updated the code

18 files changed, 93 insertions, 86 deletions

diff --git a/3428/CH21/EX14.21.1/Ex14_21_1.sce b/3428/CH21/EX14.21.1/Ex14_21_1.sce
index 5147f322f..0fdcd9a5e 100644
--- a/3428/CH21/EX14.21.1/Ex14_21_1.sce
+++ b/3428/CH21/EX14.21.1/Ex14_21_1.sce
@@ -2,10 +2,10 @@
 //To find the temperature at which  pressure of gas will reach the bursting  value.

 clc;

 //PV=nRT

-P=10

-V=(10^-3)*(1/10^-3)

-n=((5*10^-3)/30)

-R=0.0821

+P=10                   //atm

+V=(10^-3)*(1/10^-3)              //L

+n=((5*10^-3)/30)                  //mol

+R=0.0821          //(LatmK^-1mol^-1)

 T=((P*V)/(n*R))

 disp(T,'Required temperature(K)')

- 

+//Answer given in the book T=730.9 K is wrong.

diff --git a/3428/CH21/EX14.21.10/Ex14_21_10.sce b/3428/CH21/EX14.21.10/Ex14_21_10.sce
index 5ddefab2b..a14acf6e2 100644
--- a/3428/CH21/EX14.21.10/Ex14_21_10.sce
+++ b/3428/CH21/EX14.21.10/Ex14_21_10.sce
@@ -3,8 +3,8 @@
 clc;

 //v_mp=sqrt((2*K*T)/m)

 //sqrt((2*K*T_1)/m)=2*((2*K*T_2)/m)

-T_2=293

-T_1=2^2*(T_2)

+T_2=293                   //K

+T_1=2^2*(T_2)                     //K

 disp(T_1,'Required temperature(K)')

-T_1_deg=T_1-273

+T_1_deg=T_1-273                     //K

 disp(T_1_deg,'Required temperature(degree celius)')

diff --git a/3428/CH21/EX14.21.11/Ex14_21_11.sce b/3428/CH21/EX14.21.11/Ex14_21_11.sce
index fc08418a4..a5cfdbb08 100644
--- a/3428/CH21/EX14.21.11/Ex14_21_11.sce
+++ b/3428/CH21/EX14.21.11/Ex14_21_11.sce
@@ -1,9 +1,9 @@
 //Section-14,Example-7,Page no.-PC.17

 //To find the ratio of the rates of effusion of neon gas to that of helium gas at same temperature and pressure.

 clc;

-M_He=4

-M_Ne=20

+M_He=4                 //g/mol(molecular weight of He gas)

+M_Ne=20                 //g/mol(molecular weight of Ne gas)

 //r=r_Ne/r_He

 r=sqrt((M_He)/(M_Ne))

-disp(r,'Required ratio')

+disp(r,'Required ratio of the rates of effusion')

 

diff --git a/3428/CH21/EX14.21.12/Ex14_21_12.sce b/3428/CH21/EX14.21.12/Ex14_21_12.sce
index 7576cce46..f61e1e028 100644
--- a/3428/CH21/EX14.21.12/Ex14_21_12.sce
+++ b/3428/CH21/EX14.21.12/Ex14_21_12.sce
@@ -2,9 +2,9 @@
 //To find Molecular formula of Hydrocarbon.

 clc;

 //(r(hydrocarbon)/r(CH_4))=(M(CH_4)/M(hydrocarbon))

-M_CH4=16

-r_hc=1

-r_CH4=2

+M_CH4=16               //g/mol(molecular weight of CH4

+r_hc=1                 //rate of effusion of hydrocarbon

+r_CH4=2                 //rate of effusion of CH4

 M_hc=(16/(1/2)^2)

 disp(M_hc,'Molecular weight of hydrocarbon(g/mol)')

 //The hydrocarbon formula may be (C_4H_10) which has a molecular weight of 58g/mol.

