Question

(a) In an industrial process, nitrogen is heated to 500 K at a constant volume of 1.000 m³. The gas enters the container at 300 K and 100 atm. The mass of the gas is 92.4 kg. Use the van der Waals equation to determine the approximate pressure of the gas at its working temperature of 500 K. For nitrogen, a = 1.352 dm⁶ atm mol⁻², b = 0.0387 dm³ mol⁻¹. (b) Cylinders of compressed gas are typically filled to a pressure of 200 bar. For oxygen, what would be the molar volume at this pressure and 25 ∘C based on (i) the perfect gas equation, (ii) the van der Waals equation? For oxygen, a = 1.364 dm⁶ atm mol⁻², b=3.19×10⁻²dm³mol⁻¹.

Answer 1

Answer: a= 147.46atm

bi = 0.124Lmol1-1, bii =0.122Lmol^-1

Explanation:

Given

m, mass of the gas = 94.2kg = 94200g

T, Temperature = 500k

Volume of gas = 1m^3 = 1000L

Constant a for nitrogen = 1.352dm^6atm*mol^-2

= 1.352L^2atm*mol^-2

Constant b for nitrogen= 0.0387dm^3atm*mol^-1

= 0.0387Lmol^-1

Molar mass of nitrogen, M, = 28g

Number of moles, n, of nitrogen= mass/ molar mass

= 94200÷ 28

= 3364.28 moles

The Vanderwall equation, making pressure the subject of the formula

p= [(n*R*T/V - n*b)] - [(a - n^2)/V^2]

p = [(3364.28*0.0821*500)÷(1000-3364.28*0.0387)]-

[(1.352 -3364.28^2)÷ 1000^2]

Therefore p= 147.46 atm

Using ideal gas equation

P*V = n*R*T

Molar volume Vm = V/n = R*T/P

Given pressure 200 bar = 197.385atm

Temperature T 25°c = 298.15K

Vm = (0.0821 x 298.15)/ 197.385

Molar volume= 0.124Lmol^-1

If not an ideal gas,

p= [(R*T)/(Vm - b)] - [a/V^2m]

197.385= [(0.0821x298.15)/(Vm-0.0319)] - [1.364/V^2m]

Vm = 0.122Lmol^-1