Presentation on theme: "Gases and the Kinetic-Molecular Theory Chapter 12."— Presentation transcript:
Gases and the Kinetic-Molecular Theory Chapter 12
Common Properties of Gases Gases can be compressed into smaller volumes by applying increased pressure Gases exert pressure on their surroundings Gases expand without limits –Occupy the volume of any container Gases diffuse into one another –Gases are miscible unless they react The amounts and properties of gases are described in terms of temperature, pressure, the volume occupied, and the number of molecules (moles) present.
Pressure Force per unit area –lb/in 2 or psi Barometer – device for measuring the pressure of the atmosphere –Measures in millimeters of Hg (mm Hg) –The Hg will rise until the pressure exerted by the atmosphere equals the pressure exerted by the column of Hg 1 atmosphere = 760 Torr = 760 mm Hg (pressure at sea level) –1 Torr = 1 mm Hg Note: Pressure in Rexburg is less. Why?
Pressure A manometer measures the pressure of a gas in a sealed flask (page 437) –In this case, one side of the tube is open to the atmosphere –If the gas pressure is greater than the atmospheric pressure, then P gas (torr) = P atm (torr) + h torr –If the gas pressure is lower than the the atmospheric pressure, then P gas (torr)=P at m(torr) - h torr Problems: On chalkboard Illustration: Pressure apparatus
Boyle’s Law At constant temperature and mass, the volume of a gas decreases when the pressure increases –V 1/P (Syringe demonstration) Boyle illustrated that the product of pressure and volume were constant –PV = k (temperature and mass are constant) If the amount of gas and temperature do not change, two conditions of pressure and volume will be equal –P 1 V 1 = k and P 2 V 2 = k, therefore, P 1 V 1 =P 2 V 2 –Demo: syringe
Boyle’s Law Let’s do some problems with Boyle’s Law –At 25 0 C a sample of He has a volume of 400 mL under a pressure of 760 torr. What volume would it occupy under a pressure of 2.00 atm at the same T? –A sample of oxygen occupies 15.8 L under a pressure of 285 torr. At what pressure would it occupy 27.9 L?
Charles’s Law When the pressure and the mass of a gas are constant, the volume of a gas is proportional to the temperature –V T –Charles and Gay-Lussac perfomed expansion studies on different gases as a function of temperature A quantitative relationship is not obvious on the celsius scale. The Kelvin scale was devised.
Charles’s Law Lord Kelvin noticed that the extension of the volume to 0, produced a value of –273.15 C. This was defined as absolute zero on the Kelvin scale. Using this scale, a quantitative relationship was established.
Charles’s Law At constant pressure and mass, the volume occupied of a gas is directly proportional to its absolute volume –V T and V=kT (constant n and P) V/T = k Since V/T is constant, it can be expressed that –V 1 /T 1 = V 2 /T 2 (V 1 /T 1 = k, and V 2 /T 2 = k) –This is assuming that the pressure and amount of gas do not change upon changing the conditions (volume or temperature). Demo: Balloon with liquid nitrogen
Charles’s Law Problems with Charles’s Law –A sample of hydrogen, H 2, occupies 100 mL at 25 0 C and 1.00 atm. What volume would it occupy at 50 0 C under the same pressure? –At 112 C, a sample of O 2 occupies 154 mL. What temperature would be required in C to increase the volume to 215 mL? Remember to use the Kelvin scale with the gas laws. You will get the wrong answer if you do not.
Standard Temperature and Pressure Standard temperature and pressure –Temperature = 273.15 K (exactly 0 C) –Pressure = 1 atmosphere (760. Torr) This serves as a reference point for discussing gases. Abbreviated STP
Combined Gas Law Boyle’s and Charles’ Laws combined into one statement Boyle’s Charles’s P 1 V 1 =P 2 V 2 V 1 /T 1 =V 2 /T 2 For a given sample of gas. From this relationship,. This is the combined gas law.
Combined Gas Law Problems –A sample of nitrogen gas, N 2, occupies 750 mL at 75 0 C under a pressure of 810 torr. What volume would it occupy at standard conditions? –A sample of methane, CH 4, occupies 260 mL at 32 o C under a pressure of 0.500 atm. At what temperature would it occupy 500 mL under a pressure of 1200 torr?
Avogadro’s Law At the same temperature and pressure, equal volumes of all gases contain the same number of molecules. Therefore, at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles. –V n or V/n = k Demo: Expanding a balloon (assume P and T constant)
Avogadro’s Law Since V/n is always equal to the same constant, at constant T and P A balloon with a volume of 2.35 L is filled with 1.82 moles of He gas. How many more moles of He gas would have to be added to the balloon to bring the volume to 5.49 L? Assume the T and P are constant.
Standard Molar Volume One mole of any gas has the same V at STP. This would be the standard molar volume –Standard molar volume = 22.4 L (this is one mole) What would be the volume of 0.25 moles of O 2 gas be at STP? –Deviations from the standard molar volume (Table 12-3) indicate that the gases do not behave ideally.
