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+// Theory and Problems of Thermodynamics
+// Chapter 2
+// First Law of Thermodynamics
+// Example 8
+
+clear ;clc;
+
+//Given data
+V1 = 1            // one compartment volume in m^3 
+T1 = 300          // first compartment temperature in K
+P1 = 0.1          // first compartment pressure in MPa
+V2 = 2            // second compartment volume in m^3 
+T2 = 1000         // second compartment temperature in K
+P2 = 1            // second compartment pressure in MPa
+R = 8.314         // gas constant in J/K
+del_U = 0;        // tank is rigid and insulated W = 0 & Q = 0
+
+// determine final temperature and pressure of gas if the partion between 
+//               the two compartment is ruptured
+P1 = P1 * 1e6      // units conversion MPa to Pa
+P2 = P2 * 1e6      // units conversion MPa to Pa
+
+N1 = P1 * V1 / (R * T1);        // quantity of gas in first compartment 
+N2 = P2 * V2 / (R * T2);        // quantity of gas in second compartment
+
+T_f = (T1 * N1 + T2 * N2)/ (N1 + N2)        //Final state Temperature
+
+P_f = (N1+N2)*R*T_f/(V1+V2);                //Final stste Pressure
+P_f = P_f/1000000                           // Units conversion Pa to MPa
+
+// Results
+mprintf('Final state Temperature = %3.0f K', T_f)
+mprintf('\n Final state pressure = %2.1f MPa', P_f)
+
+