Question

Two moles of a van der Waals fluid are maintained at a temperature T = 0.95T_C in a volume of 200cm³. At this temperature, the reduced molar volume of the liquid phase ν_l ~ 0.68 and that of the gas phase ν_g ~ 1.7. Assuming the substance to be oxygen (i.e., using the van der Waals constants for oxygen a = 0.138 Pa-m⁶ and b = 32.6x10^-6m³), find the mole number and volume of each phase.

Answer 1

IN HE INIIAL SAGES,

HE FIUID ACS AS AN IDEAL GAS.

WHERE n= 2

V = 200 ml

= 0.95⁰ C = 273.95⁰ K

R = 8.314

P V =n R

P = n R / V

= 2 * 8.314 * 273.95 / 200

= 22.77

APPLYING HE VALUE OF 'P' IN VANDERWAALS EQUAIONS OF BOH LIQUID PHASE AND GASEOUS PHASES

GIVEN HA

V_l = 0.68

V_g = 1.7

a = 0.138

b = 32.6 * 10^-6 = 0.0000326

LIQUID PHASE ---

( P-a ) * ( V_l -b ) = n* R *

PRESSURE IS = ( P-a ) = 22.77 - 0.138 = 22.632

VOLUME IS = ( V_l -b ) = 0.68 - 0.000326 = 0.679674

n = ( P-a ) * ( V_l -b ) / R *

= 22.632 * 0.679674 / 8.314 * 273.95

= 0.006753

GASEOUS PHASE ---

( P-a ) * ( V_g -b ) = n* R *

PRESSURE IS = ( P-a ) = 22.77 - 0.138 = 22.632

VOLUME IS = (  V_gl -b ) = 1.7 - 0.000326 = 1.699674

n = ( P-a ) * (  V_gl -b ) / R *

= 22.632 * 1.699674 / 8.314 * 273.95

= 0.16889