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The Ideal Gas Law

The Ideal Gas Law. What is an ideal gas?. They do not condense to liquids at low temperatures They do not have forces of attraction or repulsion between the particles They are composed of particles that have no volume. So, there is no such thing as an ideal gas

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The Ideal Gas Law

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  1. The Ideal Gas Law

  2. What is an ideal gas? • They do not condense to liquids at low temperatures • They do not have forces of attraction or repulsion between the particles • They are composed of particles that have no volume. • So, there is no such thing as an ideal gas • That is why they are ideal, but they do give us fairly close estimates, especially when we are close to STP.

  3. The Ideal Gas Law • It combines these four basic gas laws:

  4. So, the final formula looks like this • PV = nRT, where • P = Pressure in kPa or atm • V = Volume in L • n = moles in mol • T = temperature in Kelvin (C+273=K) • R = constant • Two values of R = 8.314 (L.kPa)/(mol.K) and R = 0.0821 (L.atm)/(mol.K)

  5. As mentioned before, real gases will deviate from the ideal conditions. But they do give us a good estimation

  6. So, ideal gas problems require algebra to solve the equation • How many moles of air molecules are contained in a 2.00 L flask at 98.8 kPa and 25.0C? • Calculate the pressure exerted by 43 mol of nitrogen in a 65 L cylinder at 5C. • What will be the volume of 111 mol of nitrogen in the stratosphere, where the temperature is -57C and the pressure is 7.30 kPa?

  7. Answers • 0.0798 moles • 15.1 atm or 1529 kPa • 27306 L

  8. Ideal Gas Laws and Stoichiometry • What is the key to Stoichiometry? We need to get to what? • Moles • Remember this is n in the ideal gas law formula • So, you will either have to use the given information to calculate n to plug into the ideal gas formula to solve for a certain condition (P, V, or T) or you will need to calculate n from the conditions in order to change it into another reactant or product.

  9. Examples • What will be the volume, at 115 kPa and 355 K, of the nitrogen from the decomposition of 35.8 g of sodium azide, NaN3? 2NaN3 (s)  2Na (s) + 3N2 (g) • How many grams of sodium are needed to produce 2.24 L of hydrogen, collected at 23C and 92.5 kPa? 2 Na (s) + 2H2O (l)  2NaOH (aq) + H2 (g)

  10. Answers 35.8/65 = 0.551 mol x 3/2 = 0.826 V = (0.826)(8.314)(355)/(115) = 21.2 L (92.5)(2.24)/(8.314)(296) = 0.0842 mol 0.0842 x 2/1 = 0.1684 mol x 23 g/1mol = 3.87 g

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