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22 files changed, 337 insertions, 0 deletions

diff --git a/3718/CH3/EX3.1/Ex3_1.sce b/3718/CH3/EX3.1/Ex3_1.sce
new file mode 100644
index 000000000..654db409a
--- /dev/null
+++ b/3718/CH3/EX3.1/Ex3_1.sce
@@ -0,0 +1,12 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 1

+clc;

+

+//Declaration of Variables

+q = 120       // Heat from surrounding, cal

+W = 70        // Work done, cal

+

+// Solution

+delta_E = q - W

+

+mprintf("Change in Internal Energy :%d cals", delta_E)

diff --git a/3718/CH3/EX3.10/Ex3_10.sce b/3718/CH3/EX3.10/Ex3_10.sce
new file mode 100644
index 000000000..54fb995d8
--- /dev/null
+++ b/3718/CH3/EX3.10/Ex3_10.sce
@@ -0,0 +1,19 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 10

+clc;

+

+//Declaration of Constant

+R = 8.314                   //in J / K

+

+//Declaration of Variables

+V_O2 = 2.8                  //in litres

+V_H2 = 19.6                 //in litres

+

+// Solution

+na = V_O2 / 22.4            //in mol

+nb = V_H2 / 22.4            //in mol

+Xa = na / (na + nb)

+Xb = nb / (na + nb)

+d_s = (- R) * (na * log(Xa) + nb * log(Xb))

+

+mprintf("The increase in entropy on mixing is : %.3f J /K",d_s)

diff --git a/3718/CH3/EX3.12/Ex3_12.sce b/3718/CH3/EX3.12/Ex3_12.sce
new file mode 100644
index 000000000..ba805e88f
--- /dev/null
+++ b/3718/CH3/EX3.12/Ex3_12.sce
@@ -0,0 +1,15 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 12

+clc;

+

+//Declaration of Variables

+d_g_25 = - 85.77        // k J, Free Energy at 25 C

+d_g_35 = - 83.68        // k J, Free Energy at 35 C

+Ti = 273 + 25               // K

+Tf = 273 + 35               // K

+

+// Solution

+mprintf("Equating the entropy change at both the temperatures.\n")

+mprintf(" (d_h + d_g_25) / Ti = (d_h + d_g_35) / Tf\n")

+d_h = - 148

+mprintf(" The change in enthalpy for the process at 30C is %d kJ", d_h)

diff --git a/3718/CH3/EX3.13/Ex3_13.sce b/3718/CH3/EX3.13/Ex3_13.sce
new file mode 100644
index 000000000..66fcd25d4
--- /dev/null
+++ b/3718/CH3/EX3.13/Ex3_13.sce
@@ -0,0 +1,20 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 13

+clc;

+

+//Declaration of Constants

+l_v = 101            //in cal /g, Latent headt of vap.

+mwt = 78            // molecular weight of benzene

+

+//Declaration of Variable

+m = 2

+Tb = 80.2          // C, boiling point of benzene

+

+// Solution

+Tb = Tb + 273      // K

+d_h = l_v * mwt

+d_s = d_h / Tb

+d_g = d_h - Tb * d_s

+

+mprintf("d_s = %.2f cal / K\n",d_s)

+mprintf(" d_g = d_a = %d", d_g)

diff --git a/3718/CH3/EX3.14/Ex3_14.sce b/3718/CH3/EX3.14/Ex3_14.sce
new file mode 100644
index 000000000..f4f228331
--- /dev/null
+++ b/3718/CH3/EX3.14/Ex3_14.sce
@@ -0,0 +1,28 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 14

+clc;

+

+//Declaration of Variables

+V1 = 6            //in dm cube

+V2 = 2            //in dm cube

+T1 = 27           //in C

+m = 5

+

+// Solution

+mprintf("T1*V1 ^ (gamma - 1) = T2 * V2 ^ (gamma - 1)\n")

+

+T1 =T1 + 273       // K

+T2 = T1 * (V1 / V2) ** (8.314 / 20.91)

+

+mprintf(" The Final Temperature is %.1f K\n",T2)

+

+q = 0             //For Adiabatic process

+d_E = - m * 20.91 * (T2 - T1)