diff --git a/3428/CH21/EX14.21.13/Ex14_21_13.sce b/3428/CH21/EX14.21.13/Ex14_21_13.sce
index 2da8e03a3..828158df5 100644
--- a/3428/CH21/EX14.21.13/Ex14_21_13.sce
+++ b/3428/CH21/EX14.21.13/Ex14_21_13.sce
@@ -1,18 +1,22 @@
 //Section-14,Example-1,Page no.-PC.21

 //To calculate collision number,collision frequency and mean free path for oxygen at 298K and 1 atm pressure.

 clc;

-M=((32*10^-3)/(6.023*10^23))

+M_w=32*10^-3                  //(kg/mol) molecular weight of oxygen

+N_A=6.023*10^23              //  Avogadro no.(mol^-1)

+M=((M_w)/(N_A))

 disp(M,'Mass of one oxygen molecule(kg)')

 //N=P/(R*T)

-N=(1*6.023*10^23*10^3)/(0.0821*298)

+P=1          //atm

+R=0.0821                //litreatmK^-1mol^-1

+T=298              //K

+N=(P*N_A*10^3)/(R*T)

 disp(N,'No.of O_2 molecules per m^3')

-R=8.314

-T=298

-m=32*10^-3

-v_avg=sqrt((8*R*T)/(%pi*m))

+R_1=8.314                      //kgm^2K^-1mol^-1

+m=32*10^-3                    //kgmol^-1

+v_avg=sqrt((8*R_1*T)/(%pi*m))           //ms^-1

 disp(v_avg,'Average velocity of O2 molecule(ms^-1)')

-sig=3.6*10^-10

-Z_1=sqrt(2)*pi*(sig)^2*v_avg*N

+sig=3.6*10^-10              //m

+Z_1=sqrt(2)*%pi*(sig)^2*v_avg*N 

 disp(Z_1,'Collision number(collisions per sec)')

 Z_11=(1/2)*(Z_1*N)

 disp(Z_11,'Collision frequency(collisions s^-1 m^-3)')

diff --git a/3428/CH21/EX14.21.14/Ex14_21_14.sce b/3428/CH21/EX14.21.14/Ex14_21_14.sce
index 7cf843fe6..f1a7c6028 100644
--- a/3428/CH21/EX14.21.14/Ex14_21_14.sce
+++ b/3428/CH21/EX14.21.14/Ex14_21_14.sce
@@ -2,11 +2,11 @@
 //To calculate the pressure exerted using the Vanderwalls equation.

 clc;

 //(P+((a*n^2)/V^2)*(V-(n*b))=n*R*T

-n=10

-R=8.314

-T=298

-V=25*10^-3

-b=0.037*10^-3

-a=0.417

+n=10                 //moles

+R=8.314                        //Nmk^-1mol^-1

+T=298                         //K

+V=25*10^-3                        //m^3

+b=0.037*10^-3                       //m^3mol^-1

+a=0.417                       //Nm^4mol^-2

 P=((n*R*T)/(V-(n*b))-((a*n^2)/(V^2)))

 disp(P,'Required pressure(Nm^-2)')

diff --git a/3428/CH21/EX14.21.15/Ex14_21_15.sce b/3428/CH21/EX14.21.15/Ex14_21_15.sce
index 8e2672cdb..56cfb492b 100644
--- a/3428/CH21/EX14.21.15/Ex14_21_15.sce
+++ b/3428/CH21/EX14.21.15/Ex14_21_15.sce
@@ -1,13 +1,13 @@
 //Section-14,Example-2,Page no.-PC.30

 //To calculate pressure exerted using ideal gas equation and Vanderwalls equation.

 clc;

-n=5

-R=8.314

-T=300

-V=1*10^-3

+n=5                     //moles 

+R=8.314                    //NmK^-1mol^-1

+T=300                     //K

+V=1*10^-3                  //m^3

 P_1=((n*R*T)/V)

 disp(P_1,'Required pressure using ideal gas equation(Nm^-2)')

