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Ideal Gas Law. Ideal vs. Real Gases Ideal gases obey every law perfectly – They have fully elastic collisions – They lose no energy as they move and collide.

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Presentation on theme: "Ideal Gas Law. Ideal vs. Real Gases Ideal gases obey every law perfectly – They have fully elastic collisions – They lose no energy as they move and collide."— Presentation transcript:

1 Ideal Gas Law

2 Ideal vs. Real Gases Ideal gases obey every law perfectly – They have fully elastic collisions – They lose no energy as they move and collide – Molecules do not interact Real gases do not obey perfectly – But they are close enough that we can assume they show ideal behavior

3 Ideal Gas Law We have seen the combined gas law, but we have left out (until now) one variable The mole Avogadros Law – V α n Adding this to the combined gas Law, we have the Ideal Gas Law

4 Ideal Gas Law PV = nRT P is pressure V is volume n is number of moles T is temperature (in Kelvin) R is our Ideal Gas Law Constant – 8.314 (L kPa)/(mol K) –.0821 (L atm)/(mol K)

5 Ideal Gas Law How many moles of air molecules are contained in a 2.00L flask at 98.8kPa and 25.0 o C? – First convert to Kelvin 25.0 o C + 273 = 298 K – PV=nRT – (98.8kPa)(2.00L)=n(8.314 LkPa/molK)(298K) – n=0.0798 mol = 7.98x10 -2 mol

6 Ideal Gas Law and Stoichiometry Just combine the stoichiometry that you know with the ideal gas law How many liters of hydrogen gas will be produced at 280.0K and 96.0kPa if 40.0g of sodium react with excess water according to the following equation – 2Na(s) + 2H 2 O(l) 2NaOH(aq) + H 2 (g) – Convert from grams of sodium to moles of H 2 – (40.0g Na)(1 mol Na/22.99g Na)(1 mol H 2 /2 mol Na) – 0.870 mol H 2 – Use the ideal gas law – PV=nRT – (96.0kPa) V =(0.870 mol)(8.314LkPa/molK)(280.0K) – 21.1 L H 2


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