Gaseous The State
Properties of Gases
All gases behave similarly: Expand to fill the volume of the container Highly compressible Form homogeneous mixtures (no phase separation) Molecules are far apart Act as if the other molecules are not there Most gases are comprised of small molecules of nonmetallic elements (e.g. HCl, Cl2, F2, N2, NH3, etc.) All gases are molecular
Pressure
Pressure: is the force exerted per unit area P = Force/unit area Units of Pressure: 1 atmosphere = 14.7 psi 1 atmosphere = 760 mm Hg 1 atmosphere = 101,325 Pascals 1 Pascal = 1 kg/m.s2
CPS: Pressure Units
Mount Everest, at 29,028 feet above sea level, is the world's tallest mountain. The normal atmospheric pressure at this altitude is about 0.308 atm. Converting this pressure to mm Hg, 0.308 atm is: 117 mm Hg. 175 mm Hg. 209 mm Hg. 234 mm Hg.
A. B. C. D.
Solution: Pressure Units
Mount Everest, at 29,028 feet above sea level, is the world's tallest mountain. The normal atmospheric pressure at this altitude is about 0.308 atm. Converting this pressure to mm Hg, 0.308 atm is:
PHg = 0.308 atm = 234 mmHg 760 mmHg 1 atm
Measuring Pressure: Barometers
Atmospheric pressure is balanced by the pressure of the mercury column in the glass tube. The pressure units torr (OR mmHg) are determined by measuring the height difference (h) between the Hg in the column and the Hg pooled beneath the column.
Some Terms To Memorize
Standard temperature and pressure (STP): 0C (273.15 K) and 1 atm Room temperature and pressure (RTP): 25C (298.15 K) and 1 atm
The Gas Laws
Boyle's Law: The volume of a gas maintained at constant temperature is inversely proportional to the pressure:
V 1/P
(constant moles and T) or
Boyle's Experiment:
Charles's Law: The volume of a gas maintained at constant pressure is directly proportional to the absolute temperature
V Tabs
(constant moles and P) or
Vf Tf
=
Vi Ti
Blue balloon dipped in liquid nitrogen
Blue balloon allowed to return to room temperature
CPS: Charles Law
At a laboratory party, a helium-filled balloon with a volume of 2.00 L at 22.0C is dropped into a large container of liquid nitrogen (T = 196.0C). The final volume of the gas in the balloon is: A. B. C. D. 0.522 L. 0.734 L. 1.044 L. 17.82 L.
Solution: Charles Law
At a laboratory party, a helium-filled balloon with a volume of 2.00 L at 22.0C is dropped into a large container of liquid nitrogen (T = 196.0C). The final volume of the gas in the balloon is: Vi = 2L; Ti=295.0 K; Tf=77.0 K
2.00L = 77K 295K 2.00L 77K Vf = 295K V f = 0.522L Vf
Avogadro's Law: Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules V = const. x n At STP, the molar volume, Vm, that is, the volume occupied by one mole of any gas, is 22.4 L/mol So, the volume of a sample of gas is directly proportional to the number of moles of gas, n.
V
n
The Ideal Gas Law
The ideal gas law combines all the previous observations into one equation: PV = const.; V = const. x T; V = const. x n PV = nRT The last equation is called the ideal gas equation and it relates the pressure (P), volume (V), composition (n), and temperature (T) of a gas R is the gas constant: R = 8.314 J K-1mol-1 = 0.08206 L atm mol-1K-1
Example
Compare the volumes 1 mol of H2O (l) (M=18.02 g mol-1; =0.9584 g cm-3) and H2O (g) at 100C and 1 atm pressure. Assume that H2O (g) is ideal. ANS:
= 0.01880 L
= 30.6 L
What does this imply about the volume occupied by a gas?
CPS: Ideal Gas Law
An empty aerosol spray-paint can has a volume of 0.406 L and contains 0.025 mol of a propellant gas such as CO2. The pressure of the gas at 25.0C is: A. B. C. D. 0.13 atm. 1.3 atm. 1.5 atm. 1.5X102 atm.
