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diff --git a/647/CH2/EX2.12/Example2_12.sce b/647/CH2/EX2.12/Example2_12.sce
new file mode 100755
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+clear;
+clc;
+
+// Example: 2.12
+// Page: 57
+
+printf("Example: 2.12 - Page: 57\n\n");
+
+// Solution
+
+//*****Data*****//
+m = 1;// [kg]
+P1 = 8;// [atm]
+T1 = 50 + 273;// [K]
+// V1 = V;// [L]
+// V2 = 5V;// [L]
+V1_by_V2 = 1/5;
+gama = 1.4;
+R = 0.082;// [L-atm]
+//***************//
+
+// Adiabatic process:
+printf("Adiabatic Process \n");
+P2 = P1*V1_by_V2^gama;// [atm]
+printf("Final Pressure is %.2f atm\n",P2);
+T2 = T1*V1_by_V2^(gama - 1);// [K]
+printf("Final Temperature is %f K\n",T2);
+Wad = R*(T2 - T1)/(1 - gama);// [L-atm]
+printf("Adiabatic Work done is %.3f L-atm\n",Wad);
+printf("\n")
+
+// Isothermal Process:
+printf("Isothermal Process\n")
+// In an isothermal Process, the temperature remans constant:
+T2 = T1;// [K]
+printf("Final temperature is %d K\n",T2);
+// From the ideal gas:
+// (P2*V2/T2) = (P1*V1/T1)
+// Since T2 = T1
+// P2*V2 = P1*V1
+P2 = P1*V1_by_V2;// [atm]
+printf("Final pressure is %.1f atm\n",P2);
+W = R*T1*log(1/V1_by_V2);// [L-atm]
+printf("Work done during the isothermal process is %.2f L-atm\n",W);
+\ No newline at end of file