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| author | prashantsinalkar | 2017-10-10 12:38:01 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2017-10-10 12:38:01 +0530 |
| commit | f35ea80659b6a49d1bb2ce1d7d002583f3f40947 (patch) | |
| tree | eb72842d800ac1233e9d890e020eac5fd41b0b1b /3428 | |
| parent | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (diff) | |
| download | Scilab-TBC-Uploads-f35ea80659b6a49d1bb2ce1d7d002583f3f40947.tar.gz Scilab-TBC-Uploads-f35ea80659b6a49d1bb2ce1d7d002583f3f40947.tar.bz2 Scilab-TBC-Uploads-f35ea80659b6a49d1bb2ce1d7d002583f3f40947.zip | |
updated the code
Diffstat (limited to '3428')
68 files changed, 301 insertions, 283 deletions
diff --git a/3428/CH10/EX5.10.10/Ex5_10_10.sce b/3428/CH10/EX5.10.10/Ex5_10_10.sce index 1c2399be9..22526d316 100644 --- a/3428/CH10/EX5.10.10/Ex5_10_10.sce +++ b/3428/CH10/EX5.10.10/Ex5_10_10.sce @@ -3,8 +3,8 @@ clc;
R_0=1 //(let)
t_1=30
-R_t=0.75*R1_0
-k_1=(2.303/t_1)*(log10(R1_0/R_t))
+R_t=0.75*R_0
+k_1=(2.303/t_1)*(log10(R_0/R_t))
t_2=35
R1_t=(R_0)/(10^((k_1*t_2)/2.303))
R1_tpr=100*R1_t
diff --git a/3428/CH10/EX5.10.8/Ex5_10_8.sce b/3428/CH10/EX5.10.8/Ex5_10_8.sce index 51eef2d7e..4922ab0a8 100644 --- a/3428/CH10/EX5.10.8/Ex5_10_8.sce +++ b/3428/CH10/EX5.10.8/Ex5_10_8.sce @@ -2,11 +2,11 @@ //To show that the given reaction is a 2nd order reaction and calculate the fraction of ester decomposed in 30 minutes.
clc;
k_5=(1/5)*((1/10.2)-(1/16))
-kbar_5=k*10^2
+kbar_5=k_5*10^2
k_25=(1/25)*((1/4.3)-(1/16))
-kbar_25=k*10^2
+kbar_25=k_25*10^2
k_55=(1/55)*((1/2.3)-(1/16))
-kbar_55=k*10^2
+kbar_55=k_55*10^2
k_120=(1/120)*((1/1.1)-(1/16))
-kbar_120=k*10^2
+kbar_120=k_120*10^2
//Constant value of kbar shows that the given reaction is a 2nd order reaction
diff --git a/3428/CH11/EX5.11.3/Ex5_11_3.sce b/3428/CH11/EX5.11.3/Ex5_11_3.sce index 6f52e2fa8..69b05815b 100644 --- a/3428/CH11/EX5.11.3/Ex5_11_3.sce +++ b/3428/CH11/EX5.11.3/Ex5_11_3.sce @@ -1,6 +1,7 @@ //Section-5,Example-3,Page no.-D.124 //To calculate the number of molecules of HCl produced per joule of radiant energy absorbed clc; +h=6.626*10^-34 v=3*10^15 //Frequency c=3*10^8 lm=c/v diff --git a/3428/CH14/EX8.14.2/Ex8_14_2.sce b/3428/CH14/EX8.14.2/Ex8_14_2.sce index 301dabbe9..ad05d76bd 100644 --- a/3428/CH14/EX8.14.2/Ex8_14_2.sce +++ b/3428/CH14/EX8.14.2/Ex8_14_2.sce @@ -6,5 +6,5 @@ dl_Hsub=25.9 H=25.5 //((1/2)H)I_2 I=118.4 U=-165.4 -E=dl_Hf-dl_Hsub-I_2-I-U +E=dl_Hf-dl_Hsub-H-I-U disp(E,'Electron affinity of iodine(kCal/mol)') diff --git a/3428/CH15/EX9.15.15/Ex9_15_15.sce b/3428/CH15/EX9.15.15/Ex9_15_15.sce index dedaeb658..ced43f011 100644 --- a/3428/CH15/EX9.15.15/Ex9_15_15.sce +++ b/3428/CH15/EX9.15.15/Ex9_15_15.sce @@ -1,6 +1,7 @@ //Section-9,Example-1,Page no.-E.16 //To calculate dl(G),dl(H) and dl(S). clc; +T=-298 E=1.02 d_ETP=-5*10^-5 n=2 @@ -8,7 +9,7 @@ F=96500 dl_G=-(n*F*E) disp(dl_G,'(in Jmol^-1)') dl_S=n*F*d_ETP -disp,(dl_S,'(in JK^-1mol^-1)') +disp(dl_S,'(in JK^-1mol^-1)') dl_H=dl_G+(T*dl_S) disp(dl_H,'(in Jmol^-1)') diff --git a/3428/CH16/EX10.16.2/Ex10_16_2.sce b/3428/CH16/EX10.16.2/Ex10_16_2.sce index a851a1ea2..48c1543af 100644 --- a/3428/CH16/EX10.16.2/Ex10_16_2.sce +++ b/3428/CH16/EX10.16.2/Ex10_16_2.sce @@ -6,4 +6,4 @@ dl_HFAl2O3=-399.1 //dlH_f(Al_2O_3) dl_HAl=0 //dlH_f(Al)
dl_HFe2O3=-196.51 //dlH_f(Fe_2O_3)
dl_H=(2*dl_HFe+dl_HFAl2O3)-(2*dl_HAl+dl_HFe2O3)
-disp (dl_H,'Enthalpy change for the reduction of 1 mol Fe_2O_3(kCal)').
