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+//example 12
+//determining specific using diffenet laws
+clear
+clc
+T=100 //given temp.in 100 celsius
+P=3 //given pressure in MPa
+v1=0.0065 //specific volume in m^3/kg using table
+printf("\n hence,the specific volume for R-134a using R-134a tables is v1 = %.3f m^3/kg. \n",v1)
+M=102.3 //molecular mass in kg
+R=8.3145 //in kJ/K
+Ru=R/M //in kJ/K-kg
+v2=Ru*(T+273)/(P*1000) //specific volume assuming R-134a to be ideal gas in m^3/kg
+printf("\n hence,the specific volume for R-134a using R-134a the ideal gas laws is v2 = %.3f m^3/kg. \n",v2)
+Tr=373.2/374.2 //reduced temperature using generalized chart
+Pr=3/4.06 //reduced pressure using generalized chart
+Z=0.67 //compressibility factor 
+v3=Z*v2 // specific volume using generalized chart in m^3/kg
+printf("\n hence,the specific volume for R-134a using the generalized chart is v3 = %.3f m^3/kg. \n",v3)