Question

1. The Redlich-Kwong equation of state is given by P=_RT___ A _ _ V-RI2,, - 0.0866 - where 4-0.42748RT - B - P (The R-K constants can be calculated from the critical temperature and pressure of the gas.) This EOS was introduced in 1949 and is adequate for calculations of gas phase properties when P, </2 T, a) Derive an expression for the work associated with an isothermal reversible volume change of a R-K gas between two volumes V, and V2. You may find it helpful to know that Jxla+bx) dx=a" In a+bx x b) Pick your favorite gas molecule and calculate the values of A and B. (Critical constants can be found online.) Use the relationship you derived in part a) to calculate the work in expanding the gas from 6 V. to 7 V. at a temperature that is twice the critical temperature. (Please provide the critical constants for your gas so the grader can check your work.) c) We showed that y =0 for an ideal gas. Evaluate this partial differential for a R-K laU OVT gas. d) Knowing that A and B are positive constants for a given R-K gas, does the internal energy increase or decrease for an isothermal expansion? What does this tell you about the intermolecular interactions for the R-K gas? e) Calculate the change in internal energy for the same process for which you calculated the work. f) Calculate the heat for this same process.

Answer 1

1a) Derivation for work is as follows:

Peld to in CR7d 10 t Adr Vm (Vm +B) IT o E er, (V2 +B) " BTh LV₂ (V,+B) = -RT ln

b)

CO₂ gas:

T_c = 304.2,

P_c = 79.2 atm,

V_c = 94 cm³/mole = 0.094 liter/mole

R = 0.0821 Liter atm K^-1 Mole^-1

A = 0.42748*(0.0821)²(304.2)^2.5/79.2

A = 63.722939

B = 0.08664*0.0821*304.2/79.2

B = 0.029682

W = -0.0821*604.2*ln((0.658-0.029682)/(0.564-0.029682))

+ 63.792939*ln(0.564*(0.658+0.029682)/(0.658*(0.564+0.029682))/(2*0.0292682*(608.4)^0.5)

W = - 8.719 J

c) The derivation is as follows:

2m (+B) + 3/2 -B) RT B + 2 vor (Vante) The T (2) v ces + 20 C +8) The 2 mm ( 7/29) -P- But -6) Vonk+8) T A = 3A 7 2 m (m +B) T12

d)

239AVV5) L Vz lv HB) 2BT

$\Delta$U = (-3*63792939)ln(0.564*(0.658+0.029682)/(0.658*(0.564+0.029682))/(2*0.0292682*(608.4)^0.5)

$\Delta$U = 0.9368 J

Internal energy increases in isothermal expansion. This indicates that there are attractive interactions between molecules in the gas.

f)

q = $\Delta$U - w

= 0.9368 + 8.719

q = 9.656 J