+disp (dl_H,'Enthalpy change for the reduction of 1 mol Fe_2O_3(kCal)')
diff --git a/3428/CH17/EX10.17.10/Ex10_17_10.sce b/3428/CH17/EX10.17.10/Ex10_17_10.sce index e6c49e77f..339dda65b 100644 --- a/3428/CH17/EX10.17.10/Ex10_17_10.sce +++ b/3428/CH17/EX10.17.10/Ex10_17_10.sce @@ -14,7 +14,7 @@ C_v=(3/2)*R y=C_p/C_v T_2=T_1*(V_1/V_2)^(y-1) dl_E=n*C_v*8.314*(T_2-T_1) -disp(dl_H,'Internal energy change(Joules)') +disp(dl_E,'Internal energy change(Joules)') W=dl_E disp(W,'Joules') dl_H=n*C_p*8.314*(T_2-T_1) diff --git a/3428/CH17/EX10.17.12/Ex10_17_12.sce b/3428/CH17/EX10.17.12/Ex10_17_12.sce index e6a70010a..7ac174e46 100644 --- a/3428/CH17/EX10.17.12/Ex10_17_12.sce +++ b/3428/CH17/EX10.17.12/Ex10_17_12.sce @@ -1,12 +1,12 @@ //Section-10,Example-2,Page no.-CT.41 //To calculate Entropy change(dl_S). clc; +R=8.314 n=10 C_v=(3/2)*R T_2=323 T_1=298 V_2=2 V_1=1 -R=8.314 dl_S=n*((C_v*log(T_2/T_1))+(R*log(V_2/V_1))) disp(dl_S,'Entropy change(JK^-1)') diff --git a/3428/CH19/EX12.19.1/Ex12_19_1.sce b/3428/CH19/EX12.19.1/Ex12_19_1.sce index add11634a..9cda13a82 100644 --- a/3428/CH19/EX12.19.1/Ex12_19_1.sce +++ b/3428/CH19/EX12.19.1/Ex12_19_1.sce @@ -1,9 +1,9 @@ //Section-12,Example-1,Page no.-SS.58
//To calculate the mobility of electrons in copper.
clc;
-res=1.76*10^-6
-e=1.6*10^-19
-n=(6.023*10^23)/((63.54)/(8.96))
+res=1.76*10^-6 //ohm/cm
+e=1.6*10^-19 //coulombs
+n=(6.023*10^23)/((63.54)/(8.96)) //no. of free electrons per unit volume
u_e=(1/(res*n*e))
-disp(u_e,'Mobility of electron(cm^2/volt.sec.)
+disp(u_e,'Mobility of electron(cm^2/volt.sec)')
//mobility of electron u_e=43.28 cm^2/(volt*sec) is wrong in the book.
diff --git a/3428/CH19/EX12.19.2/Ex12_19_2.sce b/3428/CH19/EX12.19.2/Ex12_19_2.sce index 4969e6f5d..31c69291e 100644 --- a/3428/CH19/EX12.19.2/Ex12_19_2.sce +++ b/3428/CH19/EX12.19.2/Ex12_19_2.sce @@ -1,10 +1,11 @@ //Section-12,Example-2,Page no.-SS.59
//To calculate the conductivity of pure silicon at room temperature.
clc;
-u_e=1500
-u_h=500
-e=1.6*10^-19
-n_i=1.6*10^-10
+u_e=1500 //cm^2/volt-sec
+u_h=500 //cm^2/volt-sec
+e=1.6*10^-19 //Coulombs
+n_i=1.6*10^-10 //per cm^3
C_i=e*n_i*(u_e+u_h)
disp(C_i,'Conductivity of pure silicon at room temperature(mho/cm)')
+//Answer given in the book C_i=5.12*10^-6 is wrong.
diff --git a/3428/CH19/EX12.19.3/Ex12_19_3.sce b/3428/CH19/EX12.19.3/Ex12_19_3.sce index 6d42e7250..7c7482a29 100644 --- a/3428/CH19/EX12.19.3/Ex12_19_3.sce +++ b/3428/CH19/EX12.19.3/Ex12_19_3.sce @@ -1,12 +1,12 @@ //Section-12,Example-3,Page no.-SS.59
//To calculate the current produced in a small germanium plate.
clc;
-n_i=2*10^19
-e=1.6*10^-19
-u_e=0.36
-u_h=0.17
-V=2
-l=(0.3*10^-3)
-A=1*10^-4
+n_i=2*10^19 //per m^3
+e=1.6*10^-19 //coulombs
+u_e=0.36 //m^2/Vsec
+u_h=0.17 //m^2/Vsec
+V=2 //V
+l=(0.3*10^-3) //m
+A=1*10^-4 //m^2
I=(n_i*e*(u_e+u_h)*V*A)/l
disp(I,'Current produced in a small germanium plate(Amp)')
diff --git a/3428/CH19/EX12.19.4/Ex12_19_4.sce b/3428/CH19/EX12.19.4/Ex12_19_4.sce index 499084179..478802d9f 100644 --- a/3428/CH19/EX12.19.4/Ex12_19_4.sce +++ b/3428/CH19/EX12.19.4/Ex12_19_4.sce @@ -1,12 +1,12 @@ //Section-12,Example-4,Page no.-SS.59
//To calculate concentration of holes and electrons in an n-type silicon.
clc;
-C_n=0.1
-e=1.6*10^-19
-u_e=1300
-N_D=C_n/(e*u_e)
-n=N_D
-n_i=1.5*10^10
+C_n=0.1 //(ohm-cm)^-1
+e=1.6*10^-19 //Coulombs
+u_e=1300 //cm^2/Vsec
+N_D=C_n/(e*u_e) //atoms/cm^3
+n=N_D //electrons/cm^3
+n_i=1.5*10^10 //per cm^3
disp(n,'Concentration of electrons(per cm^3)')
p=((n_i)^2)/n
disp(p,'Concentration of holes(per cm^3)')
diff --git a/3428/CH19/EX12.19.5/Ex12_19_5.sce b/3428/CH19/EX12.19.5/Ex12_19_5.sce index 1c8c16152..e783b7aaf 100644 --- a/3428/CH19/EX12.19.5/Ex12_19_5.sce +++ b/3428/CH19/EX12.19.5/Ex12_19_5.sce @@ -2,14 +2,14 @@ //To find the Intrinsic and Extrinsic conductivity.