Solution: Ideal Gas Law
An empty aerosol spray-paint can has a volume of 0.406 L and contains 0.025 mol of a propellant gas such as CO2. The pressure of the gas at 25.0C is:
nRT 0.025 mol .08206 L atm mol-1 K -1 298K P= = V 0.406L = 1.5 atm
Solving Ideal Gas Law (IGL) Problems
Many of the IGL calculations in this course will involve one or more of the variables remaining fixed. In these cases, we can use the fixed variable(s) as a "handle" for solving the problem e.g. The 3.29 mol of ideal gas are contained inside a cylinderand-piston system under an external pressure 1.01 atm pressure (which is balanced by the pressure of the gas). Calculate the volume change as the system is heated from 300 K to 500 K.
n and P are constant V increases as T (Charles's Law)
V = Vf - Vi Vi = nRTi/P = (3.29)(0.08206)(300) (1.01) = 80.2 L by Charles's Law, Vf = Vi(Tf/Ti) Vf = (80.2 L)(500/300) = 134.6 L V = Vf - Vi = 134.6 - 80.2 = 53 L
1.01 atm 1.01 atm 1.01 atm 1.01 atm 1.01 atm 1.01 atm 1.01 atm 1.01 atm 1.01 atm 1.01 atm
T = 300 K 450 400 350 500
e.g. A weather balloon ascends to an altitude at which T=250K and P=0.45 atm. If the balloon had an initial volume of 7.2 L at sea level (T=293 K; P=1.0 atm), what is the volume at the high altitude? ANS: three of the variables change, but one (n) is constant: ni = nf PiVi/(RTi) = PfVf/(RTf) Vf = ViPiTf/(PfTi) Vf = (7.2 L)(1.0 atm)(250 K) (0.45 atm)(293 K) Vf = 14 L
Application: Finding Molar Mass (M)
A 536-mL glass bulb is filled with a pure gas at 294 K and 1.01 atm pressure. The mass of the gas is 1.59 g. What is the molar mass? ANS: PV = nRT and M = m/n (or n=m/M) PV = mRT/M Rearranging for M: M = mRT/(PV)
M = (1.59 g)(0.08206 Latm/mol/K)(294 K) (1.01 atm)(0.536 L) = 70.9 g mol-1
CPS: Finding Molar Mass (M)
A 25.0-mL glass bulb is filled with a pure gas at 25C and 380 mmHg pressure. The mass of the gas is 42.8 mg. What is the molar mass? A. B. C. D. 41.9 g/mol 83.7 g/mol 167 g/mol 7.02x103 g/mol
Solution: Finding Molar Mass (M)
A 25.0-mL glass bulb is filled with a pure gas at 25C and 380 mmHg pressure. The mass of the gas is 42.8 mg. What is the molar mass? P = 0.500 atm; T = 298 K; V=0.025 L; m =0.0428 g
M = (0.0428 g)(0.08206 Latm/mol/K)(298 K) (0.500 atm)(0.0250 L) = 83.7 g mol-1
Finding M from Gas Density
In the previous example we substituted the mass and molar mass for n to get M = mRT/(PV) Note that m/V = (the density of the gas) M = RT/P Molar data mass can be converted to density data and vice versa
Example
Air is a mixture of N2, O2, Ar, and other gases; therefore, it does not have a well-defined molar mass. However, the average molar mass of air is approximately 29.0 g/mol. What is the density of air at room temperature and pressure? = MP/(RT) = (29.0 g/mol)(1.00 atm) . (0.08206 L atm mol-1 K-1)(298 K) = 1.19 g/L
Dalton's Law of Partial Pressures
The total pressure of a mixture of gases is equal to the sum of the pressures that the components would exert individually if separated. For a mixture of N components, P = P1 + P2 + P3 + ... + PN For ideal gases, P1 = n1RT/V, etc. P = (n1+n2+n3+...+nN)RT/V
Example
What is the total pressure of 10.0 g of O2 and 6.0 g of C2H4 in a 12.0 L bulb at 298 K? ANS: MO2 = 32.0 g/mol; MC2H4 = 28.1 g/mol nO2 = (10.0 g)/(32.0 g/mol) = 0.312 mol nC2H4 = (6.0 g)/(28.1 g/mol) = 0.213 mol P = (nO2 + nC2H4)RT/V P = (0.312 + 0.213)(0.08206)(298)/(12.0) = 1.1 atm
CPS: Partial Pressures
What is the total pressure of a mixture of 1.570 mol of CO and 0.870 mol NO2 in a 25.0 L container at 25 C? A. B. C. D. 0.200 atm 1.54 atm 2.39 atm 242 atm
Solution: Partial Pressures
What is the total pressure of a mixture of 1.570 mol of CO and 0.870 mol NO2 in a 25.0 L container at 25 C?
