Answered Essay: Ideal gas tow: PV = nRT P = pressure (kPa) V = volume (m^3) n = number of kmoles of gas in sample R = ideal gas constant (8.314 kP

Matllab question:

ideal gas law: PV nRT P pressure (kPa) V-volume (m) n=number of kmoles of gas in sample R ideal gas constant (8.314 kPam/kmol k) T- temperature (K) Find the mass of air in the male patient. Find the mass of air in the female patient. Variables Male lung capacity 6000 ml Female lung capacity 4200 ml Temperature 300K Pressure-30 cmH,o Average air molecular mass = 29 kg/kmol Conversion to kPo: 1 cmH,0·0.098 kPa Conversion to ㎡, 1 ml+104 ㎡ Relationship between mass and moles: n m- mass (kg) MW molar mass in kg/mol m/MW Problem 2 (25 points) In class, we found the mass of air present in the lung of a male patient and female patient. Assuming that the temperature of the air is 300 K and pressure 30 cmH20, calculate the mass of air (in grams) for 10 equally spaced total lung capacity values, with the minimum value of 4200 mL and the maximum value of 6000 mL Recall from class that the average molar mass of air is 29 kg/kmol.

I need a solution for matlab (code) not calculation, Thank you

Ideal gas tow: PV = nRT P = pressure (kPa) V = volume (m^3) n = number of kmoles of gas in sample R = ideal gas constant (8.314 kPa middot m^3/kmol middot K) T = temperature (K) Conversion to kPa: 1 cm H_2O = 0.098 kPa Conversion to m^3: 1 mL = 10^-6 m^3 Relationship between moss and moles: n = m/MW m = mass (kg) MW = molar mass in kg/mol Find the mass of air in the male patient. Find the mass of air in the female patient. Variables Male lung capacity = 6000 mL Female lung capacity = 4300 mL Temperature = 300 K Pressure = 30 cmH_2O Average air molecular mass = 29 kg/kmol In class, we found the mass of air present in the lung of a male patient and female patient. Assuming that the temperature of the air is 300 K and pressure 30 cmH20, calculate the mass of air (in grams) for 10 equally spaced total lung capacity values, with the minimum value of 4200 mL and the maximum value of 6000 mL. Recall from class that the average molar mass of air is 29 kg/kmol.

Expert Answer

 

Matlab code:

P=30*0.098; %1cm H2O=0.098 kPa here 30cm H2O so 30*0.098
male_lung_capacity=6000*10^-6; %1mL=10^-6 m^3
female_lung_capacity=4200*10^-6;
temp=300; %in kelvin
MW=29; %molar mass in kg/kmol
R=8.314; %ideal gas constant in Kpa.m^3/kmol.K
n1=(P*(male_lung_capacity))/(R*temp); % PV=nRT that gives n=PV/RT
n2=(P*(female_lung_capacity))/(R*temp); % PV=nRT that gives n=PV/RT
m1=n1*MW; %n=m/MW relationship between mass and moles
m2=n2*MW; %n=m/MW relationship between mass and moles
disp(‘mass of air(in grams) in male patient is: ‘);
disp(m1*10^3); %m1 is in kg for conversion 1kg=1000 gms multiplied with 10^3
disp(‘mass of air(in grams) in female patient is: ‘);
disp(m2*10^3); %m2 is in kg for conversion 1kg=1000 gms multiplied with 10^3

Output:

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