clc;
n_i=2.5*10^13
-u_n=3800
-u_e=1800
-N_Ge=4.41*10^22
-e=1.6*10^-19
+u_n=3800 //cm^2/V sec
+u_e=1800 //cm^2/V sec
+N_Ge=4.41*10^22 //no. of Ge atoms per cm^3
+e=1.6*10^-19 //Coulombs
C_i=n_i*e*(u_n+u_e)
disp(C_i,'Intrinsic conductivity(mho per cm)')
N_D=N_Ge/10^7
-n=N_D //concentration of electrons
-p=((n_i)^2)/N_D //concentration of holes
+n=N_D //concentration of electrons per cm^3
+p=((n_i)^2)/N_D //concentration of holes per cm^3
C_n=(e*(N_D)*(u_n))
disp(C_n,'Conductivity of n-type germanium semiconductor(mho/cm)')
diff --git a/3428/CH19/EX12.19.6/Ex12_19_6.sce b/3428/CH19/EX12.19.6/Ex12_19_6.sce index e2356d188..f5c0ddc98 100644 --- a/3428/CH19/EX12.19.6/Ex12_19_6.sce +++ b/3428/CH19/EX12.19.6/Ex12_19_6.sce @@ -1,10 +1,10 @@ //Section-12,Example-6,Page no.-SS.60
//To calculate the no. of charge carriers and the conductivty of the doped material.
clc;
-a=5.431*10^-8
-u_e=1900
-e=1.6*10^-19
-V=a^3
-N_D=(8/V)/10^6
+a=5.431*10^-8 //cm
+u_e=1900 //cm^2V^-1sec^-1
+e=1.6*10^-19 //coulombs
+V=a^3 //cm^3
+N_D=(8/V)/10^6 //cm^3
C_n=e*N_D*u_e
disp(C_n,'Conductivity of P-doped Si(N- type semiconductor)(ohm^-1cm^-1)')
diff --git a/3428/CH19/EX12.19.7/Ex12_19_7.sce b/3428/CH19/EX12.19.7/Ex12_19_7.sce index 265ee50bb..3bfe06170 100644 --- a/3428/CH19/EX12.19.7/Ex12_19_7.sce +++ b/3428/CH19/EX12.19.7/Ex12_19_7.sce @@ -1,14 +1,14 @@ //Section-12,Example-7,Page no.-SS.61
//To find the no.of charge carriers essential to get te given conductivity and the no. of Antimony dopant atoms to be added to germanium.
clc;
-C=100
-e=1.6*10^-19
-u_e=2800
+C=100 //ohm^-1cm^-1
+e=1.6*10^-19 //C
+u_e=2800 //cm^-1V^-1sec^-1
N_D=C/(e*u_e)
disp(N_D,'No.of charge carriers essential to get the given conductivity(per cm^3)')
-a=5.658*10^-8
-V=a^3
-N_Sb=2.23*10^17
-N_Ge=8/V
+a=5.658*10^-8 //cm
+V=a^3 //cm^3
+N_Sb=2.23*10^17 //No. of Sb atoms per cm^3
+N_Ge=8/V //No. of atoms of Ge
N=N_Sb/N_Ge
-disp(N,'No. of Antimony dopant atoms to be added to germanium(ppm)').
+disp(N,'No. of Antimony dopant atoms to be added to germanium(ppm)')
diff --git a/3428/CH2/EX1.2.17/Ex1_2_17.sce b/3428/CH2/EX1.2.17/Ex1_2_17.sce index 10b219ce8..624bf6681 100644 --- a/3428/CH2/EX1.2.17/Ex1_2_17.sce +++ b/3428/CH2/EX1.2.17/Ex1_2_17.sce @@ -1,7 +1,6 @@ //Section-1,Example-4,Page no.-AC.205 //To find the air required for the perfect combustion of 1 m^3 of the given gas. clc; -H_2= T=0.22 L_O2=0.02 Net_O2=0.2 diff --git a/3428/CH2/EX1.2.18/Ex1_2_18.sce b/3428/CH2/EX1.2.18/Ex1_2_18.sce index c0f02bab3..5d859832b 100644 --- a/3428/CH2/EX1.2.18/Ex1_2_18.sce +++ b/3428/CH2/EX1.2.18/Ex1_2_18.sce @@ -19,5 +19,5 @@ P_N2=(W_N2/T_W)*100 disp(P_N2,'Percentage composition of N_2') P_O2=(W_O2/T_W)*100 disp(P_O2,'Percentage composition of O_2') -P_O2=(W_SO2/T_W)*100 +P_SO2=(W_SO2/T_W)*100 disp(P_SO2,'Percentage composition of SO_2') diff --git a/3428/CH20/EX13.20.1/Ex13_20_1.sce b/3428/CH20/EX13.20.1/Ex13_20_1.sce index 9afb77671..d9eb05232 100644 --- a/3428/CH20/EX13.20.1/Ex13_20_1.sce +++ b/3428/CH20/EX13.20.1/Ex13_20_1.sce @@ -1,12 +1,12 @@ //Section-13,Example-1,Page no.-MA.2 -//To express the composition of the compound alloy Cu_3Au in carats. -W_1=63.5 +//To express the composition of the compound alloy Cu_3Au in Karats. +W_1=63.5 //gm W_11=3*W_1 //Weight of 3 moles of copper(gm) -W_2=197 +W_2=197 //gm W_22=1*W_2 //Weight of 1 mole of Gold(gm) F_w=W_11+W_22 //Formula weight of Cu_3Au(gm) -P_Cu=(W_11/F_w)*100 -P_Au=(W_22/F_w)*100 +P_Cu=(W_11/F_w)*100 //% +P_Au=(W_22/F_w)*100 //% x=(P_Au/100)*24 -disp(x,'Composition of Cu_3Au in carats') +disp(x,'Composition of Cu_3Au in Karats') 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)')
diff --git a/3428/CH22/EX14.22.1/Ex14_22_1.sce b/3428/CH22/EX14.22.1/Ex14_22_1.sce index fc988b757..74de685f0 100644 --- a/3428/CH22/EX14.22.1/Ex14_22_1.sce +++ b/3428/CH22/EX14.22.1/Ex14_22_1.sce @@ -1,12 +1,12 @@ //Section-14,Example-1,Page no.-PC.48
//To calculate the surface tension of ethyl alcohol and the no. of times a water drop is heavier than a drop of ethyl alcohol.
clc;
-y_r=7.2*10^-2
+y_r=7.2*10^-2 //N/m
n_r=30
n_e=30
-d_e=0.865*10^3
-d_r=0.996*10^3
-y_e=(((y_r)*(n_r)*(d_e))/((n_r)*(d_r)))
+d_e=0.865*10^3 //g/cm^3
+d_r=0.996*10^3 //g/cm^3
+y_e=(((y_r)*(n_r)*(d_e))/((n_r)*(d_r))) //N/m
disp(y_e,'Surface tension of ethyl alcohol(N/m)')
//m=m_r/m_e=y_r/y_e
m=(y_r/y_e)
diff --git a/3428/CH22/EX14.22.2/Ex14_22_2.sce b/3428/CH22/EX14.22.2/Ex14_22_2.sce index 6e720ed52..d20925cbe 100644 --- a/3428/CH22/EX14.22.2/Ex14_22_2.sce +++ b/3428/CH22/EX14.22.2/Ex14_22_2.sce @@ -2,10 +2,10 @@ //Calculate surface tension of liquid.
