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+// Example 7_5
+clc;funcprot(0);
+// Given data
+m=0.035;// kg
+p_1=0.100;// MPa
+T_1=20.0;// °C
+p_2=5.00;// MPa
+k=1.4;// The specific heat ratio for air
+R_air=0.286;// kJ/kg.K
+
+// Solution
+T_2=((T_1+273.15)*(p_2/p_1)^((k-1)/k))-273.15;// °C
+v_1=(m*R_air*(T_1+273.15))/(p_1*10^3);// m^3/kg
+v_2=v_1*((T_2+273.15)/(T_1+273.15))^(1/(1-k));// m^3/kg
+printf('\nThe final temperature,T_2=%3.0f°C \nThe specific volume of the air,v_2=%0.5f m^3/kg',T_2,v_2);