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+// Example 11_13
+clc;funcprot(0);
+// Given data
+m=8.20;// lbm
+V=1.00;// ft^3
+T=-78.0;// °F
+
+// Solution
+// From Table C.12a, we find that
+T_c=240;// R
+p_c=507;// psia
+v_c=1.49/28.011;// ft^3/lbm
+// Also, from Table C.13a, we find that
+R=0.0709;// Btu/lbm.R
+T_R=(T+460)/T_c;
+v=V/m;// ft^3/lbm
+// Assume 'a' instead of '
+v_ca=(R*(T_c*778.16))/(p_c*144);// ft^3/lbm
+v_Ra=v/v_ca;
+// Using T_R = T/Tc = 1.60 and v′R = v/v′c = 0:67, we find from Figure 11.6 that
+p_R=2.10;
+Z=0.850;
+p=p_c*p_R;// psia
+printf("\nThe pressure exerted by 8.20 lbm of the carbon monoxide,p=%4.0f psia",p);
+// The answer is different due to round off error.