Question

A 10.77 mol sample of xenon gas is maintained in a 0.7572 L container at 300.2 K. What is the pressure in atm calculated using the van der Waals' equation for Xegas under these conditions? For Xe, a = 4.194 L²atm/mol² and b = 5.105×10^-2 L/mol.

According to the ideal gas law, a 0.9160 mol sample of oxygen gas in a 1.668 L container at 268.4 K should exert a pressure of 12.10 atm. By what percent does the pressure calculated using the van der Waals' equation differ from the ideal pressure? For O₂ gas, a = 1.360 L²atm/mol² and b = 3.183×10^-2 L/mol.

According to the ideal gas law, a 9.698 mol sample of krypton gas in a 0.8235 L container at 495.7 K should exert a pressure of 479.0 atm. By what percent does the pressure calculated using the van der Waals' equation differ from the ideal pressure? For Kr gas, a = 2.318 L²atm/mol² and b = 3.978×10^-2 L/mol.

Answer 1

The Van der Waals equation is a description of real gases, it includes all those interactions which we previously ignore in the ideal gas law.

It considers the repulsion and collision, between molecules of gases. They are no longer ignored and they also are not considered a"point" particle.

The idel gas law:

PV = nRT

P(V/n) = RT ; let V/n = v; molar volume

P*v = RT

now, the van der Waals equation corrects pressure and volume

(P+ a/v^2) * (v - b) = RT

substitute for Xe

v = V/mol = 0.7572/10.77 = 0.0703

(P+ a/v^2) * (v - b) = RT

(P + 4.194/0.0703) * (0.0703 - (5.105*10^-2)) = 0.082*300.2

(P + 4.194/0.0703) =   0.082*300.2 /  (0.0703 - (5.105*10^-2))

P = 1278.77 - 4.194/0.0703

P = 1219.11 atm