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QUESTION 5 2 P Calculate the pressure of a 0.00223 mol of CCl4 vapor that occupies 38.5 L at 19.6 °C if the vapor is treated

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Answer #1

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

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