-a=0.1378

-b=0.0318*10^-3

+a=0.1378                   //Nm^4mol^-2

+b=0.0318*10^-3                          //m^3mol^-1

 P_2=(((n*R*T)/(V-n*b))-((a*n^2)/(V^2)))

 disp(P_2,'Required pressure using vanderwalls equation(Nm^-2)')

diff --git a/3428/CH21/EX14.21.16/Ex14_21_16.sce b/3428/CH21/EX14.21.16/Ex14_21_16.sce
index 58c29edc0..888ee8a49 100644
--- a/3428/CH21/EX14.21.16/Ex14_21_16.sce
+++ b/3428/CH21/EX14.21.16/Ex14_21_16.sce
@@ -4,12 +4,12 @@ clc;
 //(P+(a-V^2))*(V-b)=R*T

 v_1=(2520+sqrt(((2520)^2)-4*(10^6)*0.2279))/(2*10^6)

 v_2=(2520-sqrt(((2520)^2)-4*(10^6)*0.2279))/(2*10^6)

-R=8.314

-T=298

-P=10^6

+R=8.314                    //NmK^-1mol^-1

+T=298                          //K

+P=10^6                         //N/m^2

 V=((R*T)/P)

 disp(V,'Volume occupied according to Vanderwalls equation(m^3)')

-a=0.2279

-b=0.0428*10^-3

+a=0.2279                       //Nm^4mol^-2

+b=0.0428*10^-3                         //m^3mol^-1

 T_B=(a/(R*b))

 disp(T_B,'Boyle`s temperature for methane gas(K)')

diff --git a/3428/CH21/EX14.21.17/Ex14_21_17.sce b/3428/CH21/EX14.21.17/Ex14_21_17.sce
index 5de637d7f..511388b60 100644
--- a/3428/CH21/EX14.21.17/Ex14_21_17.sce
+++ b/3428/CH21/EX14.21.17/Ex14_21_17.sce
@@ -1,14 +1,14 @@
 //Section-14,Example-4,Page no.-PC.31

 //To calculate pressure using ideal gas equation and vanderwall`s gas equation.

 clc;

-n=12

-R=0.0821

-T=298

-V=10.0

+n=12                    //moles

+R=0.0821                        //Latm/molK

+T=298                             //K

+V=10.0                          //L

 P_1=((n*R*T)/V)

 disp(P_1,'Pressure from ideal gas equation(atm)')

-a=1.49

-b=0.0399

+a=1.49                        //atm/mol^2

+b=0.0399                         //L/mol

 P_2=(((n*R*T)/(V-(n*b)))-((a*n^2)/(V^2)))

 disp(P_2,'Pressure from Vander walls gas equation(atm)')

 

diff --git a/3428/CH21/EX14.21.18/Ex14_21_18.sce b/3428/CH21/EX14.21.18/Ex14_21_18.sce
index d8f1e016b..33b79814e 100644
--- a/3428/CH21/EX14.21.18/Ex14_21_18.sce
+++ b/3428/CH21/EX14.21.18/Ex14_21_18.sce
@@ -1,15 +1,17 @@
 //Section-14,Example-5,Page no.-PC.31

 //To calculate volume using Ideal gas equation and vander walls equation

 clc;

-n=3

-R=0.0821

-T=373

-P=50

+n=3                    //mol

+R=0.0821                        //Latm/molK

+T=373                    //K

+P=50                           //atm

 V_1=((n*R*T)/P)

 disp(V_1,'Volume according to Ideal gas equation(L)')

-a=1.36

-b=0.0318

-V_2=((n*R*T)/(P+((a*n^2)/V^2)))

+a=1.36                            //L^2atm/mol^2

+b=0.0318                          //L/mol

+V_2=((n*R*T)/(P+((a*n^2/V_1^2))))

 disp(V_2,'Volume according to Vanderwall`s gas equation(L)')

 

 

+//Answer in the book(V_2=1.81 L)is wrong.