clc;
//surface tension(y)=(r*h*d*g)/(2*cosA)
-r=1*10^-4
-h=8*10^-2
-d=0.9*10^3
-g=9.8
-A=0
+r=1*10^-4 //m
+h=8*10^-2 //m
+d=0.9*10^3 //kg/m^3
+g=9.8 //ms^-1
+A=0 //degree
y=(r*h*d*g)/(2*cos(A))
disp(y,'Surface tension(N/m)')
diff --git a/3428/CH22/EX14.22.3/Ex14_22_3.sce b/3428/CH22/EX14.22.3/Ex14_22_3.sce index c6fda3f7e..ff8281422 100644 --- a/3428/CH22/EX14.22.3/Ex14_22_3.sce +++ b/3428/CH22/EX14.22.3/Ex14_22_3.sce @@ -2,8 +2,8 @@ //To calculate the height to which liquid B rises.
clc;
//(y_B/y_A)=(h_B*d_B)/(h_A*d_A)
-h_A=0.01
+h_A=0.01 //m
d=1/2 //d=d_A/d_B
y=1/2 //y=y_A/y_B
-h_B=h_A*d*y
+h_B=h_A*d*y //m
disp(h_B,'The height to which liquid B rises(m)')
diff --git a/3428/CH22/EX14.22.4/Ex14_22_4.sce b/3428/CH22/EX14.22.4/Ex14_22_4.sce index 1f3747cc5..c3da4809e 100644 --- a/3428/CH22/EX14.22.4/Ex14_22_4.sce +++ b/3428/CH22/EX14.22.4/Ex14_22_4.sce @@ -1,15 +1,15 @@ //Section-14,Example-3,Page no.-PC.51
//To calculate the energy required.
clc;
-y=73*10^-3
-r_S=2*10^-6
-r_B=2*10^-3
-V_B=(4/3)*pi*(r_B)^3
-V_S=(4/3)*pi*(r_S)^3
+y=73*10^-3 //Nm^-1
+r_S=2*10^-6 //m
+r_B=2*10^-3 //m
+V_B=(4/3)*%pi*(r_B)^3 //m^3
+V_S=(4/3)*%pi*(r_S)^3 //m^3
N=(V_B/V_S)
-SA_S=4*%pi*(r_S)^2
-SA_B=4*%pi*(r_B)^2
-TSA_S=(10^9)*SA_S
-SA_in=TSA_S-SA_B //Increase in surface area
-E=y*(SA_in)
+SA_S=4*%pi*(r_S)^2 //m^2
+SA_B=4*%pi*(r_B)^2 //m^2
+TSA_S=(10^9)*SA_S //m^2
+SA_in=TSA_S-SA_B //Increase in surface area(m^2)
+E=y*(SA_in) //J
disp(E,'Energy required(J)')
diff --git a/3428/CH22/EX14.22.5/Ex14_22_5.sce b/3428/CH22/EX14.22.5/Ex14_22_5.sce index a757ec245..1ef127370 100644 --- a/3428/CH22/EX14.22.5/Ex14_22_5.sce +++ b/3428/CH22/EX14.22.5/Ex14_22_5.sce @@ -1,8 +1,8 @@ //Section-14,Example-1,Page no.-PC.53
//To calculate the surface tension of experimental liquid.
clc;
-y_r=72.8
-F_ur=520
-F_ue=125
-y_e=(y_r*F_ue)/F_ur
-disp(y_e,'Surface tension of experimental liquid(dynes/cm)').
+y_r=72.8 //dynes/cm
+F_ur=520 //dynes
+F_ue=125 //dynes
+y_e=(y_r*F_ue)/F_ur //dynes/cm
+disp(y_e,'Surface tension of experimental liquid(dynes/cm)')
diff --git a/3428/CH22/EX14.22.6/Ex14_22_6.sce b/3428/CH22/EX14.22.6/Ex14_22_6.sce index 962a1787b..862f94a04 100644 --- a/3428/CH22/EX14.22.6/Ex14_22_6.sce +++ b/3428/CH22/EX14.22.6/Ex14_22_6.sce @@ -1,7 +1,7 @@ //Section-14,Example-2,Page no.-PC.54
//To calculate force necessary to lift a ring of 1.0 cm radius from liquid water.
clc;
-y=72.8
-r=1
-F=2*(2*%pi*r)*y
-disp(F,'Force necessary to lift a ring of radius r from a liquid of surface tension y)
+y=72.8 //dynes/cm
+r=1 //cm
+F=2*(2*%pi*r)*y //dynes
+disp(F,'Force necessary to lift a ring of radius r from a liquid of surface tension y(dynes)')
diff --git a/3428/CH22/EX14.22.7/Ex14_22_7.sce b/3428/CH22/EX14.22.7/Ex14_22_7.sce index e8d88c9e5..2e2eb1cff 100644 --- a/3428/CH22/EX14.22.7/Ex14_22_7.sce +++ b/3428/CH22/EX14.22.7/Ex14_22_7.sce @@ -1,10 +1,10 @@ //Section-14,Example-1,Page no.-PC.59
//To calculate coefficient of viscosity of experimental liquid.
//(n_1/n_2)=(t_1*d_1)/(t_2*d_2)
-n_2=1 //Coefficient of viscosity of reference liquid.
