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##### Gas Law Example Problems

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**Gas Law Example Problems**Mrs. Diksa/Miss Santelli**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.**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?**Collecting Over Water**Ammonium nitrate, NH4NO2, decomposes upon heating to form N2 gas: NH4NO2→ N2 + 2H2O 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?**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.**Notice, did both balloons effuse at the same rate?**Speculate why.**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 mv2 • 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.**Example**What is the rms speed of an atom of He at 25°C?**Graham’s Law of Effusion**• Discovered that effusion rate of a gas is inversely proportional to the square root of its molar mass: • r1/r2 = √(M2/M1) = √[(3RT/M1)/(3RT/M2)]**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?**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**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.**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.**Van der Waals Equation**[P + (n2a)/V2](V – nb) = nRT Where nb corrects for the volume of the particles and (n2a)/V2 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)**Example**Consider a sample of 1.000 mol of CO2 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.