Given:
V = 38.5 L
n = 0.00223 mol
R = 0.08206 atm.L/mol.K
T = 19.6 oC
=(19.6 + 273.1) K
= 292.7 K
a = 20.4 atm.L^2/mol^2
b = 0.1383 L/mol
use:
(P+an^2/V^2)*(V-nb) = n*R*T
(P + 20.4*0.00223^2/38.5^2)*(38.5-0.00223*0.1383) = 0.00223*0.08206*292.7
(P + 0)*(38.4997) = 0.0536
P + 0 = 1.392*10^-3
P =1.392*10^-3 atm
= 1.392*10^-3 * 760 torr
= 1.06 torr
Answer: 1.06 torr
QUESTION 5 2 P Calculate the pressure of a 0.00223 mol of CCl4 vapor that occupies...
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