Science, Tech, Math › Science Ideal Gas vs Non-Ideal Gas Example Problem Van Der Waals Equation Example Problem Share Flipboard Email Print At low temperatures, real gases behave as ideal gases. Tetra Images - Jessica Peterson, Getty Images Science Chemistry Basics Chemical Laws Molecules Periodic Table Projects & Experiments Scientific Method Biochemistry Physical Chemistry Medical Chemistry Chemistry In Everyday Life Famous Chemists Activities for Kids Abbreviations & Acronyms Biology Physics Geology Astronomy Weather & Climate By Todd Helmenstine Todd Helmenstine is a science writer and illustrator who has taught physics and math at the college level. He holds bachelor's degrees in both physics and mathematics. our editorial process Todd Helmenstine Updated January 06, 2020 This example problem demonstrates how to calculate the pressure of a gas system using the ideal gas law and the van der Waal's equation. It also demonstrates the difference between an ideal gas and a non-ideal gas. Van der Waals Equation Problem Calculate the pressure exerted by 0.3000 mol of helium in a 0.2000 L container at -25 °C usinga. ideal gas lawb. van der Waals equationWhat is the difference between the non-ideal and ideal gases?Given:aHe = 0.0341 atm·L2/mol2bHe = 0.0237 L·mol How to Solve the Problem Part 1: Ideal Gas LawThe ideal gas law is expressed by the formula:PV = nRTwhereP = pressureV = volumen = number of moles of gasR = ideal gas constant = 0.08206 L·atm/mol·KT = absolute temperatureFind absolute temperatureT = °C + 273.15T = -25 + 273.15T = 248.15 KFind the pressurePV = nRTP = nRT/VP = (0.3000 mol)(0.08206 L·atm/mol·K)(248.15)/0.2000 LPideal = 30.55 atmPart 2: Van der Waals EquationVan der Waals equation is expressed by the formulaP + a(n/V)2 = nRT/(V-nb)whereP = pressureV = volumen = number of moles of gasa = attraction between individual gas particlesb = average volume of individual gas particlesR = ideal gas constant = 0.08206 L·atm/mol·KT = absolute temperatureSolve for pressureP = nRT/(V-nb) - a(n/V)2To make the math easier to follow, the equation will be broken into two parts whereP = X - YwhereX = nRT/(V-nb)Y = a(n/V)2X = P = nRT/(V-nb)X = (0.3000 mol)(0.08206 L·atm/mol·K)(248.15)/[0.2000 L - (0.3000 mol)(0.0237 L/mol)]X = 6.109 L·atm/(0.2000 L - .007 L)X = 6.109 L·atm/0.19 LX = 32.152 atmY = a(n/V)2Y = 0.0341 atm·L2/mol2 x [0.3000 mol/0.2000 L]2Y = 0.0341 atm·L2/mol2 x (1.5 mol/L)2Y = 0.0341 atm·L2/mol2 x 2.25 mol2/L2Y = 0.077 atmRecombine to find pressureP = X - YP = 32.152 atm - 0.077 atmPnon-ideal = 32.075 atmPart 3 - Find the difference between ideal and non-ideal conditionsPnon-ideal - Pideal = 32.152 atm - 30.55 atmPnon-ideal - Pideal = 1.602 atmAnswer:The pressure for the ideal gas is 30.55 atm and the pressure for van der Waals equation of the non-ideal gas was 32.152 atm. The non-ideal gas had a greater pressure by 1.602 atm. Ideal vs Non-Ideal Gases An ideal gas is one in which the molecules don't interact with each other and don't take up any space. In an ideal world, collisions between gas molecules are completely elastic. All gases in the real world have molecules with diameters and which interact with each other, so there's always a bit of error involved in using any form of the Ideal Gas Law and van der Waals equation. However, noble gases act much like ideal gases because they don't participate in chemical reactions with other gases. Helium, in particular, acts like an ideal gas because each atom is so tiny. Other gases behave much like ideal gases when they are at low pressures and temperatures. Low pressure means few interactions between gas molecules occur. Low temperature means the gas molecules have less kinetic energy, so they don't move around as much to interact with each other or their container.