15.0 moles of gas are in a 4.00 l tank at 23.9 ∘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 and b=0.0430 l/mol.

14.8 atm  

Step-by-step explanation:

(a) Ideal Gas Law

The equation for the Ideal Gas Law is

pV = nRT     Divide each side by V

 p = (nRT)/V

Data:

n = 15.0 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 23.9 °C

V = 4.00 L

Calculations:

(i) Convert temperature to Kelvins

T = (23.9 + 273.15) K

  = 297.05 K

(ii) Calculate the pressure

p = (15.0 × 0.082 06 × 297.05)/4.00

  = 91.4 atm

The pressure predicted by the Ideal Gas Law is 91.4 atm.

(b) van der Waals gas

The van der Waals equation is

[p + (n²a)/V²)](V - nb) = nRT                   Divide each side by (V - nb)

             p + (n²a)/V² = (nRT)/(V - nb)     Subtract (n²a)/V² from each side

                                         p = (nRT)/(V - nb) - (n²a)/V²

(i) Data:

a=2.300 L²⋅atm·mol⁻²

b=0.0430 L·mol⁻¹

(ii) Calculation:

p = (15.0 × 0.082 06 × 297.05)/(4.00 – 15.0 × 0.0430)

      - (15.0² × 2.300)/4.00²

  = 365.6/(4.00 - 0.645) - 517.5/16

  = 365.6/3.355 - 32.34

  = 109.0 - 32.34

  = 76.6 atm

The pressure predicted by the van der Waals equation is 76.6 atm.

(iii) The pressure difference

p(Ideal) - p(vdW) = 91.4 - 76.6

                           = 14.8 atm

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