Properties of Gases Notes Properties of Gases: The Air We Breathe Intended Learning Outcomes (ILOs) Apply the Ideal Gas Law to determine properties of gases. Properties of Gases How are Gases unique?A gas expands spontaneously to fill its container. Gases are highly compressible. Gases form homogenous mixtures with other gases, regardless of identity or composition of the gas. At the same temperature and pressure:Equal volumes = equal number of molecules gas. These properties are unique to gases and do not apply to liquids or solids. Properties of gases:Volume is inversely proportional to pressure. Pressure is directly proportional to temperature. Pressure is directly proportional to quantity of gas. Gases are miscible. Densities are directly proportional to molar mass. Gases expand to occupy entire volume of their container. Gases effuse at rates inversely proportional to molar mass. Pressure of Gases Pressure of a gas is the collective force particles exert on the walls of the container. More collisions cause more pressure. More particles cause more pressure. Smaller volume causes more pressure. Higher temperature causes more pressure. Kinetic Molecular Theory (KMT) The behavior of gases can be described using kinetic molecular theory (KMT), which is based upon four assumptions:Gas molecules have tiny volumes compared with volume the gas occupies.Gases are mostly empty space Gas particles do not interact with each other.We assume there are NO interactions between gas particles, no IMFs Gas molecule collisions are elastic.Gas particles collide with each other constantly, but the collisions are elastic, meaning no loss of energy The average kinetic energy of the molecules in a gas are proportional to the absolute temperature.No matter the gas, if the temperature is the same, the average kinetic energy is the same. Particle Speed u m u_m u m u < e m > a v g u<em>{avg} u < e m > a v g u < / e m > m u</em>m u < / e m > m u r m s u_{rms} u r m s KE and u r m s u_{rms} u r m s KE: Kinetic Energy m: mass u r m s u_{rms} u r m s At the same temperature, all gas particles, regardless of element, have the same K E a v g KE_{avg} K E a v g Heavier molecules have slower average speeds. K E < e m > a v g = 1 2 m u < / e m > r m s 2 KE<em>{avg} = \frac{1}{2} m u</em>{rms}^2 K E < e m > a v g = 2 1 m u < / e m > r m s 2 Comparing Particle Speeds Let’s compare He and N2. According to KMT, all gases at the same temp have the same KE, but not the same speed K E < e m > a v g = 1 2 m u < / e m > r m s 2 KE<em>{avg} = \frac{1}{2} m u</em>{rms}^2 K E < e m > a v g = 2 1 m u < / e m > r m s 2 K E < e m > H e = K E < / e m > N 2 KE<em>{He} = KE</em>{N_2} K E < e m > H e = K E < / e m > N 2 1 2 m < e m > H e u < / e m > r m s , H e 2 = 1 2 m < e m > N < / e m > 2 u < e m > r m s , N < / e m > 2 2 \frac{1}{2} m<em>{He} u</em>{rms,He}^2 = \frac{1}{2} m<em>{N</em>2} u<em>{rms,N</em>2}^2 2 1 m < e m > H e u < / e m > r m s , H e 2 = 2 1 m < e m > N < / e m > 2 u < e m > r m s , N < / e m > 2 2 u < e m > r m s , H e u < / e m > r m s , N < e m > 2 = m < / e m > N < e m > 2 m < / e m > H e \frac{u<em>{rms,He}}{u</em>{rms,N<em>2}} = \sqrt{\frac{m</em>{N<em>2}}{m</em>{He}}} u < / e m > r m s , N < e m > 2 u < e m > r m s , H e = m < / e m > H e m < / e m > N < e m > 2 u < e m > r m s , H e u < / e m > r m s , N < e m > 2 = M < / e m > N < e m > 2 M < / e m > H e \frac{u<em>{rms,He}}{u</em>{rms,N<em>2}} = \sqrt{\frac{\mathcal{M}</em>{N<em>2}}{\mathcal{M}</em>{He}}} u < / e m > r m s , N < e m > 2 u < e m > r m s , H e = M < / e m > H e M < / e m > N < e m > 2 M = m o l a r m a s s \mathcal{M} = molar \, mass M = m o l a r ma ss He travels 2.646 times faster than N2. u < e m > r m s , H e u < / e m > r m s , N < e m > 2 = M < / e m > N < e m > 2 M < / e m > H e = 28.02 g / m o l 4.003 g / m o l = 2.646 \frac{u<em>{rms,He}}{u</em>{rms,N<em>2}} = \sqrt{\frac{\mathcal{M}</em>{N<em>2}}{\mathcal{M}</em>{He}}} = \sqrt{\frac{28.02 \, g/mol}{4.003 \, g/mol}} = 2.646 u < / e m > r m s , N < e m > 2 u < e m > r m s , H e = M < / e m > H e M < / e m > N < e m > 2 = 4.003 g / m o l 28.02 g / m o l = 2.646 