Question

1. A 9.352 mol sample of nitrogen gas is maintained in a 0.7959 L container at 304.3 K. What is the pressure in atm calculated using the van der Waals' equation for N2 gas under these conditions? For N2, a = 1.390 L2atm/mol2 and b = 3.910×10-2 ___L/mol.

2. According to the ideal gas law, a 0.9738 mol sample of carbon dioxide gas in a 1.056 L container at 272.5 K should exert a pressure of 20.62 atm. What is the percent difference between the pressure calculated using the van der Waals' equation and the ideal pressure? For CO2 gas, a = 3.592 L2atm/mol2 and b = 4.267×10-2 L/mol. ___%

Answer 1

(1) vanderwall equation is -

$$(P+\frac{an^2}{V^2})(V-nb)=nRT$$

\\

$$(P+\frac{121.57}{0.6334})(0.7959-0.3656)=9.352\times 0.08205\times304.3$$

$$(P+191.93)(0.4303)=233.5\\P=350.715\ atm$$

(2)

$$P_{ideal}=20.62\ atm\\using\ vanderwall\ equation\\(P+\frac{an^2}{V^2})(V-nb)=nRT\\(P+\frac{3.406}{1.1151})(1.056-0.04155)=0.9738\times0.08205\times272.5$$

$$(P+3.0544)(1.01445)=21.773\\P=18.4085\ atm$$

Difference between actual pressue and ideal pressure =

$$20.62-18.4085=2.2115\ atm$$

percentage difference =

$$\frac{(2.2115\times100)}{20.62}=10.725\ \%$$