-t_1=45.32 //t_1and t_2 (times of flow)
-t_2=65.66
-d_1=0.8 //d_1 and d_2 (densities)
-d_2=1.0
+n_2=1 //Coefficient of viscosity of reference liquid (centipoise)
+t_1=45.32 //t_1and t_2 (times of flow) (s)
+t_2=65.66 //(s)
+d_1=0.8 //d_1 and d_2 (densities)(g/cm^3)
+d_2=1.0 //(g/cm^3)
n_1=((n_2*t_1*d_1)/(t_2*d_2))
disp(n_1,'Coefficient of viscosity of experimental liquid(centipoise)')
diff --git a/3428/CH22/EX14.22.8/Ex14_22_8.sce b/3428/CH22/EX14.22.8/Ex14_22_8.sce index d0a58c0c8..6875cc020 100644 --- a/3428/CH22/EX14.22.8/Ex14_22_8.sce +++ b/3428/CH22/EX14.22.8/Ex14_22_8.sce @@ -1,11 +1,11 @@ //Section-14,Example-1,Page no.-PC.61
//To find absolute viscosity of liquid.
clc;
-d_s=8*10^3
-d_l=2*10^3
-r=10^-3
-l=0.1
-t=20
-g=9.8
+d_s=8*10^3 //kg/m^3
+d_l=2*10^3 //kg/m^3
+r=10^-3 //m
+l=0.1 //m
+t=20 //s
+g=9.8 //m/s^2
n_l=(2*g*r^2*(d_s-d_l))/(9*(l/t))
disp(n_l,'Absolute viscosity of liquid(Pa s)')
diff --git a/3428/CH23/EX14.23.1/Ex14_23_1.sce b/3428/CH23/EX14.23.1/Ex14_23_1.sce index 19dfb324d..b1310dd1d 100644 --- a/3428/CH23/EX14.23.1/Ex14_23_1.sce +++ b/3428/CH23/EX14.23.1/Ex14_23_1.sce @@ -3,10 +3,10 @@ clc;
//K_p=K_c*((R*T)^dl_n)
dl_n=2-(1+3)
-T=673
-R=0.0821
+T=673 //Kelvin
+R=0.0821 // (dm^3K^-1mol^-1)
K_c=0.495
-P=2
+P=2 //atm
K_p=K_c*((R*T)^dl_n)
disp(K_p)
//K_p=K_x*((P)^dl_n)
diff --git a/3428/CH23/EX14.23.11/Ex14_23_11.sce b/3428/CH23/EX14.23.11/Ex14_23_11.sce index 8d68f0791..a6f0210aa 100644 --- a/3428/CH23/EX14.23.11/Ex14_23_11.sce +++ b/3428/CH23/EX14.23.11/Ex14_23_11.sce @@ -1,7 +1,8 @@ //Section-14,Example-6,Page no.-PC.84
-//To find the concentration of Ag2+ for the given conditions.
-clc;
-K_spAgI=1.5*10^-16
-K_spAgCl=1.56*10^-10 //C=[I-]/[Cl-]
-C=(K_spAgI)/(K_spAgCl)
-disp(C,'Required concentration of AgCl')
+//To find the concentration of Ag+ ions at which I- and Cl- ions will precipitate.
+clc,
+Ksp_AgI=1.5*10^-16
+Ksp_AgCl=1.56*10^-10
+C=Ksp_AgI/Ksp_AgCl
+disp(C,'concentration of Ag+ ions at which I- and Cl- ions will precipitate')
+//Answer in the book is 9.6*10^-7 approximately equal to 0.0000010.
diff --git a/3428/CH23/EX14.23.12/Ex14_23_12.sce b/3428/CH23/EX14.23.12/Ex14_23_12.sce deleted file mode 100644 index 2e4cf103f..000000000 --- a/3428/CH23/EX14.23.12/Ex14_23_12.sce +++ /dev/null @@ -1,8 +0,0 @@ -//Section-14,Example-6,Page no.-PC.84
-//To calculate the free energy change and justify the given reactions.
-clc;
-//Cu_2S + O_2 = 2Cu + SO_2
-dl_G1=88.2
-dl_G2=300.1
-dl_G=dl_G1-dl_G2
-disp(dl_G,'Free energy change(kJ)')
diff --git a/3428/CH23/EX14.23.13/Ex14_23_13.sce b/3428/CH23/EX14.23.13/Ex14_23_13.sce index 73e257841..d35067ad8 100644 --- a/3428/CH23/EX14.23.13/Ex14_23_13.sce +++ b/3428/CH23/EX14.23.13/Ex14_23_13.sce @@ -1,7 +1,8 @@ -//Section-14,Example-1,Page no.-PC.90 -//To justify the given reaction -clc, -dl_G1=88.2 -dl_G2=-300.1 -dl_G=dl_G1+dl_G2 -disp(dl_G) +//Section-14,Example-1,Page no.-PC.90
+//To calculate the free energy change and justify the given reactions.
+clc;
+//Cu_2S + O_2 = 2Cu + SO_2
+dl_G1=88.2 //kJ
+dl_G2=300.1 //kJ
+dl_G=dl_G1-dl_G2 //kJ
+disp(dl_G,'Free energy change(kJ)')
diff --git a/3428/CH23/EX14.23.15/Ex14_23_15.sce b/3428/CH23/EX14.23.15/Ex14_23_15.sce index f60b85e99..222d348ae 100644 --- a/3428/CH23/EX14.23.15/Ex14_23_15.sce +++ b/3428/CH23/EX14.23.15/Ex14_23_15.sce @@ -2,6 +2,6 @@ //To estimate fraction of Morphine protonated.
clc;
K_b=1.6*10^-6
-B=0.010
+B=0.010 //mol/L
F_pro=sqrt(K_b/B)
disp(F_pro,'Fraction of Morphine protonated')
diff --git a/3428/CH23/EX14.23.16/Ex14_23_16.sce b/3428/CH23/EX14.23.16/Ex14_23_16.sce index 7156a46a1..e46413cfe 100644 --- a/3428/CH23/EX14.23.16/Ex14_23_16.sce +++ b/3428/CH23/EX14.23.16/Ex14_23_16.sce @@ -1,8 +1,9 @@ //Section-14,Example-2,Page no.-PC.111
//To estimate the pH of 0.10 M NH_3 at 25 degree Celcius.