Effusion Process by which gas molecules escape from a container through small holes to a region of lower pressure. Graham’s Law of Effusion e f f u s i o n r a t e < e m > A e f f u s i o n r a t e < / e m > B = M < e m > B M < / e m > A \frac{effusion \, rate<em>A}{effusion \, rate</em>B} = \sqrt{\frac{\mathcal{M}<em>B}{\mathcal{M}</em>A}} e f f u s i o n r a t e < / e m > B e f f u s i o n r a t e < e m > A = M < / e m > A M < e m > B u < e m > r m s , A u < / e m > r m s , B = M < e m > B M < / e m > A \frac{u<em>{rms,A}}{u</em>{rms,B}} = \sqrt{\frac{\mathcal{M}<em>B}{\mathcal{M}</em>A}} u < / e m > r m s , B u < e m > r m s , A = M < / e m > A M < e m > B Determining Particle Speeds K E < e m > a v g = 1 2 m u < / e m > r m s 2 KE<em>{avg} = \frac{1}{2} m u</em>{rms}^2 K E < e m > a v g = 2 1 m u < / e m > r m s 2 1 2 m u < e m > r m s 2 = 3 2 k < / e m > B T \frac{1}{2} m u<em>{rms}^2 = \frac{3}{2} k</em>B T 2 1 m u < e m > r m s 2 = 2 3 k < / e m > B T K E < e m > a v g = 3 2 k < / e m > B T KE<em>{avg} = \frac{3}{2} k</em>B T K E < e m > a v g = 2 3 k < / e m > B T u < e m > r m s = 3 k < / e m > B T m u<em>{rms} = \sqrt{\frac{3k</em>BT}{m}} u < e m > r m s = m 3 k < / e m > B T u r m s = 3 R T M u_{rms} = \sqrt{\frac{3RT}{\mathcal{M}}} u r m s = M 3 R T R = 8.314 k g ⋅ m 2 s 2 ⋅ m o l ⋅ K R = 8.314 \frac{kg \cdot m^2}{s^2 \cdot mol \cdot K} R = 8.314 s 2 ⋅ m o l ⋅ K k g ⋅ m 2 Practice Calculate the root-mean-square speed of Nitrogen gas molecules at 25℃. u r m s = 3 R T M u_{rms} = \sqrt{\frac{3RT}{\mathcal{M}}} u r m s = M 3 R T R = 8.314 k g ⋅ m 2 s 2 ⋅ m o l ⋅ K R = 8.314 \frac{kg \cdot m^2}{s^2 \cdot mol \cdot K} R = 8.314 s 2 ⋅ m o l ⋅ K k g ⋅ m 2 Diffusion Spread of one substance (usually gas) through another substance. Mean Free Path: average distance a particle can travel through a gas before colliding with another particle. At 1 atm is about 6.8 x 10 − 8 6.8 x10^{-8} 6.8 x 1 0 − 8 10 10 10^{10} 1 0 10 Gases will spontaneously fill a space through diffusion and effusion, moving from high concentration to low concentration. Atmospheric Pressure Produced by Earth’s gravity pulling with force on the air molecules. At higher altitudes, there are less gas molecules, and therefore lower pressure. Atmospheric Pressure Effects Differences in pressure produces wind. Remember that gas will spontaneously move from high concentrations to lower concentrations. High pressure: usually associated with clear weather. Low pressure: usually associated with unstable weather. Measuring Pressure BarometerPressure of the atmosphere on the mercury pool pushes the mercury up the tube. Measuring that push up the tube determines the atmospheric pressure ManometerWhen the valve is closed, Hg levels are equal. Open valve and pressure will push against the atm pressure. Measure the difference between the two to determine the pressure of a sample Measuring Pressure Standard pressure is the pressure at sea level: Partial Pressures Gases create homogeneous mixtures So, when a gas exerts a pressure, all the component gases are actually individually exerting their own pressures. Dalton’s Law of Partial Pressures P < e m > T o t a l = P < / e m > 1 + P < e m > 2 + P < / e m > 3 … P<em>{Total} = P</em>1 + P<em>2 + P</em>3 … P < e m > T o t a l = P < / e m > 1 + P < e m > 2 + P < / e m > 3 … Where P < e m > 1 P<em>1 P < e m > 1 P < / e m > 2 P</em>2 P < / e m > 2 P 3 P_3 P 3 Partial Pressures Can also calculate the partial pressure for an individual gas if you know how many moles of it are present in the total mixture (mole fraction). P < e m > 1 = χ < / e m > 1 × P T o t a l P<em>1 = \chi</em>1 \times P_{Total} P < e m > 1 = χ < / e m > 1 × P T o t a l χ 1 = m o l e s o f 1 T o t a l m o l e s \chi_1 = \frac{moles \, of \, 1}{Total \, moles} χ 1 = T o t a l m o l es m o l es o f 1 Partial Pressures One way to measure the partial pressure of a gas is over water. P < e m > t o t a l = P < / e m > d r y g a s + P < e m > H < / e m > 2 O P<em>{total} = P</em>{dry \, gas} + P<em>{H</em>2O} P < e m > t o t a l = P < / e m > d r y g a s + P < e m > H < / e m > 2 O Simple Gas Laws Boyle’s Law – Pressure and Volume Charles’s Law – Volume and Temperature Amonton’s Law – Pressure and Temperature Avogadro’s Law – Volume and Moles When comparing two CHANGING properties, the others stay CONSTANT Boyle’s Law The volume of a gas is inversely proportional to the pressure. As volume goes up, pressure goes down. P < e m > 1 V < / e m > 1 = P < e m > 2 V < / e m > 2 P<em>1V</em>1 = P<em>2V</em>2 P < e m > 1 V < / e m > 1 = P < e m > 2 V < / e m > 2 Charles’s Law The volume of a gas is directly proportional to temperature. As temperature goes up, volume goes up. V < e m > 1 T < / e m > 1 = V < e m > 2 T < / e m > 2 \frac{V<em>1}{T</em>1} = \frac{V<em>2}{T</em>2} T < / e m > 1 V < e m > 1 = T < / e m > 2 V < e m > 2 Amonton’s Law The pressure of a gas is directly proportional to temperature. As temperature goes up, pressure goes up. P < e m > 1 T < / e m > 1 = P < e m > 2 T < / e m > 2 \frac{P<em>1}{T</em>1} = \frac{P<em>2}{T</em>2} T < / e m > 1 P < e m > 1 = T < / e m > 2 P < e m > 2 Combined Gas Law Boyle’s, Charles’s, and Amonton’s Laws can be combined to produce the Combined Gas Law This is useful if P, V, and T are all changing. In the individual laws, 2 things can change while one has to be held constant. If you know the Combined Gas Law, you can derive Boyle’s, Charles’s, or Amonton’s P < e m > 1 V < / e m > 1 T < e m > 1 = P < / e m > 2 V < e m > 2 T < / e m > 2 \frac{P<em>1V</em>1}{T<em>1} = \frac{P</em>2V<em>2}{T</em>2} T < e m > 1 P < e m > 1 V < / e m > 1 = T < / e m > 2 P < / e m > 2 V < e m > 2 Avogadro’s Law The volume of a gas is directly proportional to the number of moles of the gas. As the moles of a gas go up, the volume goes up. At STP: 22.4 L = 1 mole of gas Applies to ALL gases V < e m > 1 n < / e m > 1 = V < e m > 2 n < / e m > 2 \frac{V<em>1}{n</em>1} = \frac{V<em>2}{n</em>2} n < / e m > 1 V < e m > 1 = n < / e m > 2 V < e m > 2 Ideal Gas Law Boyle’s, Charles’s, Amonton’s, and Avogadro’s Laws can be combined along with a constant to produce the Ideal Gas Law For this law, if you know 3 variables, you can determine the fourth. If you know the Idea Gas Law, you can derive Boyle’s, Charles’s, Amonton’s, or Avogadro’s. P V = n R T PV = nRT P V = n R T What is an Ideal Gas 2 Main Assumptions:The volume of individual gas particles are insignificant. Gas particles DO NOT interact with each other. P V = n R T PV = nRT P V = n R T Ideal Gas Constant R is known as the gas constant. Can have many combinations of units The units in the constant you use MUST match your units in your measurements/variables You can either choose your constant, or assure that your measurements are using the same units as your constant P V = n R T PV = nRT P V = n R T Volume at STP 1 mole of ANY gas at STP is 22.4L P V = n R T PV = nRT P V = n R T V n = R T P = ( 0.0821 L ⋅ a t m m o l ⋅ K ) ( 273 K ) 1 a t m = 22.4 L 1 m o l \frac{V}{n} = \frac{RT}{P} = \frac{(0.0821 \frac{L \cdot atm}{mol \cdot K})(273 \, K)}{1 \, atm} = \frac{22.4 \, L}{1 \, mol} n V = P R T = 1 a t m ( 0.0821 m o l ⋅ K L ⋅ a t m ) ( 273 K ) = 1 m o l 22.4 L Density of an Ideal Gas Density is mass/volume For an Ideal Gas, volume is the same from gas to gas, but mass is different, so therefore, Density is as well. Can use molar mass/molar volume instead of just mass/volume D e n s i t y = m o l a r m a s s m o l a r v o l u m e = m a s s / m o l v o l u m e / m o l Density = \frac{molar \, mass}{molar \, volume} = \frac{mass/mol}{volume/mol} D e n s i t y = m o l a r v o l u m e m o l a r ma ss = v o l u m e / m o l ma ss / m o l Density of an Ideal Gas In the Ideal Gas Equation, we know that n (moles) is the same as grams/molar mass, so we can substitute it into the Ideal Gas equation to solve for density. Can also be rearranged to solve for molar mass: m V = P ( M ) R T \frac{m}{V} = \frac{P (\mathcal{M})}{RT} V m = R T P ( M ) M = d ( R T ) P \mathcal{M} = \frac{d (RT)}{P} M = P d ( R T ) Stoichiometry Involving Gases You now have a new conversion factor you can use in stoichiometry calculations. H < e m > 2 + O < / e m > 2 → H 2 O H<em>2 + O</em>2 \rightarrow H_2O H < e m > 2 + O < / e m > 2 → H 2 O