+

diff --git a/3428/CH21/EX14.21.19/Ex14_21_19.sce b/3428/CH21/EX14.21.19/Ex14_21_19.sce
index 929824528..c13f0b2a8 100644
--- a/3428/CH21/EX14.21.19/Ex14_21_19.sce
+++ b/3428/CH21/EX14.21.19/Ex14_21_19.sce
@@ -1,16 +1,16 @@
 //Section-14,Example-6,Page no.-PC.32

 //To calculate moles of ammonia.

 clc;

-P=20

-V=7.0

-R=0.0821

-T=373

-n=((P*V)/(R*T))

-a=4.17

-b=0.0371

-n_1=((P+((a*n^2)/V^2))*(V-(n*b)))/(R*T)

-n_2=((P+((a*n_1^2)/V^2))*(V-(n_1*b)))/(R*T)

-n_3=((P+((a*n_2^2)/V^2))*(V-(n_2*b)))/(R*T)

-n_4=((P+((a*n_3^2)/V^2))*(V-(n_3*b)))/(R*T)

-disp(n_4,'Moles of ammonia that wil occupy 7.0L at 20 atm and 100 degree C)

+P=20             //atm

+V=7.0                    //L

+R=0.0821                     //Latm/molK

+T=373                             //K

+n=((P*V)/(R*T))                    //mol

+a=4.17                    //L^2atm/mol^2

+b=0.0371                     //L/mol

+n_1=((P+((a*n^2)/V^2))*(V-(n*b)))/(R*T)                    //mol

+n_2=((P+((a*n_1^2)/V^2))*(V-(n_1*b)))/(R*T)                    //mol

+n_3=((P+((a*n_2^2)/V^2))*(V-(n_2*b)))/(R*T)                         //mol

+n_4=((P+((a*n_3^2)/V^2))*(V-(n_3*b)))/(R*T)                             //mol

+disp(n_4,'Moles of ammonia that wil occupy 7.0L at 20 atm and 100 degree C')

 

diff --git a/3428/CH21/EX14.21.2/Ex14_21_2.sce b/3428/CH21/EX14.21.2/Ex14_21_2.sce
index 0c4b82950..53754e614 100644
--- a/3428/CH21/EX14.21.2/Ex14_21_2.sce
+++ b/3428/CH21/EX14.21.2/Ex14_21_2.sce
@@ -2,11 +2,11 @@
 //To calculate the number of gas molecules left.

 clc;

 //PV=nRT

-P=(10^-5*(1/760))

-V=(10^-3*(1/1000))

-R=0.0821

-T=298

-n=((P*V)/(R*T))

+P=(10^-5*(1/760))                   //atm

+V=(10^-3*(1/1000))                  //L

+R=0.0821                          //LatmK^-1mol^-1

+T=298                              //K

+n=((P*V)/(R*T))                       //moles

 N_a=6.023*10^23                          //1 mole gas=6.023*10^23 molecules

 N=n*N_a

 disp(N,'No. of gas molecules left')

diff --git a/3428/CH21/EX14.21.3/Ex14_21_3.sce b/3428/CH21/EX14.21.3/Ex14_21_3.sce
index 9b11d0c0a..a37af92fb 100644
--- a/3428/CH21/EX14.21.3/Ex14_21_3.sce
+++ b/3428/CH21/EX14.21.3/Ex14_21_3.sce
@@ -1,10 +1,10 @@
 //Section-14,Example-3,Page no.-PC.8

 //To find whether the tank will blow up before it melts.

 clc;

-P_1=200

-T_1=298

-T_2=1808

+P_1=200                    //atm

+T_1=298                   //K

+T_2=1808                           //K

 //(P_1/T_1)=(P_2/T_2)

-P_2=(P_1/T_1)*T_2

+P_2=(P_1/T_1)*T_2                       //atm

 disp(P_2,'Final pressure in the tank(atm)') 

 //since P_2>700 atm,tank will blow up.

diff --git a/3428/CH21/EX14.21.4/Ex14_21_4.sce b/3428/CH21/EX14.21.4/Ex14_21_4.sce
index 9c15a02c9..a277d2e65 100644
--- a/3428/CH21/EX14.21.4/Ex14_21_4.sce
+++ b/3428/CH21/EX14.21.4/Ex14_21_4.sce
@@ -1,8 +1,8 @@
 //Section-14,Example-4,Page no.-PC.8