+d_E = d_E / 1000

+

+mprintf(" q =%d \n", q)

+mprintf(" Change is Energy is %.2f kJ / mol\n",d_E)

+

+W = - d_E

+

+mprintf(" W = %.2f kJ /mol",d_E)

diff --git a/3718/CH3/EX3.15/Ex3_15.sce b/3718/CH3/EX3.15/Ex3_15.sce
new file mode 100644
index 000000000..6072caddb
--- /dev/null
+++ b/3718/CH3/EX3.15/Ex3_15.sce
@@ -0,0 +1,27 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 15

+clc;

+

+//Declaration of Constant

+R = 8.314          //in J / K mol

+

+//Declaration of Variables

+m = 1

+V1 = 5             // dm cube

+V2 = 10            // dm cube

+T = 300            // K

+

+// Solution

+mprintf("For isothermal and reversible process,\n")

+

+d_E = 0

+d_H = 0

+d_A = - 2.303 * m * R * T * log10(V2 / V1)

+d_G = - 2.303 * m * R * T * log10(V2 / V1)

+q = - d_G

+W = - d_G

+

+mprintf(" d_E = d_H = %d \n", d_H)

+mprintf(" d_G = d_A =%.3f J / mol\n",d_G)

+mprintf(" For isothermal and reversible expansion\n")

+mprintf(" q = W = -d_G = %.3f J / mol",W)

diff --git a/3718/CH3/EX3.16/Ex3_16.sce b/3718/CH3/EX3.16/Ex3_16.sce
new file mode 100644
index 000000000..e33f2180a
--- /dev/null
+++ b/3718/CH3/EX3.16/Ex3_16.sce
@@ -0,0 +1,18 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 16

+clc;

+

+//Declaration of Constant

+R = 8.314         //in J per K mol

+

+//Declaration of Variables

+n = 5             // moles

+T = 27            // C

+V1 = 50.0         // L, Initial Volume

+V2 = 1000         // L, Final Volume

+

+//Solution

+T = T + 273

+d_G = 2.303 * n * R * T * log10(V1 / V2)

+d_G = d_G / 1000

+mprintf("The free energy change is :%.3f k J",d_G)

diff --git a/3718/CH3/EX3.17/Ex3_17.sce b/3718/CH3/EX3.17/Ex3_17.sce
new file mode 100644
index 000000000..750c05ca7
--- /dev/null
+++ b/3718/CH3/EX3.17/Ex3_17.sce
@@ -0,0 +1,12 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 17

+clc;

+

+//Declaration of Variables

+d_H_n = - 51.46        // k J/mol, neutralization

+d_H_i = - 57.1         // k J/mol, ionization

+

+//Solution

+d_H = - d_H_i + d_H_n

+

+mprintf("The head of ionization for NH4OH is %.2f kJ / mol", d_H)

diff --git a/3718/CH3/EX3.2/Ex3_2.sce b/3718/CH3/EX3.2/Ex3_2.sce
new file mode 100644
index 000000000..e5675252e
--- /dev/null
+++ b/3718/CH3/EX3.2/Ex3_2.sce
@@ -0,0 +1,11 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 2

+clc;

+

+//Solution

+mprintf("CH4 (g) + 2O2 (g) -> CO2 (g) + 2H20 (l)\n")

+

+delta_n = 1 - (1 + 2)

+solution = - 2 * 2 * 298             // cals

+

+mprintf(" Delta H - Delta E is: %d cals", solution)

diff --git a/3718/CH3/EX3.20/Ex3_20.sce b/3718/CH3/EX3.20/Ex3_20.sce
new file mode 100644
index 000000000..ed107fe67
--- /dev/null
+++ b/3718/CH3/EX3.20/Ex3_20.sce
@@ -0,0 +1,10 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 20

+clc;

+

+// Solution

+Eq_HI = 1.56 / 2

+Eq_H2 = 0.22 / 2

+Eq_I2 = 0.22 / 2

+Kc = Eq_H2 * Eq_I2 / (Eq_HI ** 2)

+mprintf("The equilibrium constant for the dissociation reaction %.4f",Kc)

diff --git a/3718/CH3/EX3.21/Ex3_21.sce b/3718/CH3/EX3.21/Ex3_21.sce
new file mode 100644
index 000000000..35b67c388
--- /dev/null
+++ b/3718/CH3/EX3.21/Ex3_21.sce
@@ -0,0 +1,15 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 21

