 # 7.1 The compressibility factor chart provides a quick way to assess...

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7.1 The compressibility factor chart provides a quick way to assess when the ideal gas law is valid. For the following fluids. what is the minimum temperature in K where the fluid has a gas phase compressibility factor greater than 0.95 at 30 bar? (a) Nitrogen (b) Carbon dioxide (c) Ethanol 7.9 Evaluate After ap for the Redlich-Kwong equation of state P = RT a * where a and b are temperature-independent parameters. V-b 1/2 T V(V + b) 8.14 A 1 m 3 isolated chamber with rigid walls is divided into two compartments of equal vol- ume. The partition permits transfer of heat. One side contains a nonideal gas at 5 MPa and 300 K and the other side contains a perfect vacuum. The partition is ruptured, and after suf- ficient time for the system to reach equilibrium, the temperature and pressure are uniform throughout the system. The objective of the problem statements below is to find the final T and P. The gas follows the equation of state RT PV = 11(6-9)12 where b = 20 cm³/ mole; a = 40,000 em K/mole: and Cp = 41.84 + 0.084T(K) J/mol-K. (a) Set up and simplify the energy balance and entropy balance for this problem. (b) Derive formulas for the departure functions required to solve the problem. (c) Determine the final P and T. 8.15 P-V-T behavior of a simple fluid is found to obey the equation of state given in problem 8.14. (a) Derive a formula for the enthalpy departure for the fluid. (b) Determine the enthalpy departure at 20 bar and 300 K. (c) What value does the entropy departure have at 20 bar and 300 K? 8.22 1 m of CO2 initially at 150°C and 50 bar is to be isothermally compressed in a frictionless piston/cylinder device to a final pressure of 300 bar. Calculate the volume of the com- pressed gas, AU. the work done to compress the gas. and the heat flow on compression assuming (a) CO2 is an ideal gas. (b) CO2 obeys the Peng-Robinson equation of state.

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% To find final temperature (T2) and pressure (P2) after wall rupture

clc;
clear all;

a = 4e-2;
b = 20e-6;
R = 8.314;
T1 = 300;
P1 = 5e6;
V1 = R*T1/P1+b-a/T1;
fprintf('V1 = %d m^3/kg \n',V1);
V2 = 2*V1;
fprintf('V2 = %d m^3/kg \n',V2);
T1 = 300;
fun = @(T2) 33.526*(T2-T1)+0.042*(T2^2-T1^2)-R*a*(1/(V2-b+a/T2)-1/(V1-b+a/T1));
Temp = fzero(fun,300);
T2 = Temp;
fprintf('The final...

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