 //To determine how many times faster will He initially leak through a pinhole in the container.

 clc;

-M_2=28

-M_1=4

+M_2=28                     //gmol^-1

+M_1=4                        //gmol^-1

 //r=r_1/r_2

 r=sqrt(M_2/M_1)

 disp(r,'r_1/r_2')

diff --git a/3428/CH21/EX14.21.6/Ex14_21_6.sce b/3428/CH21/EX14.21.6/Ex14_21_6.sce
index 2601b01b6..4847d0b8d 100644
--- a/3428/CH21/EX14.21.6/Ex14_21_6.sce
+++ b/3428/CH21/EX14.21.6/Ex14_21_6.sce
@@ -2,9 +2,9 @@
 //To calculate temperature at which rms velocity of hydrogen gas =100 ms^-1

 clc;

 //v_rms=sqrt((3*R*T)/M)

-v_rms=100

-R=8.314

-M=2*10^-3

-T=((v_rms^2*M)/(3*R))

+v_rms=100                 //ms^-1

+R=8.314                  //JK^-1mol^-1

+M=2*10^-3                //kgmol^-1

+T=((v_rms^2*M)/(3*R))            //K

 disp(T,'Required temperature(K)')

 

diff --git a/3428/CH21/EX14.21.7/Ex14_21_7.sce b/3428/CH21/EX14.21.7/Ex14_21_7.sce
index 578eaf75a..d79e1479b 100644
--- a/3428/CH21/EX14.21.7/Ex14_21_7.sce
+++ b/3428/CH21/EX14.21.7/Ex14_21_7.sce
@@ -2,9 +2,9 @@
 //To calculate temperature at which v_mp of oxygen= v_mp of hydrogen

 clc;

 //v_mp=sqrt((2*R*T)/M)

-M_O2=32

-M_H2=2

-T_H2=298

+M_O2=32                   //mol^-1

+M_H2=2                     //mol^-1

+T_H2=298                      //K

 //v_mp(O2)/v_mp(H2)=(T_O2/M_O2)/(T_H2/M_H2)=1

-T_O2=T_H2*(M_O2/M_H2)

+T_O2=T_H2*(M_O2/M_H2)              //K

 disp(T_O2,'Required temperature(K)')

diff --git a/3428/CH21/EX14.21.8/Ex14_21_8.sce b/3428/CH21/EX14.21.8/Ex14_21_8.sce
index 3e456d40e..537915e6a 100644
--- a/3428/CH21/EX14.21.8/Ex14_21_8.sce
+++ b/3428/CH21/EX14.21.8/Ex14_21_8.sce
@@ -4,7 +4,7 @@ clc;
 //v_avg=sqrt((8*K*T)/pi*m)

 //v_avgHe=sqrt((8*K*330)/(pi*4))

 //v_avgN_2=sqrt((8*K*T_2)/(pi*28))

-T_1=330

+T_1=330                     //(K)

 K=1                       //K=1(let)

-T_2=(8*K*T_1*%pi*28)/(%pi*4*8*K)

+T_2=(8*K*T_1*%pi*28)/(%pi*4*8*K)          //(K)

 disp(T_2,'Required temperature(K)')

diff --git a/3428/CH21/EX14.21.9/Ex14_21_9.sce b/3428/CH21/EX14.21.9/Ex14_21_9.sce
index 98c865954..24f4ccd7f 100644
--- a/3428/CH21/EX14.21.9/Ex14_21_9.sce
+++ b/3428/CH21/EX14.21.9/Ex14_21_9.sce
@@ -3,6 +3,7 @@
 clc;

 //v_rms=sqrt((3*K*T)/m)

 //K=1(let)

+T=200                //(K)   Given temperature

 K=1

-T_He=(3*K*200*4)/(3*K*2)

+T_He=(3*K*T*4)/(3*K*2)

 disp(T_He,'Required temperature(K)')