+clc;

+

+//Declaration of Variables

+Kc = 0.5         // mole square litre square

+T = 400          // K

+R = 0.082        // litre atm per degree  per mole

+

+// Solution

+Kp = Kc * (R * T) ** (-2)

+

+mprintf("The given equilibrium is\n")

+mprintf(" N2(g) + 3H2(g) <--> 2NH3(g)\n")

+mprintf(" Kp is %.3e",Kp)

diff --git a/3718/CH3/EX3.22/Ex3_22.sce b/3718/CH3/EX3.22/Ex3_22.sce
new file mode 100644
index 000000000..7d1397138
--- /dev/null
+++ b/3718/CH3/EX3.22/Ex3_22.sce
@@ -0,0 +1,10 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 22

+clc;

+

+//Declaration of Variables

+solubility = 7.5 * 10 ** - 5        // mol per L

+

+// Solution

+Ksp = 4 * (solubility ** 3)

+mprintf("Solubility product of the salt is %.3e mol cube L cube",Ksp) 

diff --git a/3718/CH3/EX3.23/Ex3_23.sce b/3718/CH3/EX3.23/Ex3_23.sce
new file mode 100644
index 000000000..6c8cbba09
--- /dev/null
+++ b/3718/CH3/EX3.23/Ex3_23.sce
@@ -0,0 +1,12 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 23

+clc;

+

+//Declaration of Variables

+Ti = 25           // C

+S = 0.00179       // g / L

+

+// Solution

+S =S / 170        // mol / L

+Ksp = S ** 2

+mprintf("Solubility product at 25 C is %.4e mol square L square",Ksp)

diff --git a/3718/CH3/EX3.24/Ex3_24.sce b/3718/CH3/EX3.24/Ex3_24.sce
new file mode 100644
index 000000000..a4fd53a80
--- /dev/null
+++ b/3718/CH3/EX3.24/Ex3_24.sce
@@ -0,0 +1,13 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 24

+clc;

+

+//Declaration of Variables

+Ksp = 8 * 10 ** - 5               // Solubility product PbBr2

+diss = 80 / 100                  // % dissociation

+

+// Solution

+S = (Ksp / 4) ** (1 / 3.0)          // Solubility is 100%

+S_80 = S * (80 / 100.0)

+S_per_g = S_80 * 367 - 1.621

+mprintf("Solubility in gm per litre is %.3f gm / l",S_per_g)

diff --git a/3718/CH3/EX3.27/Ex3_27.sce b/3718/CH3/EX3.27/Ex3_27.sce
new file mode 100644
index 000000000..90c160c07
--- /dev/null
+++ b/3718/CH3/EX3.27/Ex3_27.sce
@@ -0,0 +1,13 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 27

+clc;

+

+//Declaration of Variables

+n_salt = 0.02     // mole

+n_base = 0.2      // mole

+pKb = 4.7

+

+// Solution

+pOH = pKb + log10(n_salt / n_base)

+pH = 14 - pOH

+mprintf("pH of a buffer solution is %.1f", pH)

diff --git a/3718/CH3/EX3.3/Ex3_3.sce b/3718/CH3/EX3.3/Ex3_3.sce
new file mode 100644
index 000000000..fc8f0dd6c
--- /dev/null
+++ b/3718/CH3/EX3.3/Ex3_3.sce
@@ -0,0 +1,15 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 3

+clc;

+

+//Declaration of Variables

+delta_g = -16.0        // Kelvin cal

+delta_h = -10.0        // Kelvin cal

+T = 300                // Kelvin

+

+// Solution

+delta_s = (delta_h - delta_g) * 10 ** 3 / T    // cal/deg

+new_t = 330                     // Kelvin

+new_delta_g = (delta_h * 10 ** 3) - new_t * delta_s

+

+mprintf("The free energy at 330K is: %.2e K cal",new_delta_g)

diff --git a/3718/CH3/EX3.4/Ex3_4.sce b/3718/CH3/EX3.4/Ex3_4.sce
new file mode 100644
index 000000000..8623977e9
--- /dev/null
+++ b/3718/CH3/EX3.4/Ex3_4.sce
@@ -0,0 +1,14 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 4

