Question

According to the ideal gas law, a 1.074 mol sample of oxygen gas in a 1.746 L container at 267.6 K should exert a pressure of 13.51 atm. What is the percent difference between the pressure calculated using the van der Waals' equation and the ideal pressure? For O2 gas, a = 1.360 L2atm/mol2 and b = 3.183×10-2 L/mol.

Answer 1

% differ 1.72%

Step-by-step explanation:

given data:

P_ideal = 13.51 atm

n = 1.074 mol

V = 1.746 L

T = 267.6 K

According to ideal gas law we have

`(P+ (n^2 *a)/(v^2)) (V - nb) = nRT`

`(P+ ((1.074^2 *1.360)/(1.746^2))) (1.746 - 1.074*3.183*10^(-2)) = 1.074*0.0821*267.6`

(P+0.514)(1.711) = 23.59

P_v = 13.276 atm

% differ
`= ( P_I - P_v)/(P_I) *100`

`= (13.51 - 13.27)/(13.51) *100`

= 1.72%