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+clear;

+clc;

+disp('Example 5.11');

+

+//  aim : To determine the

+//  (a) original and final volume of the gas

+//  (b) final pressure of the gas

+//  (c) final temperature of the gas

+

+//  Given values

+m = .675;//  mass of the gas,[kg]

+P1 = 1.4;//  original pressure,[MN/m^2]

+T1 = 273+280;//  original temperature,[K]

+R = .287;//gas constant,[kJ/kg K]

+

+//  solution

+

+//  (a)

+//  using characteristic equation, P1*V1=m*R*T1

+V1 = m*R*T1*10^-3/P1;//  [m^3]

+//  also Given 

+V2 =  4*V1;// [m^3]

+mprintf('\n (a) The original volume of the gas is  =  %f  m^3\n',V1);

+mprintf('\n      and The final volume of the gas is  =  %f  m^3\n',V2);

+

+//  (b)

+//  Given that gas is following the law P*V^1.3=constant

+//  hence process is polytropic with 

+n = 1.3; //  polytropic index

+P2 = P1*(V1/V2)^n;//  formula for polytropic process,[MN/m^2]

+mprintf('\n (b) The final pressure of the gas is  =  %f  kN/m^2\n',P2*10^3);

+

+//  (c)

+//  since mass is constant so,using P*V/T=constant

+//  hence

+T2 = P2*V2*T1/(P1*V1);//  [K]

+t2 = T2-273;//  [C]

+mprintf('\n (c) The final temperature of the gas is  =  %f C\n',t2);

+

+//  End