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+//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)