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Combined and ideal gas laws

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Gases have mass Gases diffuse Gases expand to fill containers Gases exert pressure Gases are compressible Pressure & temperature are dependent Gases have mass Gases diffuse Gases expand to fill containers Gases exert pressure Gases are compressible Pressure & temperature are dependent Gas Properties

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Volume (V) Units of volume (L) Amount (n) Units of amount (moles) Temperature (T) Units of temperature (K) Pressure (P) Units of pressure (mmHg) Units of pressure (KPa) Units of pressure (atm) Volume (V) Units of volume (L) Amount (n) Units of amount (moles) Temperature (T) Units of temperature (K) Pressure (P) Units of pressure (mmHg) Units of pressure (KPa) Units of pressure (atm) Gas Variables

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P 1 V 1 = P 2 V 2 Boyles law pressure & volume as P then V at constant T, n Boyles law pressure & volume as P then V at constant T, n Charles law: Temperature & volume As T then V At constant P, n Charles law: Temperature & volume As T then V At constant P, n T 1 V 2 = T 2 V 1 A Little Review

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Gay-Lussacs law: Temperature & pressure As P then T At constant V, n Gay-Lussacs law: Temperature & pressure As P then T At constant V, n P 1 T 2 = P 2 T 1 A Little Review

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Combined gas law PV = k 1 V/T = k 2 P/T = k 3 If we combine all of the relationships from the 3 laws covered thus far (Boyles, Charless, and Gay- Lussacs) we can develop a mathematical equation that can solve for a situation where 3 variables change :

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Amount is held constant Is used when you have a change in volume, pressure, or temperature Amount is held constant Is used when you have a change in volume, pressure, or temperature P1V1P1V1 P1V1P1V1 T1T1 T1T1 = K = P2V2P2V2 P2V2P2V2 T2T2 T2T2 Combined gas law

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Amount is held constant Is used when you have a change in volume, pressure, or temperature Amount is held constant Is used when you have a change in volume, pressure, or temperature P 1 V 1 T 2 = P 2 V 2 T 1 P1V1P1V1 P1V1P1V1 T1T1 T1T1 = = P2V2P2V2 P2V2P2V2 T2T2 T2T2

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A gas with a volume of 4.0L at STP. What is its volume at 2.0atm and at 30°C? Example problem - P 1 - V 1 - T 1 - P 2 - V 2 - T 2 1atm 4.0 L 273K 2.0 atm ? ? 30°C + 273 = 303K

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2.22L = V 2 P1V1P1V1 P1V1P1V1 T1T1 T1T1 = = P2V2P2V2 P2V2P2V2 T2T2 T2T2 Example problem

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So far weve compared all the variables except the amount of a gas (n). There is a lesser known law called Avogadros Law which relates V & n. It turns out that they are directly related to each other. As # of moles increases then V increases. So far weve compared all the variables except the amount of a gas (n). There is a lesser known law called Avogadros 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 Avogadros Law

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Which leads us to the ideal gas law – So far we have always held at least 1 of the variables constant. We can set up a much more powerful eqn, which can be derived by combining the proportions expressed by the previous laws. Which leads us to the ideal gas law – So far we have always held at least 1 of the variables constant. We can set up a much more powerful eqn, which can be derived by combining the proportions expressed by the previous laws. Ideal Gas Law

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

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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 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 Ideal Gas Constant (R)

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Because of the different pressure units there are 3 possibilities for our ideal gas constant R =. 0821 Latm molK If pressure is given in mmHg R = 62.4 LmmHg molK If pressure is given in kPa R = 8.314 LkPa molK If pressure is given in atm Ideal Gas Constant

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Using the Ideal Gas Law What volume does 9.45g of C 2 H 2 occupy at STP? What volume does 9.45g of C 2 H 2 occupy at STP? P P V V T T 1atm ? ? 273K R R n n =.3635 mol.0821 Latm molK 9.45 g 26 g

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

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A camping stove propane tank holds 3000g of C 3 H 8. 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? P P V V T T 303kPa ? ? 298K R R n n = 68.2 mol 8.31 LkPa molK 3000 g 44 g

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

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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? PV = nRT (1.00 atm) (1.00 L) n H2O = (.0821 L atm/mol K ) (573K) n H2O =.021257 mols

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Ideal Gas Law & Stoichiometry 2 mol H 2 O 2 mol H 2 = = 1mol H 2 22.4 L H 2 2H 2 + O 2 2H 2 O.021257 mol.476 L H 2

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Loose Ends of Gases There are a couple more laws that we need to address dealing with gases. –Daltons Law of Partial Pressures –Grahams Law of Diffusion and Effusion. There are a couple more laws that we need to address dealing with gases. –Daltons Law of Partial Pressures –Grahams Law of Diffusion and Effusion.

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Daltons 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. P T =P 1 +P 2 +P 3 +… What that means is that each gas involved in a mixture exerts an independent pressure on its containers walls

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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. 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. Daltons Law of Partial Pressure

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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. 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. Daltons Law of Partial Pressure

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Three of the primary components of air are CO 2, N 2, and O 2. In a sample containing a mixture of these gases at exactly 760 mmHg, the partial pressures of CO 2 and N 2 are given as P CO 2 = 0.285mmHg and P N 2 = 593.525mmHg. What is the partial pressure of O 2 ? Simple Daltons Law Calculation

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P T = P CO2 + P N2 + P O2 Simple Daltons Law Calculation 760mmHg =.285mmHg + 593.525mmHg + P O2 760mmHg =.285mmHg + 593.525mmHg + P O2 P O2 = 167mmHg

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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 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 Daltons Law of Partial Pressure

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P tot = P atmospheric pressure = P gas + P H 2 O Atmospheric Pressure Atmospheric Pressure The waters vapor pressure can be determined from a list and subtract- ed from the atmospheric pressure

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WATER VAPOR PRESSURES Temp (°C)Vapor pressure (kPa) 10.65176 5.87260 101.2281 151.7056 202.3388 253.1691 304.2455 355.6267 407.3814 459.5898 5012.344

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WATER VAPOR PRESSURES Temp (°C)Vapor pressure (kPa) 5515.752 60.19.932 6525.022 7031.176 7538.563 8047.373 8557.815 9070.117 9584.529 100101.32 105120.79

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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. Simple Daltons Law Calculation P H2O at 20.0°C= 2.3388 kPa We need to convert to mmHg.

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P T = P H2O + P O2 Simple Daltons Law Calculation 730mmHg = 17.5468 + P O2 P O2 = 712.5 mmHg 2.3388 kPa 760 mmHg 101.3 kPa P H2O = 17.5468 mmHg

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Grahams 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 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

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Grahams 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. Grahams 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. Grahams Law

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Kinetic energy can be calculated with the equation ½ mv 2 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: Kinetic energy can be calculated with the equation ½ mv 2 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:

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M represents molar mass v represents molecular velocity A is one gas B is another gas M represents molar mass v represents molecular velocity A is one gas B is another gas ½ M A v A 2 = ½ M B v B 2 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: 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:

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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 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 vAvA vAvA vBvB vBvB = = MBMB MBMB MAMA MAMA Rate of effusion of A = = Rate of effusion of B MBMB MBMB MAMA MAMA

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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. 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. Grahams Law

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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? Grahams Law Example Calc. Rate of effusion of A = = Rate of effusion of B MBMB MBMB MAMA MAMA

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

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Ch. 12 - Gases I. Physical Properties.

Ch. 12 - Gases I. Physical Properties.

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