+clc;

+

+//Declaration of Variables

+delta_s = -20.7        // cal per deg per mol

+delta_h = -67.37       // K cal

+T = 25                 // deg C

+

+// Solution

+T = T + 273              // K

+delta_g = delta_h - (T * delta_s * 10 ** -3)

+

+mprintf("The change in free energy at 25deg C is: %.4f K cal per mol", delta_g)

diff --git a/3718/CH3/EX3.5/Ex3_5.sce b/3718/CH3/EX3.5/Ex3_5.sce
new file mode 100644
index 000000000..a99f0aeba
--- /dev/null
+++ b/3718/CH3/EX3.5/Ex3_5.sce
@@ -0,0 +1,13 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 5

+clc;

+

+//Declaration of Variables

+wt = 1              // g

+delta_h = 149       // joules

+

+// Solution

+delta_h_f = delta_h * (10 * 12 + 8 * 1)

+delta_h_f_c=delta_h_f * 10 ** -3

+

+mprintf("Enthalpy of fusion of naphthalene:%.3f kJ/mol", delta_h_f_c)

diff --git a/3718/CH3/EX3.6/Ex3_6.sce b/3718/CH3/EX3.6/Ex3_6.sce
new file mode 100644
index 000000000..2e0af5c97
--- /dev/null
+++ b/3718/CH3/EX3.6/Ex3_6.sce
@@ -0,0 +1,13 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 6

+clc;

+

+//Declaration of Variables

+d_h_acetylene = 230       // kJ per mol

+d_h_benzene = 85          // kJ per mol

+T = 298                   // K

+

+// Solution

+d_h = d_h_benzene - 3 * d_h_acetylene

+

+mprintf("The enthalpy change for the reaction is: %d kJ/mole", d_h)

diff --git a/3718/CH3/EX3.7/Ex3_7.sce b/3718/CH3/EX3.7/Ex3_7.sce
new file mode 100644
index 000000000..6d5415a06
--- /dev/null
+++ b/3718/CH3/EX3.7/Ex3_7.sce
@@ -0,0 +1,16 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 7

+clc;

+

+//Declaration of Constant

+d_h_vap = 2.0723         // kJ per g

+Tb = 373                     // K

+

+// Solution

+d_h_vap = d_h_vap * 18            // kJ per mol

+d_s = d_h_vap / Tb

+d_g = d_h_vap - Tb * d_s

+d_s = d_s * 1000

+

+mprintf("The Entropy change is: %.1f J / mol / K\n",d_s)

+mprintf(" The Free Energy change is:%d kJ/mol", d_g)

diff --git a/3718/CH3/EX3.8/Ex3_8.sce b/3718/CH3/EX3.8/Ex3_8.sce
new file mode 100644
index 000000000..6f40f4418
--- /dev/null
+++ b/3718/CH3/EX3.8/Ex3_8.sce
@@ -0,0 +1,19 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 8

+clc;

+

+//Declaration of Constant

+R = 1.987               // cal per K per mol

+

+//Declaration of Variables

+m = 5

+Vo = 4                  //in litres, Initial Volume

+Vf = 40                 //in litres, Final Volume

+T = 27                  //in deg C

+

+// Solution

+mprintf("dS = nRln(V2 / V1)\n")

+

+dS = m * R * 2.303 * log10(Vf / Vo)

+

+mprintf(" The change in entropy is: %.2f cal / degree",dS)

diff --git a/3718/CH3/EX3.9/Ex3_9.sce b/3718/CH3/EX3.9/Ex3_9.sce
new file mode 100644
index 000000000..10b2d2fff
--- /dev/null
+++ b/3718/CH3/EX3.9/Ex3_9.sce
@@ -0,0 +1,12 @@
+//Chapter 3: Thermodynamic and Chemical Equilibrium

+//Problem: 9

+clc;

+

+//Declaration of Variables

+wt = 10            //in g

+heat_a = 4.5       //in K

+

+// Solution

+m = 10 / 100.0     // mol

+d_h = heat_a / m

+mprintf("The heat of the reaction is:%d K cal / mol", d_h)