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Units of Measure for Gases 1 atm.= 760 torr = 760 mmHg =101,325Pa, or ~1 x 10 5 Pa.

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Presentation on theme: "Units of Measure for Gases 1 atm.= 760 torr = 760 mmHg =101,325Pa, or ~1 x 10 5 Pa."— Presentation transcript:

1 Units of Measure for Gases 1 atm.= 760 torr = 760 mmHg =101,325Pa, or ~1 x 10 5 Pa

2 Unlike liquids and solids, gases take up the volume of their container. Atmospheric pressure is measured with a barometer. Atmospheric pressure is the measure of the weight of air. (not mass). It is the measure of the gravitational pull on air. At higher altitudes atm. Pressure is lower because air is thinner. Think of it as the heavier molecules sink, or there are less of them at higher altitudes.

3 P = F/A A is area, F is Force = ma, (is mass x acceleration), a = 9.8 m/s 2 ) Since Force = (mass x a), then Kg m / s 2 = units for force = 1 Newton (N) And since Area, A = m 2 then P = N / m 2 which is 1 Pascal (Pa)

4 Gas Laws Boyle’s Law – shows that pressure and volume are inversely proportional, at constant temperature This means that PV = constant, i.e., as pressure goes up volume goes down and their product is a constant. From this we get the equation:P 1 V 1 = P 2 V 2 This relationship is only true under low pressure conditions. Under these conditions a gas is referred to as an IDEAL GAS. At high pressures gasses deviate from ideal behavior and are referred to as REAL GASSES, and PV = constant.

5 Charles’ Law Charles’ Law – Volume is proportional to temperature at constant pressure From this relationship we arrive at the equation: V 1 / T 1 = V 2 / T 2

6 Avagadro’s Law At constant temperature and pressure the volume of a gas is directly proportional to the # of moles of gas i.e., at the same temperature and pressure, equal volumes of gasses have the same number of moles. From this relationship the following equation is derived: V 1 / n 1 = V 2 / n 2

7 The Ideal Gas Law From Boyle’s Law : P V = constant Charles’Law: V / T = constant Avogadro’s Law: V / n = constant The equation PV = nRT can be derived. PV = nRT R = the universal gas constant = L atm. / mol K

8 The Combined Gas Law This equation is derived from the Ideal Gas Law: P 1 V 1 / T 1 = P 2 V 2 / T 2 For all gas laws the temperature must be in Kelvins.

9 Examples: In the laboratory an instrument called a manometer is used to measure pressure. Problem 9, page 224 deals with a manometer. The difference in height of the liquid in the U-tube is added to the atmospheric pressure to determine the pressure in the flask, if the pressure in the flask is greater. It is subtracted if the pressure in the flask is less than the outside pressure.

10 Molar Mass of a Gas The Ideal gas law can also be used to calculate the molar mass of a gas as follows: g R T PV or d R T P

11 Gas Stoichiometry Gas stoichiometry is the same, except to determine volume at STP, we can use the relationship 22.4 L / mole. If the conditions are NOT STP then we must use the relationship n gas = PV/ RT and V = nRT / P See sample exercise 5.13 in text

12 Dalton’s Law of Partial Pressures The total pressure of a mixture of different gasses is the sum of the pressures that each gas would exert if it were alone. P total = P 1 + P 2 + P 3 +…= n 1 RT + n 2 RT + n 3 RT +… V V V …and since RT and V are constant: P total = n total (RT / V)

13 Mole Fraction of a Gas The mole fraction is the ratio of the moles of a particular gas in a mixture to the total number of moles of gas in the mixture. It can be calculated using the following equation: X i = n i / n total = P i / P total Therefore the partial pressure of a gas, P i can be calculated: P i = X i P total

14 The Kinetic Theory of Gases This theory is used to explain the similarities in the behavior of all gases. 1.Gases consist of particles (atoms or molecules) in continuous, random motion. They undergo frequent collisions with each other and with the walls of the container. The pressure associated with a gas is due to the forces from these wall collisions.

