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diff --git a/260/CH11/EX11.2/11_2.sce b/260/CH11/EX11.2/11_2.sce
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index 000000000..f51ef5e29
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+++ b/260/CH11/EX11.2/11_2.sce
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+//Eg-11.2
+//pg-472
+
+
+// To find the work done  pressure is to be integrated with respect to volume which is compressed from 20 to 5 litres at 300k
+// i.e integrating the function [RT/(V-b)-a/V^2] from 20 to 5. Here we take 500 intervals between the given limits 20 and 5.
+// b and a represent the upper and lower limits of integration in the code below.
+// Putting in the given values the function simplifies to [24.6/(V-0.065)-5.5/V^2].
+clc
+clear
+
+deff('out =func(in)','out = 24.6/(in-0.065)-5.5/in^2')     //V is the in and P is the out.
+b = 5;
+a = 20;
+n = 500;
+summation = 0;
+
+h = (b-a)/n;     //step size
+
+for(i = 1:499)
+    F(i) = func(a+i*h);
+end
+
+for(i = 1:499)
+    summation = summation + F(i);
+end
+
+I = (h/2)*(func(a) + 2*summation + func(b));    //Composite version of the trapezoidal rule.
+
+printf('The value of the integral and there by the work done is  %f litre atm\n\n',abs(I))
+
+printf(' Using 500 segments ,this value is same as the exact value obtained analytically.\n However, when less number of segments(say 100) are used the numerical solution\n differs from the analytical one.')
+\ No newline at end of file