Standard Molar Volume Using the standard molar volume densities at various temperatures can be converted to densities at STP One mole of a gas occupies 36.5 L and its density is 1.36 g/L at a given T & P. –What is the molecular weight (g/mol)? –What is the density at STP? Another problem?
Deriving the Ideal Gas Law Boyle’s Law - V 1/P Charles’ Law - V T Avogadro’s Law - V n combination of these three - V nT/P Ideal Gas Law V = nRT/P or PV = nRT R = proportionality constant R called the universal gas constant
Deriving R in the Ideal Gas Law One mole of an ideal gas occupies 22.414 liters at STP (1.000 atm and 273.15 K). R can has different values depending on the units. Make sure you are using the units that match the gas constant. Remember, the ideal gas law is used to describe one set of conditions.
The Ideal Gas Law Problems with the ideal gas law –What volume would 50.0 g of ethane, C 2 H 6, occupy at 140 o C under a pressure of 1820 torr? –Calculate the number of moles in, and the mass of, an 8.96 L sample of methane, CH 4, measured at standard conditions. –Calculate the pressure exerted by 50.0 g of ethane, C 2 H 6, in a 25.0 L container at 25 o C.
Determining Molecular Weights and Formulas We observed in chapter 2 that the molecular weight must be known in order to determine molecular formulas. For gases, molecular weights and formulas can commonly be determined with the help of the ideal gas law. A 1.502-gram sample of a pure gaseous compound occupies 852 mL at 34.2 C and 845 Torr. What is the molecular weight?
Determining Molecular Weights and Formulas A 250.0-mL flask contains 0.585 grams of a gaseous compound at 115 C and 0.950 atm pressure. What is the molecular weight of the compound? A compound that contains only carbon & hydrogen is 80.0% C and 20.0% H by mass. At STP 546 mL of the gas has a mass of 0.732 g. What is the molecular (true) formula for the compound?
Determining Molecular Weights and Formulas A 1.74 g sample of a compound that contains only carbon & hydrogen contains 1.44 g of C and 0.300 g of H. At STP 101 mL of the gas has a mass of 0.262 gram. What is its molecular (true) formula?
Dalton’s Law of Partial Pressure It can be shown utilizing the ideal gas law that pressure exerted by a mixture of gases is the sum of the partial pressures of the individual gases. –P total = P A + P B + P C +…. (constant T and V) A, B, and C are the individual gases that are present Look at page 456 for the derivation of Dalton’s Law
Dalton’s Law of Partial Pressures A 25.0-liter flask contains 0.450 moles of ethane, 0.350 moles of oxygen, and 0.200 moles of nitrogen at 28 C. What is the pressure, in atmospheres, inside the flask?
Mole Fractions and Partial Pressures The mole fraction of a component in a mixture can be expressed as Where X A indicates the mole fraction of A Using the ideal gas law (page 458), the gas mole fraction can be expressed in terms of P
Mole Fractions and Partial Pressures A sample of air was analyzed. The total number of moles in the sample was 0.582. The number of moles of O 2 in the sample was 0.185. If the total pressure is 692 torr. What is the partial pressure of oxygen? Example 12-18 in books
Vapor Pressure The total pressure in a sealed container containing water is affected by the pressence of water. Gaseous molecules leaving the liquid phase produces a contribution to the total pressure. This contribution is called the vapor pressure of water. The vapor pressure of water increases with increasing temperature (Table 12-4). Actually, every liquid has a unique vapor pressure. Any gas in contact with water soon becomes saturated with water vapor.
Producing Gases over Water (by displacement) The pressure inside a sealed container is the sum of the partial pressure of H 2 O(l) (or vapor pressure) and the partial pressure of the other gas(es) that is present. –A sample of hydrogen was collected over water at 25 0 C. The volume in the collection flask is 250.0 mL and the pressure is 748 torr. What is the partial pressure of the hydrogen gas in the collection flask? How many moles of H 2 and H 2 O(g) are in the container?
Producing Gases over Water Demonstration: 0.292 grams of magnesium is reacted with 14.0 mL of 1.00 M HCl. Determine the limiting reactant. The pressure inside the flask (P atm ) is 642 torr. Determine the total volume occupied. A sample of oxygen was collected by displacement of water. The volume in the collection flask was 742 mL at 27 o C. The barometric pressure was 753 torr. What volume would the dry oxygen occupy at STP?
Mass-Volume Relationships in Reactions Involving Gases Many chemical reactions produce gases. –NaHCO 3 (s) + CH 3 COOH(aq) NaCH 3 COO(aq) + CO 2 (g) + H 2 O(l) 0.790 grams of NaHCO 3 is reacted with 3.00 mL of 6.00 M CH 3 COOH. Determine the limiting reactant. How much CO 2 (g) is produced? Is this a displacement reaction?
