Chapter 5 Gases I. Pressure A. Properties of

Chapter 5 Gases I. Pressure A. Properties of

Chapter 5 Gases I. Pressure A. Properties of Gases Expand to completely fill their container Take the Shape of their container Low Density, much less than solid or liquid state Compressible Mixtures of gases are always homogeneous Fluid B. Pressure = total force applied to a certain area 1. larger force = larger pressure 2. smaller area = larger pressure 3. Gas pressure caused by gas molecules colliding with container or surface 4. More forceful or frequent collisions mean higher gas pressure F P A C. Measuring Pressure 1. Atmospheric Pressure is constantly present

a. Decreases with altitude because of less air b. Varies with weather conditions 2. Measuring pressure using a barometer a. Column of mercury supported by air pressure b. Longer mercury column = higher pressure c. Force of the air on the surface of the mercury is balanced by the pull of gravity on the column of mercury 3. Units of Gas Pressure a. atmosphere (atm) = 29.92 in Hg = 760 mm Hg b. Torr = mm Hg 1 atm = 760 torr c. Pascal (Pa) = 1 N/m2 1 atm = 101,325 Pa 4. Example: Convert 49 torr to atm and Pa 1atm 0.064atm 760torr 49torr 49torr 101,325Pa 6500Pa 760torr 5. Measuring the Pressure of a Trapped Gas Sample a. Use an manometer to compare to atmospheric pressure b. Open-end manometer i. if gas end lower than open end, Pgas = Pair + h ii. if gas end higher than open end, Pgas = Pair h II. Simple Gas Laws A.

Boyles Law = Pressure is inversely proportional to Volume (constant T, n) 1. PV = k or P1V1 = P2V2 (k = Boyles Law Constant) 2. As Pressure on a gas increases, the Volume decreases V = k(1/P) 3. Example: What is new V of 1.53L of SO2 at 5600Pa when changed to 15,000Pa? P1V1 5,600 Pa 1.53L P1V1 P2V2 V2 0.57 L 15,000Pa P2 B. Charles Law = V is directly proportional to T (constant P, n) V1 V2 1. V = bT b T2 2. Another way to write Charles Law T1 3. As the temperature decreases, the volume of a gas decreases as well

4. Absolute Zero a. Theoretical temperature at which a gas would have zero V and P b. 0 K = -273.2 C = -459 F (K = oC + 273) c. All gas law problems use Kelvin temperature scale! 5. Example: V = ? If 2.58L of gas at 15 oC is heated to 38 oC at same P? V1 V2 T2V1 38 273K 2.58L 311K 2.58L V2 2.79 L 15 273K 288K T1 T2 T1 V1 V2 a n1 n2 2 mol O 3 0.50 mol O 2 3 mol O 2 0.33 mol O 3

III. Ideal Gas Law A. By combining the constants from the 3 gas laws we can write a general equation 1. Each simple gas law holds something constant 2. To consider changes in P, V, T, n at the same time, we combine constants k Tn Tn V bT P, n an T, P kba R P T, n P P PV nRT 3. PV = nRT is called the Ideal Gas Law 4. R is called the gas constant a. The value of R depends on the units of P and V b. Generally use R = 0.08206Latm/Kmol when P in atm and V in L B. Only Ideal gases obey this law exactly 1. Most gases obey when P is low (< 1 atm) and T is high (> 0C) 2. An ideal gas is only a hypothetical substance 3. Constant conditions drop out of the equation to give simpler laws

a. If you hold n constant b. P1V1 P2 V2 PV PV nRT nR T T1 T2 C. Ideal Gas Law Problems 1. Example: How many moles of H2 gas if: 8.56L, 0 oC, 1.5atm? PV nRT n 1.5atm 8.56L 0.57mol PV RT 0.08206Latm 273K Kmol 2. Example: P = ? if 7.0 ml gas at 1.68 atm is compressed to 2.7 ml? PV nRT constant Reduces to Boyle's Law P1V1 1.68atm 7.0ml

P1V1 P2 V2 P2 4.4atm 2.7ml V2 3. Example: V = ? If 345 torr, -15 oC, 3.48L changes to 36 oC, 468torr? PV nRT PV PV PV nR constant Reduces to 1 1 2 2 T T1 T2 P1V1 P2 V2 T2 P1V1 309 K 345torr 3.48L V2 3.07 L 468torr 258K T1 T2 P2 T1 4. Example: V = ? If 0.35mol, 13oC, 568torr changes to 56oC, 897torr

PV nRT V nRT P VV V2 - V1 n 2 R 2 T 2 n 1R 1T1 P2 P1 0.35mol 0.08206Latm 329K 0.35mol 0.08206Latm 286K VV Kmol 897torr 1atm 760torr Kmol 568torr 1atm 760torr 8L 11L 3L D. Gas Stoichiometry 1. Molar volume = volume one mole of any gas occupies 2. Standard Temperature and Pressure (STP) are 0 oC and 1atm PV nRT V

nRT (1mol )(0.08206Latm / Kmol )(273.2 K ) 22.42 L P 1.000atm 3. Real gases differ only slightly from the ideal molar volume 4. Example: n = ? For 1.75L of N2 at STP? 1mol 0.0781mol 22.42 L 1.75L 5. Example: V(CO2) = ? If 152g CaCO3 decomposes to CaO+CO2? a. CaCO3(s) -------> CaO(s) + CO2(g) b. Assume ideal behavior for CO2 1molgas CaCO 3 1 mol CO 2 152 g CaCO 3 1.52 mol CaCO 3

1mol CaCO 3 100.09g CaCO 3 1.52 mol CO 2 22.42L 34.1 L CO 2 mol 1.52 mol CO 2 6. Example: a. 2.80 L of CH4 at 25 oC and 1.65atm b. 35.0L of O2 at 31 oC and 1.25 atm c. What is the volume of CO2 formed at 2.50 atm and 125 oC? d. CH4(g) + 2O2(g) -------> CO2(g) + 2H2O(g) e. Find Limiting reagent n(CH 4 ) 1.65atm 2.80 L PV 0.189 mol 0.08206 Latm / Kmol 298K RT n(O2 ) 1.25atm 35.0L

1.75mol 0.08206 Latm / Kmol 304 K f. Calculate the amount of CO2 product based on CH4 as limiting reagent 1 molCO 2 0.189mol CH 4 1 mol CH 4 0.189 mol CO 2 g. Use the ideal gas law to find the volume of this amount of CO 2 PV nRT V nRT (0.189mol)(0.08206Latm / Kmol)(398K ) 2.47 L P 2.50atm 7. Calculating Molar Mass of a Gas from density Example: m grams a. Molar Mass d

V mol b. Example: d = 1.95g/L at 1.50atm and 27 oC. Molar Mass = ? c. Assume we have 1.0L of this gas 1.95 g m 1L 1.95 g m dV V L PV (1.50atm)1L PV nRT n 0.0609mol RT (0.08206Latm / Kmol)300K d d. Calculate Molar Mass from its definition Molar Mass grams 1.95 g 32.0 g / mol mol 0.0609mol

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