15 2.Collisions between gas particles are elastic. That is, when two particles collide, no energy is lost, nor converted to heat. One particle may speed up while the other slows down, but no energy is lost. 3.The average Energy of Translational Motion of a gas particle is directly proportional to temperature. That is, the energy of motion of a particle from one place to another is related to its speed by: E t = mu 2 / 2, where E = kinetic energy of translation, m = mass of particle, u = velocity.(equation from KE=1mv 2 /2) 4.Volume of gas particles is negligible compared to the container it is in. 5.Attractive forces between gas particles are neglected. Note: KE of a gas is directly proportional to its Kelvin Temp.

16 Gas Pressure Relationships At constant V,T – Pressure increases as n increases (more particles = more collisions) At constant n,T – pressure increases as volume decreases (smaller volume = particles striking walls more often) At constant n,V - Pressure increases as Temperature increases (higher temperature = faster movement of particles, thus more collisions and with greater force)

17 The Root Mean Square Velocity for a gas particle U = (3RT / M) 1/2 units = m/s, cm/s, etc. M = molar mass of the gas Derivation for this equation is in Appendix 2 and on pages Note: look at your table for R values and you will see that for cm/s, R = 8.31 x 10 7 g cm 2 /s 2 mol K

18 Example: Find the average speed of an oxygen molecule in air at room temperature, (25 o C). 25 o C = T = 298 K, MM O 2 = 32.0 g/mol U = (3RT/MM) 1/2 U = ( ( 3(8.31 x 10 7 g cm 2 /s 2 mol K)x 298K ) ) 1/ g / mol U = 4.82 x 10 4 cm / s = 482 m / s Convert this velocity to miles / hour

19 Graham’s Law of Effusion Effusion is the flow of gas particles through a small opening or pinhole. Rate of effusion is directly proportional to the average velocity of the particles, so: Rate of effusion of A = u A Rate of effusion of B u B

20 Therefore, u A / u B = (3RT/M A ) 1/2 / (3RT/M B ) 1/2 And so rate of effusion of A = M B 1/2 = time B rate of effusion of B M A time A Note: Heavy molecules take longer to effuse.

21 Example: In an effusion experiment it required 45 s for an unknown gas, X, to pass through a small opening into a vacuum. Under the same conditions, it took 28 s for the same number of moles of Ar to effuse. Find the molar mass of the unknown gas. Time B / Time A = (MM B / MM A ) 1/2 28 s / 45 s = (39.9 g/mol / MM x ) 1/2 (28 s / 45 s) 2 =39.9 g/mol / MM x MM x = 1.0 x 10 2 g / mol

22 Real Gases and the van der Waals Equation Real gases approach “ideal” behavior at low pressures and high temperatures. The van der Waals equation allows us to calculate for real gas conditions. As a partial derivation, we start with the Ideal Gas Law. We then have to make some corrections.

23 To get the Real Volume, we have to realize that the volume of the container is not truly the volume of the gas. The gas takes up some of that volume. The correction factor for Volume is nb, where n = # of moles of gas and b is a constant that has been calculated from experimental results for individual gases. To make the correction for Volume: V-nb is used instead of V

24 The pressure must also have a correction factor to allow for small attractive forces that occur in a real gas. P observed = P- a (n/V) 2 where n/V = moles of gas particles / Liter and ‘a’ is a proportionality constant. This correction of ‘n / V’ is because pressure depends on the concentration of the particles, i.e., moles / volume.

25 The final corrected equation is: P = nRT– a n 2 V-nb V There is actually a longer version of this equation but it is not addressed in your book.

26 Example: # 55 Calculate the pressure exerted by mol N 2 in a L container at 25 o C. a N = 1.39 atm L 2 / mol 2 b N = L / mol T = 298 K n = mol V = 1.000L


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