Use both the ideal gas law and the Van der Waals’ equation to calculate the temperature of water vapor (steam) for the following conditions: P = 220 bar, n = 2 mol, V = 1 L, a = 5.536 L2 bar/mol2, b = 0.03049 L/mol and R = 0.08314472 L bar/K mol.
Use both the ideal gas law and the Van der Waals’ equation to calculate the temperature of water vapor (steam) for the following conditions: P = 220 bar, n = 2 mol, V = 1 L, a = 5.536 L2 bar/mol2, b = 0.03049 L/mol and R = 0.08314472 L bar/K mol.
Matlab:
p = 220
n = 2
v = 1
a = 5.536
b = 0.03049
R = 0.08314472
T1=(p*v)/(n*R)
%T1 is the temperature measure using Ideal gas law
T2= ((p+((n^2*a)/v^2))*(v-(n*b)))/(n*R)
%T2 is the temperature using Van der waals correction
p =
220
n =
2
v =
1
a =
5.5360
b =
0.0305
R =
0.0831
T1 =
1.3230e+003
T2 =
1.3674e+003
Part b)
Find the value of temperature (T), for 11 values of pressure from 0 bar to 400 bar for volume of 1 L using the ideal gas law and then using the Van der Waals’ modification. Use the values of a, b, n, R, and V given in part (b) above. Present your results as a table of “Pressure, Temp_ideal, Temp_real
Matlab:
p = linspace(0,400,11)
T1=(v.*p)/(n*R)
%T1 is the temperature measure using Ideal gas law
T2=((((n^2*a)/v^2)+p)*(v-(n*b)))/(n*R)
%T2 is the temperature using Van der waals correction
p =
Columns 1 through 7
0 40 80 120 160 200 240
Columns 8 through 11
280 320 360 400
T1 =
1.0e+003 *
Columns 1 through 4
0 0.2405 0.4811 0.7216
Columns 5 through 8
0.9622 1.2027 1.4433 1.6838
Columns 9 through 11
1.9244 2.1649 2.4054
T2 =
1.0e+003 *
Columns 1 through 4
0.1250 0.3509 0.5768 0.8027
Columns 5 through 8
1.0285 1.2544 1.4803 1.7062
Columns 9 through 11
1.9321 2.1579 2.3838
