Combined & Ideal Gas Laws

Combined &Ideal Gas Laws

Avogadro’s Law • So far we’ve compared all the gas variables except amount (n). • The relationship of amount and volume is called Avogadro’s Law • If you increase the moles of a gas the volume will increase • Directly proportional V ___ = k n

PV = R nT Ideal Gas Law • If we combine all of the laws together including Avogadro’s Law mentioned earlier we get: Where R is the universal gas constant Normally written as PV = nRT

Ideal Gas Law Liters atm Kelvin moles kPa PV = nRT mmHg L•mmHg L•kPa L•atm _____ _____ ______ .0821 62.4 8.31 mol•K mol•K mol•K match up pressure units

Ideal Gas Law Example 1 How many moles of a gas at 100.0°C does it take to fill a 1.00 L flask to a pressure of 1.50 atm? PV = nRT (1.50atm) (1.00L) = n (373K) L•atm _____ .0821 mol•K n = 0.0490 mols

Ideal Gas Law Example 2 What volume would be occupied by 100.0 g of oxygen gas at a pressure of 1140 mmHg and a temp of 25.0°C? PV = nRT 1 mol _____ 100 g = 3.13 mols 32 g L•mmHg _____ 62.4 (V) (1140 mmHg) mol•K = 3.13 mols (298K) V = 51.1 L

Classroom Practice 4 Use the Ideal Gas Law to complete the following table for ammonia gas (NH3).

Solving for Molar Mass • We can use the ideal gas law to derive an equation to solve for molar mass (MM) of an unknown gas. • Remember moles = mass ÷ molar mass, and then if we substitute… PV = nRT n = m/MM

Solving for Density • We can then use the MM equation to derive an equation that solves for the density of a gas. • Remember thatD = m/V

Classroom Practice 5 A gas consisting of only carbon and hydrogen has an empirical formula of CH2. The gas has a density of 1.65 g/L at 27C and 734 mmHg. Determine the molar mass and the molecular formula of the gas. Silicon tetrachloride (SiCl4) and trichlorosilane (SiHCl3) are both starting materials for the production of electro-nics-grade silicon. Calculate the densities of pure SiCl4 and pure SiHCl3 vapor at 85C and 758 mmHg.

2 mol H2 2.016 g H2 = 1mol H2 2 mol H2O Ideal Gas Law & Stoichiometry What mass of H2 gas must be burned to form 1.00 L of water vapor at 1.00 atm pressure & 300°C? PV = nRT (1.00 atm) (1.00 L) nH2O= = 0.021257 mols (573K) (.0821L atm/mol K) 2H2 + O2 2H2O 0.021257 mol 0.0429 gH2

Classroom Practice 6 To find the formula of a transition metal carbonyl, one of a family of compounds having the general formula Mx(CO)y, you can heat the solid compound in a vacuum to produce solid metal and CO gas. You heat 0.112 g of Crx(CO)y Crx(CO)y(s)  x Cr(s) + y CO(g) and find that the CO evolved has a pressure of 369 mmHg in a 155 ml flask at 27C. What is the empirical formula of Crx(CO)y?

Loose Ends of Gases • There are a couple more laws that we need to address dealing with gases. • Dalton’s Law of Partial Pressures • Graham’s Law of Diffusion and Effusion.

Dalton’s Law of Partial Pressures • States that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. • PT=P1+P2+P3+… • Which means that each gas in a mixture exerts it’s own pressure on its container’s walls, and the total pressure is the sum of all of the individual pressures.

Example Dalton’s Law Three of the primary components of air are CO2, N2, and O2. In a sample containing a mixture of these gases at exactly 760 mmHg, the partial pressures of CO2 and N2 are given as 0.285 mmHg and 593.525 mmHg respectively. What is PO2? PT = PCO2 + PN2 + PO2 760mmHg = .285mmHg + 593.525mmHg + PO2 PO2= 167mmHg

Dalton’s Law of Partial Pressure • Partial pressures are also important when a gas is collected by bubbling through water. • Any time a gas is collected through water the gas is “contaminated” with water vapor. • You can determine the pressure of the dry gas by subtracting out the pressure exerted by water vapor

Atmospheric Pressure Ptot = Patmospheric pressure = Pgas + PH2O • The water’s vapor pressure can be determined from a list and subtract-ed from the atmospheric pressure

Simple Dalton’s Law Calculation • Determine the partial pressure of O2 collected by water displacement if the water temp is 20.0°C and the total pressure of the gases in the collection bottle is 730 mmHg. PH2O at 20.0°C= 17.5 mmHg PT = PH2O + PO2 PT = 730 mmHg = 17.5 + PO2 PO2= 712.5 mmHg

Classroom Practice 7 A mixture of 1.00 g H2 and 1.00 g He is placed in a 1.00 L container at 27C. Calculate the partial pressure of each gas and the total pressure in atm. Helium is collected over water @ 25C and 1.00 atm total pressure. What total volume of He must be collected to obtain 0.586 g of He? 1) PH2 = 12.3 atm; PHe = 6.16 atm 2) 3.7 L

Graham’s Law • Thomas Graham studied the effusion and diffusion of gases. • Diffusion is the mixing of gases through each other. • Effusion is the process where the gas escapes from its container through a tiny hole

Graham’s Law Diffusion Effusion

Graham’s Law • Graham’s Law states that the rates of effusion and diffusion of gases at the same temperature and pressure is dependent on the size of the molecule. • The bigger the molecule the slower it moves the slower it mixes and escapes.

Graham’s Law • Graham’s law is derived from the kinetic energy equation • Kinetic energy can be calculated with the equation ½ mv2 • m is the mass of the object • v is the velocity. • If we work with two different gases at the same temp their energies would be equal and the eqn can be rewritten as:

Graham’s Law • “MM” represents molar mass • “v” represents molecular velocity • “A” is Gas #1 • “B” is Gas #2 • We can compare both gases’ velocities, to determine which gas moves faster, by writing the ratio of their velocities.

Graham’s Law • This shows that the velocities of two different gases areinversely propor-tionalto the square roots of their molar masses. • This can be expanded to deal with rates of diffusion or effusion • To finally get Graham’s Law

Graham’s Law • The equation tells us how much faster gas A is than gas B. • Note that the equation, shows us that the velocity of gas A compared to gas B is inversely proportional to the molar masses of A and B • So if A is half the size of B than it effuses or diffuses 1.4 times faster.

Graham’s Law Example Calc. If equal amounts of helium and argon are placed in a porous container and allowed to escape, which gas will escape faster and how much faster? Helium is 3.16 times faster than Argon.

Real Vs. Ideal • All of our calcs with gases have been assuming ideal conditions and behaviors. • We assumed that there was no attraction established between particles. • We assumed that each particle has no volume of its own • Under normal atmospheric conditions gases tend to behave as we expect and as predicted by the KMT.

Real Vs. Ideal • However, under high pressures and low temperatures, gases tend to deviate from ideal behaviors. • Under extreme conditions we tend to see a tendency of gases to not behave as independently as the ideal gas law predicts. • Attractive forces between gas particles under high pressures or low temperature cause the gas not to behave predictably.

Classroom Practice 8 Calculate the average rate of effusion of a H2 molecule at 0° C if the average rate of effusion of an O2 molecule at this temperature is 500 m/s. The rate of effusion of a gas was meas-ured to be 24.0 ml/min. Under the same conditions, the rate of effusion of pure CH4 gas is 47.8 ml/min. What is the molar mass of the unknown gas? 1) 2000 m/s 2) 63.5 g/mol

Combined & Ideal Gas Laws