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

Answer:

% differ  1.72%

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

$(P+ \frac{n^2 *a}{v^2}) (V - nb) = nRT$

$(P+ (\frac{1.074^2 *1.360}{1.746^2})) (1.746 - 1.074*3.183*10^{-2}) = 1.074*0.0821*267.6$

(P+0.514)(1.711) = 23.59

P_v = 13.276 atm

% differ $= \frac{ P_I - P_v}{P_I} *100$

             $= \frac{13.51 - 13.27}{13.51} *100$

             = 1.72%