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diff --git a/647/CH2/EX2.13/Example2_13.sce b/647/CH2/EX2.13/Example2_13.sce
new file mode 100755
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+clear;
+clc;
+
+// Example: 2.13
+// Page: 58
+
+printf("Example: 2.13 - Page: 58\n\n");
+
+// Solution
+
+//*****Data*****//
+m = 5;// [kg]
+M = 29;// [kg/mol]
+T1 = 37 + 273;// [K]
+P1 = 101.33;// [kPa]
+T2 = 237 + 273;// [K]
+Cp = 29.1;// [J/mol K]
+Cv = 20.78;// [J/mol K]
+R = 8.314;// [J/K mol]
+//*****************//
+
+n = m/M;
+// From ideal gas equation:
+V1 = n*R*T1/P1;// [cubic m]
+
+// Isochoric Process:
+printf("Isochoric Process\n");
+// Volume = constant
+V2 = V1;// [cubic m]
+deltaU = n*Cv*(T2 - T1);// [kJ]
+// Since Volume is constant
+W = 0;
+Q = deltaU + W;// [kJ]
+// deltaH = deltaU + P*deltaV
+// deltaH = deltaU + n*R*deltaT
+deltaH = deltaU + n*R*(T2 - T1);// [kJ]
+printf("Change in Internal Energy is %.2f kJ\n",deltaU);
+printf("Heat Supplied is %.2f kJ\n",Q);
+printf("Work done is %d kJ\n",W);
+printf("Change in Enthalpy is %.2f kJ\n",deltaH);
+printf("\n");
+
+// Isobaric Process
+printf("Isobaric Process\n");
+// Since Pressure is constant.
+P2 = P1;// [kPa]
+deltaH = n*Cp*(T2 - T1);// [kJ]
+Qp = deltaH;// [kJ]
+// deltaU = deltaH - P*deltaV
+// From ideal gas equation:
+deltaU = deltaH - n*R*(T2 - T1);// [kJ]
+W = Qp - deltaU;// [kJ]
+printf("Change in Internal Energy is %.2f kJ\n",deltaU);
+printf("Heat Supplied is %.2f kJ\n",Qp);
+printf("Work done is %.2f kJ\n",W);
+printf("Change in Enthalpy is %.2f kJ\n",deltaH);
+\ No newline at end of file