Question

15.0 moles of gas are in a 8.00 L tank at 22.3 ∘C∘C . Calculate the difference in pressure between methane and an ideal gas under these conditions. The van der Waals constants for methane are a=2.300L2⋅atm/mol2 a=2.300L2⋅atm/mol2 and b=0.0430L/molb=0.0430L/mol.

Answer 1

Answer:

$\Delta P=4.10atm$

Explanation:

Hello!

In this case, since the ideal gas equation is used under the assumption of no interaction between molecules and perfectly sphere-shaped molecules but the van der Waals equation actually includes those effects, we can compute each pressure as shown below, considering the temperature in kelvins (22.3+273.15=295.45K):

$P^{ideal}=\frac{nRT}{V}=\frac{15.0mol*0.08206\frac{atm*L}{mol*K}*295.45K}{8.00L}=45.5atm$

Next, since the VdW equation requires the molar volume, we proceed as shown below:

$v=\frac{8.00L}{15.0mol}=0.533\frac{L}{mol}$

Now, we use its definition:

$P^{VdW}=\frac{RT}{v-b} -\frac{a}{v^2}$

Thus, by plugging in we obtain:

$P^{VdW}=\frac{0.08206\frac{atm*L}{mol*K}*295.45K}{0.533mol/L-0.0430L/mol} -\frac{2.300L^2*atm/mol^2}{(0.533L/mol)^2}\\\\P^{VdW}=49.44atm-8.09atm\\\\P^{VdW}=41.4atm$

Thus, the pressure difference is:

$\Delta P=45.5atm-41.4atm\\\\\Delta P=4.10atm$

Best regards!