From f35ea80659b6a49d1bb2ce1d7d002583f3f40947 Mon Sep 17 00:00:00 2001 From: prashantsinalkar Date: Tue, 10 Oct 2017 12:38:01 +0530 Subject: updated the code --- 3428/CH10/EX5.10.10/Ex5_10_10.sce | 4 ++-- 3428/CH10/EX5.10.8/Ex5_10_8.sce | 8 ++++---- 3428/CH11/EX5.11.3/Ex5_11_3.sce | 1 + 3428/CH14/EX8.14.2/Ex8_14_2.sce | 2 +- 3428/CH15/EX9.15.15/Ex9_15_15.sce | 3 ++- 3428/CH16/EX10.16.2/Ex10_16_2.sce | 2 +- 3428/CH17/EX10.17.10/Ex10_17_10.sce | 2 +- 3428/CH17/EX10.17.12/Ex10_17_12.sce | 2 +- 3428/CH19/EX12.19.1/Ex12_19_1.sce | 8 ++++---- 3428/CH19/EX12.19.2/Ex12_19_2.sce | 9 +++++---- 3428/CH19/EX12.19.3/Ex12_19_3.sce | 14 +++++++------- 3428/CH19/EX12.19.4/Ex12_19_4.sce | 12 ++++++------ 3428/CH19/EX12.19.5/Ex12_19_5.sce | 12 ++++++------ 3428/CH19/EX12.19.6/Ex12_19_6.sce | 10 +++++----- 3428/CH19/EX12.19.7/Ex12_19_7.sce | 16 ++++++++-------- 3428/CH2/EX1.2.17/Ex1_2_17.sce | 1 - 3428/CH2/EX1.2.18/Ex1_2_18.sce | 2 +- 3428/CH20/EX13.20.1/Ex13_20_1.sce | 12 ++++++------ 3428/CH21/EX14.21.1/Ex14_21_1.sce | 10 +++++----- 3428/CH21/EX14.21.10/Ex14_21_10.sce | 6 +++--- 3428/CH21/EX14.21.11/Ex14_21_11.sce | 6 +++--- 3428/CH21/EX14.21.12/Ex14_21_12.sce | 6 +++--- 3428/CH21/EX14.21.13/Ex14_21_13.sce | 20 ++++++++++++-------- 3428/CH21/EX14.21.14/Ex14_21_14.sce | 12 ++++++------ 3428/CH21/EX14.21.15/Ex14_21_15.sce | 12 ++++++------ 3428/CH21/EX14.21.16/Ex14_21_16.sce | 10 +++++----- 3428/CH21/EX14.21.17/Ex14_21_17.sce | 12 ++++++------ 3428/CH21/EX14.21.18/Ex14_21_18.sce | 16 +++++++++------- 3428/CH21/EX14.21.19/Ex14_21_19.sce | 24 ++++++++++++------------ 3428/CH21/EX14.21.2/Ex14_21_2.sce | 10 +++++----- 3428/CH21/EX14.21.3/Ex14_21_3.sce | 8 ++++---- 3428/CH21/EX14.21.4/Ex14_21_4.sce | 4 ++-- 3428/CH21/EX14.21.6/Ex14_21_6.sce | 8 ++++---- 3428/CH21/EX14.21.7/Ex14_21_7.sce | 8 ++++---- 3428/CH21/EX14.21.8/Ex14_21_8.sce | 4 ++-- 3428/CH21/EX14.21.9/Ex14_21_9.sce | 3 ++- 3428/CH22/EX14.22.1/Ex14_22_1.sce | 8 ++++---- 3428/CH22/EX14.22.2/Ex14_22_2.sce | 10 +++++----- 3428/CH22/EX14.22.3/Ex14_22_3.sce | 4 ++-- 3428/CH22/EX14.22.4/Ex14_22_4.sce | 20 ++++++++++---------- 3428/CH22/EX14.22.5/Ex14_22_5.sce | 10 +++++----- 3428/CH22/EX14.22.6/Ex14_22_6.sce | 8 ++++---- 3428/CH22/EX14.22.7/Ex14_22_7.sce | 10 +++++----- 3428/CH22/EX14.22.8/Ex14_22_8.sce | 12 ++++++------ 3428/CH23/EX14.23.1/Ex14_23_1.sce | 6 +++--- 3428/CH23/EX14.23.11/Ex14_23_11.sce | 13 +++++++------ 3428/CH23/EX14.23.12/Ex14_23_12.sce | 8 -------- 3428/CH23/EX14.23.13/Ex14_23_13.sce | 15 ++++++++------- 3428/CH23/EX14.23.15/Ex14_23_15.sce | 2 +- 3428/CH23/EX14.23.16/Ex14_23_16.sce | 5 +++-- 3428/CH23/EX14.23.17/Ex14_23_17.sce | 20 +++++++++++--------- 3428/CH23/EX14.23.18/Ex14_23_18.sce | 22 +++++++++++----------- 3428/CH23/EX14.23.2/Ex14_23_2.sce | 6 +++--- 3428/CH23/EX14.23.20/Ex14_23_20.sce | 8 ++++---- 3428/CH23/EX14.23.21/Ex14_23_21.sce | 13 +++++++------ 3428/CH23/EX14.23.22/Ex14_23_22.sce | 7 +++++-- 3428/CH23/EX14.23.23/Ex14_23_23.sce | 2 +- 3428/CH23/EX14.23.25/Ex14_23_25.sce | 2 +- 3428/CH23/EX14.23.26/Ex14_23_26.sce | 4 ++-- 3428/CH23/EX14.23.27/Ex14_23_27.sce | 16 ++++++++-------- 3428/CH23/EX14.23.30/Ex14_23_30.sce | 10 +++++----- 3428/CH23/EX14.23.4/Ex14_23_4.sce | 19 +++++++++++++------ 3428/CH23/EX14.23.7/Ex14_23_7.sce | 10 +++++----- 3428/CH23/EX14.23.9/Ex14_23_9.sce | 4 ++-- 3428/CH4/EX1.4.3/Ex1_4_3.sce | 2 +- 3428/CH5/EX1.5.3/Ex1_5_3.sce | 4 ++-- 3428/CH9/EX4.9.5/Ex4_9_5.sce | 4 ++-- 3428/CH9/EX4.9.6/Ex4_9_6.sce | 1 + 68 files changed, 301 insertions(+), 283 deletions(-) delete mode 100644 3428/CH23/EX14.23.12/Ex14_23_12.sce (limited to '3428') 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 -- cgit