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Normal atmospheric pressure at sea level is equal to: 1.00 atm 760 torr or 760 mm Hg 101.325 kPa 14.7 Psi mm Hg or torr: literally the difference in the.

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Presentation on theme: "Normal atmospheric pressure at sea level is equal to: 1.00 atm 760 torr or 760 mm Hg 101.325 kPa 14.7 Psi mm Hg or torr: literally the difference in the."— Presentation transcript:

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2 Normal atmospheric pressure at sea level is equal to: 1.00 atm 760 torr or 760 mm Hg kPa 14.7 Psi mm Hg or torr: literally the difference in the heights measured in mm (h) of two connected columns of mercury.

3 The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone. In other words, P total = P 1 + P 2 + P 3 + …

4 Gases consist of large numbers of molecules that are in continuous, random motion. The combined volume of all the molecules of the gas is negligible relative to the total volume in which the gas is contained. V of molecules=0 Attractive forces between gas molecules are negligible. In real life, there is always an attraction force between two objects having mass. Under standard condition (STP, P= 1 atm, T= room tempreture 298K), we may assume all the gases as ideal gas.

5 PV = nRT P = pressure of gas V = volume of gas n = moles of total gas molecules R = proportionality constant = L atm/ mol· T = temperature in Kelvins

6 The constant of proportionality is known as R, the gas constant. Its value depends on the units of P and V.

7 If P and T are fixed, the volume of gas is directly proportional to the number of moles of the gas. If all the reactants and products are gas, ratio of coefficients = ratio of V V = kn

8 If reactants and products are at the same conditions of temperature and pressure, then mole ratios of gases are also volume ratios. 3 H 2 (g) + N 2 (g) 2NH 3 (g) 3 moles H mole N 2 2 moles NH 3 3 liters H liter N 2 2 liters NH 3 How many liters of ammonia can be produced when 12 liters of hydrogen react with an excess of nitrogen?

9 V 1/P (If n and T are constant) V T (If n and P are constant) V n (If P and T are constant) Combining these, we get V nT P PV = nRT

10 We know that moles molecular mass = mass So multiplying both sides by the molecular mass ( ) gives n = m P RT mVmV = nVnV P RT =

11 Mass volume = density. So, Note: One only needs to know the molecular mass, the pressure, and the temperature to calculate the density of a gas. P RT mVmV = d =

12 We can manipulate the density equation to enable us to find the molecular mass of a gas: Becomes P RT d = dRT P =

13 Page 231: 37, 47, 53, 55, 57, 63, 71, 73,

14 The attractions between molecules are not nearly as strong as the bonds inside of compounds. But they are strong enough to control physical properties such as boiling and melting points, vapor pressures, and viscosities.

15 These intermolecular forces (the attraction forces) as a group are referred to as van der Waals forces.

16 Dipole-dipole interactions Hydrogen bonding London dispersion forces

17 A fourth type of force, ion-dipole interactions are an important force in solutions of ions. The strength of these forces are what make it possible for ionic substances to dissolve in polar solvents.

18 Molecules that have permanent dipoles are attracted to each other. The positive end of one is attracted to the negative end of the other and vice-versa. These forces are only important when the molecules are close to each other.

19 The more polar the molecule, the higher is its boiling point.

20 At that instant, then, the helium atom is polar Another helium nearby, then, would have a dipole induced in it, as the electrons on the left side of helium atom 2 repel the electrons in the cloud on helium atom 1.

21 London dispersion forces are attractions between an instantaneous dipole and an induced dipole. Synonyms: London forces, dispersion forces, and dispersion-interaction forces

22 These forces are present in all molecules, whether they are polar or nonpolar. But it plays a major role in the nonpolar molecules. The tendency of an electron cloud to distort in this way is called polarizability.

23 The shape of the molecule: long, skinny molecules (like n-pentane tend to have stronger dispersion forces than short, fat ones (like neopentane). This is due to the increased surface area in n-pentane.

24 The molecular weight: strength of dispersion forces tends to increase with increased MW. Larger atoms have larger electron clouds, which are easier to polarize.

25 The nonpolar series (SnH 4 to CH 4 ) follow the expected trend. The polar series follows the trend from H 2 Te through H 2 S, but water is quite an anomaly.

26 The dipole-dipole interactions experienced when H is bonded to N, O, or F are unusually strong. We call these interactions hydrogen bonds.

27 Hydrogen bonding arises in part from the high electronegativity of nitrogen, oxygen, and fluorine. Also, when hydrogen is bonded to one of those very electronegative elements, the hydrogen nucleus is exposed.

28 The types of bonding forces vary in their strength as measured by average bond energy. Ionic bonds Covalent bonds (400 kcal/mol) Ion- dipole interactions Hydrogen bonding (12-16 kcal/mol ) Dipole-dipole interactions (2-0.5 kcal/mol) London forces (less than 1 kcal/mol) Strongest Weakest

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32 Resistance of a liquid to flow is called viscosity. It is related to the ease with which molecules can move past each other. Viscosity increases with stronger intermolecular forces and decreases with higher temperature. Higher T faster molecules move

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34 Surface tension results from the net inward force experienced by the molecules on the surface of a liquid. Water surface due to surface tension Hg surface lack of surface tension

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36 Heat of Fusion: Energy required to change a solid at its melting point to a liquid. Heat of Vaporization: Energy required to change a liquid at its boiling point to a gas.

37 The heat added to the system at the melting and boiling points goes into pulling the molecules farther apart from each other. The temperature of the substance does NOT rise during the phase change.

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40 At any temperature, some molecules in a liquid have enough energy to escape. As more molecules escape the liquid, the pressure they exert increases. The liquid and vapor reach a state of dynamic equilibrium: liquid molecules evaporate and vapor molecules condense at the same rate.

41 As the temperature rises, the fraction of molecules that have enough energy to escape increases. VP increases.

42 The boiling point of a liquid is the temperature at which its vapor pressure = atmospheric pressure. The normal boiling point is the temperature at which its vapor pressure is 760 torr / 1 atm.

43 Phase diagrams display the state of a substance at various atmosphere pressures and temperatures and the places where equilibria exist between phases.

44 The AB line is the liquid-vapor interface. It starts at the triple point (A), the point at which all three states are in equilibrium.

45 It ends at the critical point (B); above this critical temperature and critical pressure the liquid and vapor are indistinguishable from each other. Each point along this line is the boiling point of the substance at that pressure. Supercritical substance

46 The AD line is the interface between liquid and solid. The melting point at each pressure can be found along this line.

47 Below A the substance cannot exist in the liquid state. Along the AC line the solid and gas phases are in equilibrium; the sublimation point at each pressure is along this line.

48 Note the high critical temperature and critical pressure: These are due to the strong van der Waals forces between water molecules.

49 The slope of the solid–liquid line is negative. This means that as the pressure is increased at a temperature just below the melting point, water goes from a solid to a liquid.

50 Carbon dioxide cannot exist in the liquid state at pressures below 5.11 atm; CO 2 sublimes at normal pressures. The low critical temperature and critical pressure for CO 2 make supercritical CO 2 a good solvent for extracting nonpolar substances (such as caffeine).

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