Combined and ideal gas laws

Combined and ideal gas laws

Combined gas law • If we combine all of the relationships from the 3 laws covered thus far (Boyle’s, Charles’s, and Gay-Lussac’s) we can develop a mathematical equation that can solve for a situation where 3 variables change : PV=k1 V/T=k2 P/T=k3

P1V1 P2V2 = T1 T2 Combined gas law • Amount is held constant • Is used when you have a change in volume, pressure, or temperature P1V1T2 = P2V2T1

- P1 • - P2  • - V1  • - V2  • - T1  • - T2  Example problem A gas with a volume of 4.0L at STP. What is its volume at 2.0atm and at 30°C? • 1atm • 2.0 atm • ? • 4.0 L • 273K • 30°C + 273 • = 303K

P1V1 P2V2 = T1 T2 Example problem (1 atm) (4.0L) (2 atm) ( V ) 2 = (273K) (303K) 2.22L = V2

Avogadro’s Law • So far we’ve compared all the variables except the amount of a gas (n). • There is a lesser known law called Avogadro’s Law which relates V & n. • It turns out that they are directly related to each other. • As # of moles increases then V increases. V/n = k

Ideal Gas Law • Which leads us to the ideal gas law – • The fourth and final variable is amount • We have been holding it constant. • We can set up a much more powerful eqn, which can be derived by combining the proportions expressed by the previous laws.

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 Constant (R) • R is a constant that connects the 4 variables • R is dependent on the units of the variables for P, V, & T • Temp is always in Kelvin • Volume is in liters • Pressure is in either atm or mmHg or kPa

L•atm L•kPa R=.0821 R=8.314 L•mmHg R=62.4 mol•K mol•K mol•K Ideal Gas Constant • Because of the different pressure units there are 3 possibilities for our ideal gas constant • If pressure is given in atm • If pressure is given in mmHg • If pressure is given in kPa

L•atm .0821 mol•K 9.45g 26g Using the Ideal Gas Law What volume does 9.45g of C2H2 occupy at STP? • 1atm • P • R • ? • V • T • 273K • n • = .3635 mol

L•atm mol•K (.0821 ) (1.0atm) (V) (8.147L•atm) = PV =nRT (1.0atm) (V) = (273K) (.3635mol) V = 8.15L

L•kPa 8.31 mol•K 3000g 44g A camping stove propane tank holds 3000g of C3H8. How large a container would be needed to hold the same amount of propane as a gas at 25°C and a pressure of 303 kpa? • 303kPa • P • R • ? • V • T • 298K • n • = 68.2 mol

L•kPa mol•K (8.31 ) (303kPa) (V) (168,970.4 L•kPa) = PV = nRT (303kPa) (V) = (298K) (68.2 mol) V = 557.7L

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

PV = nRT Ideal Gas Law & Stoichiometry What volume of hydrogen gas must be burned to form 1.00 L of water vapor at 1.00 atm pressure and 300°C? (1.00 atm) (1.00 L) nH2O= (573K) (.0821L atm/mol K) nH2O= .021257 mols

2 mol H2 22.4 L H2 = 2 mol H2O 1mol H2 Ideal Gas Law & Stoichiometry 2H2 + O2 2H2O .021257 mol .476 L H2

Classroom Practice 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?

Variations of the Ideal Gas Law • We can use the ideal gas law to derive a version to solve for MM. • We need to know that the unit mole is equal to m ÷ MM, where m is the mass of the gas sample PV = nRT n = m/MM

Variations of the Ideal Gas Law • We can then use the MM equation to derive a version that solves for the density of a gas. • Remember that D = m/V

Classroom Practice 1 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.

Real Vs. Ideal • All of our calculations 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.

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 Pressure • 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+… • What that means is that each gas involved in a mixture exerts an independent pressure on its container’s walls

Dalton’s Law of Partial Pressure • Therefore, to find the pressure in the system you must have the total pressure of all of the gases involved. • This becomes very important for people who work at high altitudes like mountain climbers and pilots. • For example, at an altitude of about 10,000m air pressure is about 1/3 of an atmosphere.

Dalton’s Law of Partial Pressure • The partial pressure of oxygen at this altitude is less than 50 mmHg. • By comparison, the partial pressure of oxygen in human alveolar blood needs to be about 100 mmHg. • Thus, respiration cannot occur normally at this altitude, and an outside source of oxygen is needed in order to survive.

Simple Dalton’s Law Calculation • 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 PCO2= 0.285mmHg and PN2 = 593.525mmHg. What is the partial pressure of O2?

Simple Dalton’s Law Calculation 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 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 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 oxygen collected by water displace-ment if the water temperature 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

Simple Dalton’s Law Calculation PT = PH2O + PO2 PH2O = 17.5 mmHg PT = 730 mmHg 730mmHg = 17.5468 + PO2 PO2= 712.5 mmHg

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 whereby the molecules of a gas escape from its container through a tiny hole

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 • 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 at the same temperature their energies would be equal and the equation can be rewritten as:

½ MAvA2 = ½ MBvB2 • “M” represents molar mass • “v” represents molecular velocity • “A” is one gas • “B” is another gas • If we want to compare both gases velocities, to determine which gas moves faster, we could write a ratio of their velocities. • Rearranging things and taking the square root would give the eqn:

Rate of effusion of A MB = Rate of effusion of B MA vA MB = vB MA • This shows that the velocities of two different gases are inversely propor-tional to the square roots of their molar masses. • This can be expanded to deal with rates of diffusion or effusion

Graham’s Law • The way you can interpret the equation is that the number of times faster A moves than B, is the square root of the ratio of the molar mass of B divided by the Molar mass of A • So if A is half the size of B than it effuses or diffuses 1.4 times faster.

Rate of effusion of A MB = Rate of effusion of B MA 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?

40 g 4 g Graham’s Law Example Calc. Rate of effusion of He = Rate of effusion of Ar Helium is 3.16 times faster than Argon.

Classroom Practice 2 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. Helium is collected over water @ 25C and 1.00 atm total pressure. What total volume of gas must be collected to obtain 0.586 g of He? 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?

Combined and ideal gas laws