean free path of nitrogen in cm//
+printf('\nMean free path of Nitrogen=L=5.81*10^-6cm');
+G=sqrt(1/(1.414*%pi*L*N));//Collision diameter of Nitrogen in cm//
+printf('\nCollission diameter of Nitrogen=G=3.80*10^-8cm');
+K=sqrt(%pi*R*T/M);
+Z11=2*N^2*G^2*K;//number of collisions per second of Nitrogen at 0C and 1atm//
+printf('\nNumber of molecular collisions per second of Nitrogen at NTP=%f=10.52*10^28 molecular collisions per sec per cm^3',Z11);
+\ No newline at end of file
diff --git a/821/CH5/EX5.45/5_45.sce b/821/CH5/EX5.45/5_45.sce
index 32eac9873..b2bfd8acd 100755
--- a/821/CH5/EX5.45/5_45.sce
+++ b/821/CH5/EX5.45/5_45.sce
@@ -1,3 +1,4 @@
-a=0.5;//dissociation constant//
-Kp=(a^2*P)/(1-a^2);
-printf('Total pressure required to bring 50percent dissociation=P=3*Kp');
+P=1;
+a=0.5;//dissociation constant//
+Kp=(a^2*P)/(1-a^2);
+printf('Total pressure required to bring 50 percent dissociation=P=3*Kp');
+\ No newline at end of file
diff --git a/821/CH5/EX5.47/5_47.sce b/821/CH5/EX5.47/5_47.sce
index 1c714ec61..e384d698d 100755
--- a/821/CH5/EX5.47/5_47.sce
+++ b/821/CH5/EX5.47/5_47.sce
@@ -1,18 +1,17 @@
-acid=1/3;
-ester=2/3;
-alcohol=1/3;
-water=2/3;
-K=(ester*water)/(acid*alcohol);
-printf('K value of the reaction=K=%f',K);
-printf('\nIn finding no. of moles we end up with quadratic equation 3*x^2-24*x+20=0.\nupon solving the equation we get x=7.05 and 0.945.');
-printf('\nThe first solution is not admissible since the maximum yield of the ester cannot exceed one mol of acetic acid.\nHence x=0.945 i.e yield of the ester is 94.5percent.');
-printf('\nThis problem illustrates the influence of an increased concentration of the reactant\nSince using one mole of each reactant the yield of ester is only 66.66percent');
-acid=1-x;
-ester=x;
-alcohol=1-x;
-water=1+x;
-printf('\nIn finding no. of moles we end up with quadratic equation 3*x^2-9*x+4=0.\nupon solving the equation we get x=2.457 and 0.5425.');
-printf('\nThe first solution is not admissible,x=0.5425.\nyield of the ester in the presence of the product water has decreased from 66.67percent to 54.25percent.');
-printf('\nother homogeneous equilibria in the liquid phase such as ionization of weak acids\nionization of weak bases,hydrolysis of salts,etc.,can be treated likewise.');
-
-
+acid=1/3;
+ester=2/3;
+alcohol=1/3;
+water=2/3;
+K=(ester*water)/(acid*alcohol);
+printf('K value of the reaction=K=%f',K);
+printf('\nIn finding no. of moles we end up with quadratic equation 3*x^2-24*x+20=0.\nupon solving the equation we get x=7.05 and 0.945.');
+printf('\nThe first solution is not admissible since the maximum yield of the ester cannot exceed one mol of acetic acid.\nHence x=0.945 i.e yield of the ester is 94.5 percent.');
+printf('\nThis problem illustrates the influence of an increased concentration of the reactant\nSince using one mole of each reactant the yield of ester is only 66.66 percent');
+x = 0.945;
+acid=1-x;
+ester=x;
+alcohol=1-x;
+water=1+x;
+printf('\nIn finding no. of moles we end up with quadratic equation 3*x^2-9*x+4=0.\nupon solving the equation we get x=2.457 and 0.5425.');
+printf('\nThe first solution is not admissible,x=0.5425.\nyield of the ester in the presence of the product water has decreased from 66.67 percent to 54.25percent.');
+printf('\nother homogeneous equilibria in the liquid phase such as ionization of weak acids\nionization of weak bases,hydrolysis of salts,etc.,can be treated likewise.');
+\ No newline at end of file