**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