Calculate the pressure exerted by 2 mol of water vapor (steam) in a 1 L containe

Question

Calculate the pressure exerted by 2 mol of water vapor (steam) in a 1 L container at 2000 Fusing the Van der Waals' equation and the ideal Gas Law equation. What is the difference between the Waals and ideal gases? Given the above values, write a MATLAB program that computes and displays the results of Van der Waals' equation, the idea Gas Law equation, and the difference between them. The Waals' gas law describes the relationships between pressure Equations: (p), absolute temperature (in K)m, volume M and the number of moles of gas (m). The See above additional symbol, R represents the ideal gas constant. Equations: See a bove C = (F - 32)/1.8 K = C = 273.15 Output: The pressure (atm) using van der Waals equation: 219.8442 The pressure (atm) using Ideal Gas Law: 227.2317 The difference between Waals and ideal conditions 7.3876

Explanation / Answer

VERSION 1

-----------------------Program----------------------------------------------------------------

clc;
clear all;
%Constant
format long;
R=0.08314472; % Ideal Gas Constant L*atm/K mol
% Inputs
% temperature=2000;
% moles=2;
% v=1;
% a=5.536;
% b=0.03049;
temperature=input('Enter Temperature in F : ');
moles =input('Enter Moles : '); % mol
v = input('Enter Volume : '); %Litre
a=input('Enter a : '); % L^2*atm/mol^2
b=input('Enter b : ');% L/mol
%Formulas
C=(temperature-32)/1.8; % Input Temperature is in degree F then conversion into centigrade
K=C + 273.15 ; % conversion of Centigrade into Kelvin
%Van der Waals Equation
Waals_P=((moles*R*K)/(v-(moles*b)))-(a*moles*2/v*v); % Waals equation of Pressure Calculations
Ideal_P=moles*R*K/v; % Ideal Gas Euqation for Pressure Calculations
fprintf('The pressure (atm) using Van der Waals equation: %.4f ',Waals_P);
fprintf('The pressure (atm) using Ideal Gas Law: %.4f ',Ideal_P);
fprintf('The difference between Waals and Ideal conditions: %.4f ',abs(Waals_P-Ideal_P));

-----------------------------Output------------------------------------------------------------

Enter Temperature in F : 2000
Enter Moles : 2
Enter Volume : 1
Enter a : 5.536
Enter b : .03049
The pressure (atm) using Van der Waals equation:
219.8442

The pressure (atm) using Ideal Gas Law:
227.2317

The difference between Waals and Ideal conditions:
7.3876
>>

VERSION 2 (No Inputs)

clc;
clear all;
%Constant
format long;
R=0.08314472; % Ideal Gas Constant L*atm/K mol
% Inputs
temperature=2000;
moles=2;
v=1;
a=5.536;
b=0.03049;
% temperature=input('Enter Temperature in F : ');
% moles =input('Enter Moles : '); % mol
% v = input('Enter Volume : '); %Litre
% a=input('Enter a : '); % L^2*atm/mol^2
% b=input('Enter b : ');% L/mol
%Formulas
C=(temperature-32)/1.8; % Input Temperature is in degree F then conversion into centigrade
K=C + 273.15 ; % conversion of Centigrade into Kelvin
%Van der Waals Equation
Waals_P=((moles*R*K)/(v-(moles*b)))-(a*moles*2/v*v); % Waals equation of Pressure Calculations
Ideal_P=moles*R*K/v; % Ideal Gas Euqation for Pressure Calculations
fprintf('The pressure (atm) using Van der Waals equation: %.4f ',Waals_P);
fprintf('The pressure (atm) using Ideal Gas Law: %.4f ',Ideal_P);
fprintf('The difference between Waals and Ideal conditions: %.4f ',abs(Waals_P-Ideal_P));

-----------------Output--------------------------------------------------------------

The pressure (atm) using Van der Waals equation:
219.8442

The pressure (atm) using Ideal Gas Law:
227.2317

The difference between Waals and Ideal conditions:
7.3876
>>

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