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Gas Law Example Problems Mrs. Diksa/Miss Santelli.

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Presentation on theme: "Gas Law Example Problems Mrs. Diksa/Miss Santelli."— Presentation transcript:

1 Gas Law Example Problems Mrs. Diksa/Miss Santelli

2 The mean molar mass of the atmosphere at the surface of Titan, Saturn’s largest moon, is 28.6 g/mol. The surface temperature is 95K, and the pressure is 1.6 atm. Assuming ideal behavior, calculate the density of Titan’s atmosphere.

3 Dalton’s Law of Partial Pressures What is the total pressure exerted by a mixture of 2.00 g of hydrogen and 8.00 g of nitrogen at 273 K in a 10.0-L vessel?

4 Collecting Over Water Ammonium nitrate, NH 4 NO 2, decomposes upon heating to form N 2 gas: NH 4 NO 2 → N 2 + 2H 2 O When a sample of ammonium nitrate is decomposed in a test tube, 511 mL of nitrogen is collected over water at 26 °C and 745 torr total pressure. How many grams of ammonium nitrate decomposed?

5 Effusion and Diffusion  Effusion – the escape of a gas through a tiny hole into an evacuated space.  Diffusion – The spread of one substance throughout a space or throughout a second substance.

6 Notice, did both balloons effuse at the same rate? Speculate why.

7 Why do diffusion rates differ?  While KMT states that the average kinetic energy of the molecules of all gaseous samples is determined by their average temperatures, all gaseous molecules are not the same size.  Recall, KE = ½ mv 2  Because gaseous samples do not all have the same mass, the cannot all have the same root mean square speed, μ = √(3RT/M)  Notice that gases with smaller molar masses will have a greater root mean square speed.

8 Example What is the rms speed of an atom of He at 25°C?

9 Graham’s Law of Effusion  Discovered that effusion rate of a gas is inversely proportional to the square root of its molar mass:  r 1 /r 2 = √(M 2 /M 1 ) = √[(3RT/M 1 )/(3RT/M 2 )]

10 Example  An unknown gas composed of homonuclear diatomic molecules effuses at a rate that is only 0.355 times that of O2 at the same temperature. What is the identity of the unknown?

11 Real Gases  Unlike Ideal Gases which behave under the assumptions of KMT, the molecules of Real Gases Have finite volumes and They do attract one another

12 Real Gases  As the pressure of a gaseous sample increases, the volume of the gaseous particles becomes less negligible and, because the molecules are crowded together, intermolecular attraction also increases  As the temperature of a gaseous sample decreases, the average KE of the particles decreases, decreasing the ability of the molecules to overcome the attraction of their neighbors.  Of these two KMT assumptions, intermolecular attraction plays a much bigger role in deviation from ideal behavior.

13 Van der Waals Equation  Johannes van der Waals recognized that for real gases (those under conditions that the ideal gas equation will not all for reasonable predictions), a correction factor would be needed to address the finite volume of the gaseous particles and the intermolecular attractions within the sample.

14 Van der Waals Equation [P + (n 2 a)/V 2 ](V – nb) = nRT Where nb corrects for the volume of the particles and (n 2 a)/V 2 corrects for molecular attractions. The values of both a and b are van der Waals constants specific to the identity of the gas (see table 10.3 of text)

15 Example Consider a sample of 1.000 mol of CO 2 confined to a volume of 3.000L at 0.0°C. Calculate the pressure of the gas using (a) the ideal gas equation and (b) the van der Waals equation.

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