(1.570 + 0.870) 0.08206L atm mol - 1K - 1 298K P= 25.0L = 2.39 atm
Collecting Gases Over Water
Collecting Gases Over Water
Solid in tube decomposes Water levels inside and outside vessel are equalised before measurement
Collection vessel contains both decomposition gas and water vapour which must be subtracted for accurate measurement
Example
A sample containing CaCO3 is decomposed under high temperature, producing CO2 which is collected over water. The volume of gas collected is 34.1 mL at 22C and 756 torr atmospheric pressure. How many moles of CO2 were produced? What mass of CaCO3 was present in the original sample? (PH2O = 19.8 torr)
Answer
We note that the total pressure of the gas is 756 torr (because water levels are equal) PH2O + PCO2 = 756 torr PH2O = 19.8 torr PCO2 = 736 torr Now, 760 torr = 1 atm PCO2 = 0.968 atm nCO2 = PCO2V/(RT) (0.968 atm)(34.1x10-3 L) . (0.08206 L atm mol-1 K-1)(295 K) = 1.36x10-3 mol =
For the second part, we need a balanced chemical equation: CaCO3 (s) CaO (s) + CO2 (g) Therefore, 1 mol of CO2 is produced from 1 mol of CaCO3 nCaCO3 = nCO2 x (1 mol CaCO3/1 mol CO2) = 1.36x10-3 mol mCaCO3 = (1.36x10-3 mol)(100.1 g/mol) = 0.136 g
CPS: Gases Over Water
A 41.4 mL sample of a unknown pure gas is collected over water at 25.0C and 769 torr atmospheric pressure. What is the partial pressure of the unknown gas? (PH2O = 23.77 torr) A. B. C. D. 0.03 atm 0.98 atm 1.01 atm 1.14 atm
Solution: Gases Over Water
A 41.4 mL sample of a unknown pure gas is collected over water at 25.0C and 769 torr atmospheric pressure. What is the partial pressure of the unknown gas? (PH2O = 23.77 torr) Pgas = (769 torr - 23.77 torr)/760 torr atm-1 = 0.98 atm
Kinetic Molecular Theory
Gases consist of molecules that are in continuous, random motion The volume of the molecules is negligible compared to the volume of the gas Intermolecular forces are negligible The average kinetic energy of the molecules does not change with time if temperature is constant The average kinetic energy of the molecules is proportional to the temperature. All gases at a given temperature have the same average kinetic energy regardless of chemical identity.
Molecular Picture of Gas Pressure
Momentum transfer: particle p = -px -px = -2px wall p = +2px Force: over a time interval, t, momentum, <p>, is transferred to the wall from large number of collisions F = <p>/t Pressure: P = F/A
px p py -px py p
Molecular Speeds
Speeds of molecules depend on Temperature: speed increases with T
Molecular Speeds
Speeds of molecules depend on Mass: speed decreases with mass
CPS: Kinetic Molecular Theory
Which of the following second period gases would have molecules traveling at the slowest average speed at 25C: A. N2 B. O2 C. F2 D. All these gases will have the same average speed
Effusion
Effusion: escape of gas molecules through a hole in the wall of a vessel into an evacuated space Rate is higher for molecules of low mass Rate increases with temperature e.g. Helium escapes through the pores of a balloon faster than air does (at a given T) Can be used to separate gases of different mass
How do we understand this? Effusion rate depends on the number molecules that "collide" with the hole in the wall of the vessel per unit time Faster molecules have more collisions per unit time
Diffusion
Diffusion is the spread of one substance throughout space (or another substance) Rate is higher for higher T, lower mass--same as for effusion. Why? Molecular speeds at room temp are ~ 500 m/s, yet diffusion takes seconds to minutes. Why? Intermolecular collisions slow diffusion Average distance between collisions is called "mean free path" (MFP) MFP is ~ 60 nm for N2 at room temp. and pressure
Example of Molecular Motion
Real Gases
Ideal gases are theoretical constructs. Their molecules have the following properties: They occupy no space. The volume of a gas is due to the random motions of the point masses They do not attract or repel one another For ideal gases,
PV =1 nRT
Real gases differ from ideal gases on all these points, but approximate IGL at high T and low P
Properties of Real Gases
Negative deviations: gas vol. lower than expected, attractive forces Positive deviations: gas vol. higher than expected; molecular volumes
Effects of Temperature on N2 Gas
Summary
Gas is a state of matter comprised of molecules that interact little and move randomly in space Most (>99%) of a gas is empty space Gases at normal temperatures and pressures (> 273K and < 10 atm) can be described (approximately) by the IGL and Dalton's Law The IGL can be used in conjunction with stoichiometry to get quantitative information about a sample (e.g. collecting gases over water) Real gases deviate from ideal behaviour because molecular interactions and volumes are non-zero