-clc;
+clc
+T=25 //degree Celsius
K_b=1.8*10^-5
-B=0.1
+B=0.1 //M
pK_b=-log10(K_b)
pK_w=14
F_pro=sqrt(K_b/B) //Fraction protonated
diff --git a/3428/CH23/EX14.23.17/Ex14_23_17.sce b/3428/CH23/EX14.23.17/Ex14_23_17.sce index 97bdadbf2..9f8d59de8 100644 --- a/3428/CH23/EX14.23.17/Ex14_23_17.sce +++ b/3428/CH23/EX14.23.17/Ex14_23_17.sce @@ -1,22 +1,24 @@ //Section-14,Example-1,Page no.-PC.112 //To calculate the pH values of the following. clc; -C_HCl=0.001 -C_1=C_HCl //Since HCl is a strong acid,[H3O+]=[HCl] +C_HCl=0.001 //(M) +C_1=C_HCl //Since HCl is a strong acid,[H3O+]=[HCl](M) pH_1=-log10(C_1) disp(pH_1,'pH of 0.001 M HCl') -C_NaOH=0.0001 -C_2=C_NaOH //Since NaOH is a strong base so [OH-]=[NaOH] +C_NaOH=0.0001 //(M) +C_2=C_NaOH //Since NaOH is a strong base so [OH-]=[NaOH](M) pOH=-log10(C_2) pH_2=14-pOH disp(pH_2,'pH of 0.0001 M NaOH') -C_BaOH2=0.001 -C_31=2*C_BaOH2 // [OH-]=2*[Ba(OH)2] +C_BaOH2=0.001 //(M) +C_31=2*C_BaOH2 // [OH-]=2*[Ba(OH)2](M) k_w=10^-14 -C_32=k_w/(C_31) +C_32=k_w/(C_31) //(M) pH_3=-log10(C_32) disp(pH_3,'pH of 0.001 M Ba(OH)2') -M=(0.049/98)/(200/1000) //Molarity of H_2SO_4 solution -C_4=2*M //[H_3O+] +M_H2SO4=0.0049 //Mass of H2SO4 (gm) +V_req=200 //Volume of solution to be prepared(ml) +Mo=(M_H2SO4/98)/(V_req/1000) //Molarity of H_2SO_4 solution(M) +C_4=2*Mo //[H_3O+](M) pH_4=-log10(C_4) disp(pH_4,'pH of the given solution') diff --git a/3428/CH23/EX14.23.18/Ex14_23_18.sce b/3428/CH23/EX14.23.18/Ex14_23_18.sce index a19fbf110..0fdda2abd 100644 --- a/3428/CH23/EX14.23.18/Ex14_23_18.sce +++ b/3428/CH23/EX14.23.18/Ex14_23_18.sce @@ -1,21 +1,21 @@ //Section-14,Example-2,Page no.-PC.112 //To calculate the pH in the following cases. clc; -V_1=150 //volume of 0.1 NaOH solution -V_2=150 //volume of 0.2 HCl solution -N_1=0.1 -N_2=0.2 -V=V_1+V_2 //Total volume of the solution -m_eq=(V_2*N_2)-(V_1*N_1) //Total milliequivalents of excess HCl -N=m_eq/V -C_1=N //Since HCl is a strong acid so[HCl]=[H3O+] +V_1=150 //volume of 0.1 NaOH solution(ml) +V_2=150 //volume of 0.2 HCl solution(ml) +N_1=0.1 //(N) +N_2=0.2 //(N) +V=V_1+V_2 //Total volume of the solution(ml) +m_eq=(V_2*N_2)-(V_1*N_1) //Total milligram equivalents of excess HCl(gm equivalents) +N=m_eq/V //(N) +C_1=N //Since HCl is a strong acid so[HCl]=[H3O+] (M) pH_1=-log10(C_1) disp(pH_1,'pH of the required solution') pH1=5 -C1=10^-5 //[H3O+] +C1=10^-5 //[H3O+] (M) pH2=3 -C2=10^-3 //[H3O+] -C_3=(C1+C2)/2 //[H3O+] +C2=10^-3 //[H3O+] (M) +C_3=(C1+C2)/2 //[H3O+] (M) pH_2=-log10(C_3) disp(pH_2,'pH of the required solution') diff --git a/3428/CH23/EX14.23.2/Ex14_23_2.sce b/3428/CH23/EX14.23.2/Ex14_23_2.sce index f4f950dd0..5bd1ce993 100644 --- a/3428/CH23/EX14.23.2/Ex14_23_2.sce +++ b/3428/CH23/EX14.23.2/Ex14_23_2.sce @@ -1,9 +1,9 @@ //Section-14,Example-1,Page no.-PC.71
//To find Q_c and predict the direction in which the reaction would proceed.
clc;
-[N_2]=0.03
-[O_2]=0.01
-[NO]=0.04
+[N_2]=0.03 //mol
+[O_2]=0.01 //mol
+[NO]=0.04 //mol
K_c=2.2*10^3
Q_c=([N_2]*[O_2])/[NO]^2
disp(Q_c)
diff --git a/3428/CH23/EX14.23.20/Ex14_23_20.sce b/3428/CH23/EX14.23.20/Ex14_23_20.sce index 4535b7238..bf7f8f1cd 100644 --- a/3428/CH23/EX14.23.20/Ex14_23_20.sce +++ b/3428/CH23/EX14.23.20/Ex14_23_20.sce @@ -2,15 +2,15 @@ //To calculate the pH in the following cases. clc; K_a=7.3*10^-6 -c_1=0.23 +c_1=0.23 //(M) alpha_1=sqrt(K_a/c_1) -C_1=c_1*alpha_1 +C_1=c_1*alpha_1 //(M) pH_1=-log10(C_1) disp(pH_1,'pH of the given weak acid') -c_2=0.2 +c_2=0.2 //(M) K_b=4.4*10^-5 alpha_2=sqrt(K_b/c_2) -C_2=c_2*alpha_2 //[OH-] +C_2=c_2*alpha_2 //[OH-] (M) pOH=-log10(C_2) pH_2=14-pOH disp(pH_2,'pH of CH_3NH_2') diff --git a/3428/CH23/EX14.23.21/Ex14_23_21.sce b/3428/CH23/EX14.23.21/Ex14_23_21.sce index 19726bf37..34102a8e6 100644 --- a/3428/CH23/EX14.23.21/Ex14_23_21.sce +++ b/3428/CH23/EX14.23.21/Ex14_23_21.sce @@ -1,12 +1,13 @@ //Section-14,Example-5,Page no.