Using the van der Waals equation, determine the pressure of 500.0 g of SO2(g) in a 6.30-L vessel at 633 K. For SO2(g), a = 6.865 L2 • atm/mol2 and b = 0.05679 L/mol. (R = 0.0821 L • atm/(K • mol))

The pressure is 58.75 atm.

Explanation:

From Vanderwaal's equation,

P = nRT/(V-nb) - n^2a/V^2

n is the number of moles of SO2 = mass/MW = 500/64 = 7.81 mol

R is gas constant = 0.0821 L.atm/mol.K

T is temperature of the vessel = 633 K

V is volume of the vessel = 6.3 L

a & b are Vanderwaal's constant = 6.865 L^2.atm/mol^2 and 0.0567 L/mol respectively.

P = (7.81×0.0821×633)/(6.3 - 7.81×0.05679) - (7.81^2 × 6.865)/6.3^2 = 69.30 - 10.55 = 58.75 atm

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3