Question

The amount 2.00 mol of a van der waals gas with a=0.245 m^6 Pa mol^-2 occupies a volume of 0.840 L if the gas is at a temperature of 85k and at a pressure of 2850 kPa. From this information, calculate the van der waals constant b and pressure p of this sample when it occupies a volume of 1680 dm^3 at T=255k.

Answer 1

use:
(P + an^2/V^2) (V-nb) = n*R*T
R = 8.314 m^3 pa K-1 mol-1

1st find b:
P = 2850 kpa = 2850000 pa
V = 0.840 L = 8.40*10^-4 m^3
(P + an^2/V^2) (V-nb) = n*R*T
(2850000 + 0.245*2^2/ (8.40*10^-4)^2) * (8.40*10^-4 -2*b) = 2*8.314*85
4238889 * (8.40*10^-4 -2*b) = 1413.38
(8.40*10^-4 -2*b) = 3.33*10^-4
b = 2.5*10^-4 mol-1
Answer: b = 2.5*10^-4 mol-1

-----------------------
P = ?
V = 1680 dm^3 = 1.68 m^3
(P + an^2/V^2) (V-nb) = n*R*T
(P + 0.245*2^2/ (1.68)^2) * (1.68 -2*2.5*10^-4 ) = 2*8.314*255
(P+0.3472) * (1.6795) = 4240.14
P = 2524.3 Pa
Answer: P = 2524.3 Pa