-PC.114 //To calculate the pH of a solution obtained in the given condition. clc; -M_1=0.1 -M_2=0.2 -V_1=10 -V_2=40 -V=V_1+V_2 +M_1=0.1 //(M) +M_2=0.2 //(M) +V_1=10 //(ml) +V_2=40 //(ml) +V=V_1+V_2 //(ml) m_eq=(M_1*V_1)+(M_2*V_2*2) //mgm. equivalents of [H3O+] -C_1=m_eq/V //[H3O+] +C_1=m_eq/V //[H3O+] //(M) pH=-log10(C_1) disp(pH,'pH of the given solution') +//Answer given in the book(pH=4.685) is wrong diff --git a/3428/CH23/EX14.23.22/Ex14_23_22.sce b/3428/CH23/EX14.23.22/Ex14_23_22.sce index 32f3b36a6..c3db5aa20 100644 --- a/3428/CH23/EX14.23.22/Ex14_23_22.sce +++ b/3428/CH23/EX14.23.22/Ex14_23_22.sce @@ -1,8 +1,11 @@ //Section-14,Example-6,Page no.-PC.114 //To calculate the pH of 10^-8 M HCl solution. clc; +C=10^-8 //(M) Concentration of HCl solution k_w=10^-14 -x=9.5*10^-8 -C_1=10^-8+x +x1=(-C+sqrt((C)^2-(4*1*(-k_w))))/2 //(M) Concentration of OH- +x2=(-C-sqrt((C)^2+(4*1*(-k_w))))/2 //(M) Concentration of OH- +//since value of x2 is complex so it is rejected. +C_1=C+x1 //(M) Concentration of (H3O+) pH=-log10(C_1) disp(pH,'pH of the given solution') diff --git a/3428/CH23/EX14.23.23/Ex14_23_23.sce b/3428/CH23/EX14.23.23/Ex14_23_23.sce index ea1a8508b..53224b6ae 100644 --- a/3428/CH23/EX14.23.23/Ex14_23_23.sce +++ b/3428/CH23/EX14.23.23/Ex14_23_23.sce @@ -5,5 +5,5 @@ pKa=3.08 Ka=10^(-pKa) x1=(-Ka+sqrt((Ka)^2+(4*0.01*Ka)))/2 x2=(-Ka-sqrt((Ka)^2+(4*0.01*Ka)))/2 //neglected -pH=-log10(x2) +pH=-log10(x1) disp(pH,'pH of the given solution') diff --git a/3428/CH23/EX14.23.25/Ex14_23_25.sce b/3428/CH23/EX14.23.25/Ex14_23_25.sce index d66e1afac..3f7a15338 100644 --- a/3428/CH23/EX14.23.25/Ex14_23_25.sce +++ b/3428/CH23/EX14.23.25/Ex14_23_25.sce @@ -2,7 +2,7 @@ //To estimate the pH of 0.01 M CH3COONa.
clc;
pK_w=14
-B=0.01
+B=0.01 //(M)
pK_a=-log10(1.8*10^-5)
pH=(1/2*pK_w)+(1/2*log10(B))+(1/2*pK_a)
disp(pH,'pH of given solution of CH3COONa')
diff --git a/3428/CH23/EX14.23.26/Ex14_23_26.sce b/3428/CH23/EX14.23.26/Ex14_23_26.sce index bd8cf4310..443e96eb1 100644 --- a/3428/CH23/EX14.23.26/Ex14_23_26.sce +++ b/3428/CH23/EX14.23.26/Ex14_23_26.sce @@ -3,7 +3,7 @@ clc;
K_a=1.75*10^-5
pK_a=-log10(K_a)
-[CH_3COOH]=(1000/(60*100))
-[CH_3COONa]=((1.5*1000)/(82*100))
+[CH_3COOH]=(1000/(60*100)) //moldm^-3
+[CH_3COONa]=((1.5*1000)/(82*100)) //moldm^-3
pH=(pK_a+ (log10([CH_3COONa]/[CH_3COOH])))
disp(pH,'pH of the given solution')
diff --git a/3428/CH23/EX14.23.27/Ex14_23_27.sce b/3428/CH23/EX14.23.27/Ex14_23_27.sce index c494a9345..8728c5de9 100644 --- a/3428/CH23/EX14.23.27/Ex14_23_27.sce +++ b/3428/CH23/EX14.23.27/Ex14_23_27.sce @@ -1,19 +1,19 @@ //Section-14,Example-2,Page no.-PC.126
clc;
-CH_3COONa_1=0.01
-CH_3COOH_1=0.1
+CH_3COONa_1=0.01 //moldm^-3
+CH_3COOH_1=0.1 //moldm^-3
K_a=1.75*10^-5
pK_a=-log10(K_a)
pH1=pK_a +(log10(CH_3COONa_1/CH_3COOH_1))
disp(pH1,'pH of the given buffer solution')
-HCl=0.0002
-CH_3COONa_2=0.01+0.0002
-CH_3COOH_2=0.1-0.002
+HCl=0.0002 //moles
+CH_3COONa_2=0.01+0.0002 // moldm^-3
+CH_3COOH_2=0.1-0.002 //moldm^-3
pH2=pK_a +(log10(CH_3COONa_2/CH_3COOH_2))
disp(pH2,'pH of the solution after addition of HCl')
-C_1=pH1-pH2 //change in pH
-CH_3COONa_3=0.01+0.002
-CH_3COOH_3=0.1-0.002
+C_1=pH1-pH2 //change in pH (M)
+CH_3COONa_3=0.01+0.002 //moldm^-3
+CH_3COOH_3=0.1-0.002 //moldm^-3
pH3=pK_a +(log10(CH_3COONa_3/CH_3COOH_3))
pH4=pH1-pH3 //change in pH
disp(pH4,'Required pH')
diff --git a/3428/CH23/EX14.23.30/Ex14_23_30.sce b/3428/CH23/EX14.23.30/Ex14_23_30.sce index 157ef4c65..47a2f1098 100644 --- a/3428/CH23/EX14.23.30/Ex14_23_30.sce +++ b/3428/CH23/EX14.23.30/Ex14_23_30.sce @@ -1,10 +1,10 @@ //Section-14,Example-2,Page no.-PC.129
//To find the Molarity of given sulphuric acid solution.
