The van der Waals constants for NO2 are a 528 L2 atmmol2 a

The van der Waals constants for NO2 are a = 5.28 L2 . atm/mol2 and b = 004424 L/mol, what is the pressure of 67.55 g of NO, in a 14.00 L container at 41.00°C when it is treated as (a) ideal and (b) real? 4.

Solution

4)
a)
Molar mass of NO2,
MM = 1*MM(N) + 2*MM(O)
= 1*14.01 + 2*16.0
= 46.01 g/mol


mass(NO2)= 67.55 g

use:
number of mol of NO2,
n = mass of NO2/molar mass of NO2
=(67.55 g)/(46.01 g/mol)
= 1.468 mol

Given:
V = 14.0 L
n = 1.4682 mol
T = 41.0 oC
= (41.0+273) K
= 314 K

use:
P * V = n*R*T
P * 14 L = 1.4682 mol* 0.08206 atm.L/mol.K * 314 K
P = 2.70 atm
Answer: 2.70 atm

b)
Given:
V = 14.0 L
n = 1.4682 mol
R = 0.0821 atm.L/mol.K
T = 314.0 K
a = 5.28 atm.L^2/mol^2
b = 0.04424 L/mol

use:
(P+an^2/V^2)*(V-nb) = n*R*T
(P + 5.28*1.4682^2/14.0^2)*(14.0-1.4682*0.04424) = 1.4682*0.0821*314.0
(P + 0.0581)*(13.935) = 37.8493
P + 0.0581 = 2.7161
P = 2.6581 atm

Answer: 2.66 atm