Mass-Volume Relationships in Reactions Involving Gases What volume of CO 2 (g) is produced when 1.519 grams of C 4 H 8 is reacted with excess O 2 (this is a combustion reaction)? The volume of gas is measured at STP. How many L of O 2 (g) is produced when 252 moles of KClO 3 is decomposed according to the reaction below? The amount of O 2 (g) was measured at 353 K and 692 Torr. –KClO 3 (s) KCl(s) + O 2 (g)
The Kinetic-Molecular Theory Assumptions Gases consists of small discrete molecules that are far apart –Molecule size is very small compared to the total volume The gas molecules are in constant straight-line motion with varying velocities The collisions experienced by the gas molecules are elastic –No energy loss There are no attractive or repulsive forces acting between the gas molecules
The Kinetic-Molecular Theory The average kinetic energy of molecules is directly proportional to the temperature. –Can be expressed as The average KE of molecules of different gases are equal at any given temperature –This means that smaller molecules will have greater velocities at a given temperature
The Kinetic-Molecular Theory Only the average temperature of the gas molecules has been addressed. There will actually be a distribution of molecule speeds that depends on both the temperature and the mass of the gaseous molecules (refer to page 465) –An excellent discussion on the kinetic- molecular theory is presented in the enrichment section and 12.11 on the CD.
The Kinetic-Molecular Theory and Boyle’s Law The pressure is a result of gas molecules colliding with the walls of the container –Rate of molecules colliding with the walls –Velocity of molecules colliding with the wall Decreasing the volume increases the pressure since there will be a higher rate of collisions in the reduced volume. –The average velocity of molecules remains the same.
The Kinetic-Molecular Theory and Dalton’s Law Molecules are far apart and do not interact with each other significantly. –Each gas acts independently of the other molecules when colliding with the container Frequency and velocity of collisions Therefore, each gas exerts a partial pressure, and the total pressure is a sum of all the molecule-wall collisions.
The Kinetic-Molecular Theory and Charles’s Law Average KE is proportional to the temperature –Increasing the temperature causes the speed of the collisions to increase. Normally, this would increase the pressure. Volume has to increase to keep the pressure constant. –The reverse occurs when the temperature decreases.
Diffusion and Effusion of Gases Diffusion – intermingling or mixing of gases Effusion – gases passing through porous (possessing small holes) containers Rate of effusion or diffusion is inversely proportional to the square roots of the molecular weights
Diffusion and Effusion of Gases Demo: U-tube and balloons (N 2 and He) Calculate the ratio of the rate of effusion of He to that of sulfur dioxide, SO 2, at the same T & P. The diffusion rate of CH 4 is 4.5 m/s in air. What is the diffusion rate of C 3 H 8 in air? Another problem on CD.
Diffusion and Effusion of Gases The kinetic-molecular theory can be used to calculate an average speed of gaseous molecules that behave ideally. –u rms is root-mean square speed of the gaseous molecule. Notice that this speed is inversely related to the square root of the molecular weight. It should be noted, however, that the u rms is not equal to the diffusion rate of a gas through air. Why?
Diffusion and Effusion of Gases What is the u rms of Xe and He? This would be the distance covered by one molecule in one second. If no collisions occurred, this would be the distance traveled by the gaseous atom from a certain starting point. Since collisions do occur on this planet, the distance traveled away form a certain starting point is much less.
Deviations from the Ideal Gas Behavior Under ordinary conditions of pressure and temperature most gases behave ideally. Nonideal gas behavior becomes significant at high pressures and/or low temperatures. –Nonideal behavior is especially evident near the conditions under which the gas turns to a liquid Johanes van der Waals studied the nonideal behavior of gases and adjusted the ideal gas equation.
Factors Contributing to Nonideal Gas Behavior Upon increasing the pressure, the volume of the gas molecules becomes important to the overall/available volume –V ideally available = V measured – nb nb corrects for the volume occupied by the molecules The value of ‘b’ increases with the size of the molecule n is equal the number of moles How will V ideally available vary with He and C 3 H 8 ?
Factors Contributing to Nonideal Gas Behavior As the pressure increases and the temperature decreases, the gaseous molecules move more slowly due to attractive forces between molecules. This will lower the predicted pressure due to the reduced number of collisions. –Large values of ‘a’ indicate strong attractive forces. More molecules and smaller volumes also produces a large correction term.
The Van der Waals Equation P, V, T, and n represent the measured values just like the ideal gas equation. The values, a and b, are experimentally determined and will be provided. They are different for each gas. –a accounts for attractive forces between gas molecules –b accounts for the volume of the gas molecule The Van der Waals equation reduces to the ideal gas law at high temperatures and low pressures. Why?
Some Problems with the Van der Waals Equation Calculate the pressure exerted by 84.0 g of ammonia, NH 3, in a 5.00 L at 200 0 C using the ideal gas law. Solve Example 12-17 using the van der Waal’s equation. –What is the difference? Another problem.