clc;
-N_T=0.1354
-V_T=42.20
-V_S=50.00
+N_T=0.1354 //(N)
+V_T=42.20 //(ml)
+V_S=50.00 //(ml)
//N_S=N_H_2SO_4(let) and M_S=M_H_2SO_4
-N_S=(N_T*V_T)/V_S
-M_S=(N_S/2)
+N_S=(N_T*V_T)/V_S //(N)
+M_S=(N_S/2) //(M)
disp(M_S,'Molarity of H_2SO_4')
diff --git a/3428/CH23/EX14.23.4/Ex14_23_4.sce b/3428/CH23/EX14.23.4/Ex14_23_4.sce index 75861f586..798ba3942 100644 --- a/3428/CH23/EX14.23.4/Ex14_23_4.sce +++ b/3428/CH23/EX14.23.4/Ex14_23_4.sce @@ -1,12 +1,19 @@ //Section-14,Example-2,Page no.-PC.75
//To calculate equilibrium composition of reaction mixture
clc;
-a_N2=1.00-x
-a_H2=3.00-(3*x)
-a_NH_3=2*x
+//a_N2=1.00-x //Equilibrium partial pressure of N_2(bar)
+//a_H2=3.00-(3*x) //Equilibrium partial pressure of H_2(bar)
+//a_NH_3=2*x //Equilibrium partial pressure of NH(bar)
//K=(2*x^2)/((1.00-x)*(3.00-3*x)^3)=977
x_1=(163.416+sqrt((163.416)^2-(4*81.208*81.208)))/(2*81.208)
x_2=(163.416-sqrt((163.416)^2-(4*81.208*81.208)))/(2*81.208)
-p_N2=1-x_2
-p_H2=3*(1-x_2)
-p_NH3=2*x_2
+disp(x_1,'1st value of K')
+disp(x_2,'2nd value of K')
+//since x_2<1,it is accepted as the value of x.
+p_N2=1-x_2 //bar
+disp(p_N2,'Equilibrium composition of N2(bar)')
+p_H2=3*(1-x_2) //bar
+disp(p_H2,'Equilibrium composition of H2(bar)')
+p_NH3=2*x_2 //bar
+disp(p_NH3,'Equilibrium composition of NH3(bar)')
+// It can be concluded that product NH3 dominate at equilibrium.
diff --git a/3428/CH23/EX14.23.7/Ex14_23_7.sce b/3428/CH23/EX14.23.7/Ex14_23_7.sce index 87f2a4e84..a2eed29b4 100644 --- a/3428/CH23/EX14.23.7/Ex14_23_7.sce +++ b/3428/CH23/EX14.23.7/Ex14_23_7.sce @@ -1,11 +1,11 @@ //Section-14,Example-1,Page no.-PC.80
//To calculate K_p at 1000 K
clc;
-T_1=925
-T_2=1000
-K_p925=18.5
-dl_H=-71.09*10^3
-R=8.314
+T_1=925 //Kelvin
+T_2=1000 //Kelvin
+K_p925=18.5 //K_p at 925 K
+dl_H=-71.09*10^3 //kJmol^-1
+R=8.314 //JK^-1mol^-1
//ln(K_p1000)/(K_p925)=(dl_H/R)*((1/T_1)-(1/T_2))
K=((dl_H)/R)*((1/T_1)-(1/T_2))
K_p1000=(%e^(K))*18.5
diff --git a/3428/CH23/EX14.23.9/Ex14_23_9.sce b/3428/CH23/EX14.23.9/Ex14_23_9.sce index 9d2925809..927bbbd96 100644 --- a/3428/CH23/EX14.23.9/Ex14_23_9.sce +++ b/3428/CH23/EX14.23.9/Ex14_23_9.sce @@ -1,6 +1,6 @@ //Section-14,Example-4,Page no.-PC.82
//To calculate the solubility of Ag_2CrO_4.
clc;
-K_sp=(9*10^-12)/4
-S=(K_sp)^(1/3)
+K_sp=(9*10^-12)
+S=(K_sp/4)^(1/3)
disp(S,'Solubility product of Ag_2CrO_4(mol/dm^3)')
diff --git a/3428/CH4/EX1.4.3/Ex1_4_3.sce b/3428/CH4/EX1.4.3/Ex1_4_3.sce index 23d83248b..74d0b1f1c 100644 --- a/3428/CH4/EX1.4.3/Ex1_4_3.sce +++ b/3428/CH4/EX1.4.3/Ex1_4_3.sce @@ -3,5 +3,5 @@ L=758
H=420
U=61
-VI=((L-U)/L-H))*100
+VI=((L-U)/(L-H))*100
disp (VI,'Viscosity index of the given oil sample')
diff --git a/3428/CH5/EX1.5.3/Ex1_5_3.sce b/3428/CH5/EX1.5.3/Ex1_5_3.sce index ec5a92ca8..a481050ba 100644 --- a/3428/CH5/EX1.5.3/Ex1_5_3.sce +++ b/3428/CH5/EX1.5.3/Ex1_5_3.sce @@ -10,5 +10,5 @@ Q_fu=3.20 e_f=Q_f/E_f //faliure strains of fibres
e_m=Q_m/E_m //faliure strains of matrix
Q_m=E_m*e_f
-Q_cu =((Q_fu*V_f)+((Q_m*(1-V_f)))
-disp(Q_cu,'stress carried by composite at failure(GPa)')
+Q_cu =((Q_fu*V_f)+((Q_m*(1-V_f))))
+disp(Q_cu,'Stress carried by composite at failure(GPa)')
diff --git a/3428/CH9/EX4.9.5/Ex4_9_5.sce b/3428/CH9/EX4.9.5/Ex4_9_5.sce index 24c2d14ea..0bd8ccae7 100644 --- a/3428/CH9/EX4.9.5/Ex4_9_5.sce +++ b/3428/CH9/EX4.9.5/Ex4_9_5.sce @@ -7,7 +7,7 @@ c=3*10^10 v_bar=2140
mu_CO=((m_c*m_o)/(m_c+m_o)) //Reduced mass of CO(kg)
//v=((1/2*pi)*(k/mu)^1/2) and v_bar=((1/2*pi*c)*(k/mu)^1/2)
-k=(4*%pi^2*c^2*v_bar^2*mu)
+k=(4*%pi^2*c^2*v_bar^2*mu_CO)
disp(k,'Force constant of the molecule(N/m)')
-//k=1853(N/m) is wrong in the book.
+
diff --git a/3428/CH9/EX4.9.6/Ex4_9_6.sce b/3428/CH9/EX4.9.6/Ex4_9_6.sce index 2ef6ae8bb..d61d6a4d3 100644 --- a/3428/CH9/EX4.9.6/Ex4_9_6.sce +++ b/3428/CH9/EX4.9.6/Ex4_9_6.sce @@ -1,6 +1,7 @@ //Section-4,Example-2,Page no.-I.22 //To calculate the force constant and Bond length of the CO bond from the given data. clc; +h=6.626*10^-34 v_Qbar=2143.26 v_0bar=v_Qbar c=3*10^10 |