Question

This problem involves solving the ideal gas law. The first three questions are tied to Problem 1, so refer back to the problem statement for conditions. Using the ideal gas equation, calculate the pressure of oxygen gas in a cylinder with a volume of 25.00 L. The oxygen masses 4.362 kg and room temperature is at 22.5oC. How many moles of oxygen are there?

2.Using the same values of volume, mass, and room temperature from Problem 1, calculate the pressure of oxygen in the cylinder using the van der Waals equation of state

van der Waals equation:(P+a[(n)/(v)]^(2))*(V-bn)=nrt

Answer 1

Given that

volume of 25.00 L.

The oxygen masses 4.362 kg

room temperature is at 22.5oC or 295.5 K

mole of O2 = amount in g / molar mass

= 4.362 *1000 / 32.0g / mole

= 136.3 mole oxygen

Ideal gas law,

PV= n RT

P = n RT / V

= 136.3 * 0.08206 * 295.5 / 25

= 132.22 atm

V= 1.628 L

R = 0.0821L atm/Kmol
a = 1.382 L^2 atm/mol^2
b = 0.03186 L/mol

The Formula for P derived from Van der Waals Equation:

P = [nRT/(V - nb)] - n²a/V²

P = [nRT/(V - nb)] - n2a/V2

Substituting values:
P = {[(136.3 *0.0821*295.5)] / [25 -( 136.3 *0.03186 )]} - [(136.3^2* 1.382) / (25 ^2)]

P = (3305.10/ 20.666) – 41.079

= 159.93– 41.079
P = 118.87 atm