KINETIC MOLECULAR DESCRIPTION OF THE STATES OF MATTER

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1 KINETIC MOLECULAR DESCRIPTION OF THE STATES OF MATTER CHAPTER 9 The Gaseous State CHAPTER 10 Solids, Liquids, and Phase Transitions CHAPTER 11 Solutions 392 Gas Liquid Solid

2 9 THE GASEOUS STATE 9.1 The Chemistry of Gases 9.2 Pressure and Temperature of Gases 9.3 The Ideal Gas Law 9.4 Mixtures of Gases 9.5 The Kinetic Theory of Gases 9.6 Real Gases: Intermolecular Forces 395 Li + H 2 O LiOH + H 2

3 9.1 THE CHEMISTRY OF GASES CaCO 3 (s) + 2HCl(g) CaCl 2 (s) + CO 2 (g) + H 2 O(l) K 2 SO 3 (s) + 2HCl(g) 2KCl(s) + SO 2 (g) + H 2 O(l)

9.2 PRESSURE AND TEMPERATURE OF GASES Pressure 398 Torricelli s

399 Pressure : F mg mg m P gh gh A A V / h V = 13.

80665 m s 2 gravitational acceleration h = 76 cm height of mercury

4 9.2 PRESSURE AND TEMPERATURE OF GASES Pressure 398 Torricelli s barometer Force exerted by the mercury column at its base F = mg 399 Pressure : F mg mg m P gh gh A A V / h V = g cm 3 density of Hg(l) at 0 o C g = m s 2 gravitational acceleration h = 76 cm height of mercury column P kg m s Pa bar 1 atm 760 torr (at any temperature) o 760 mmhg (at 0 C)

5 Pressure and Boyle s Law 400 ~ Experiments on the compression and expansion of air. The spring of the Air and Its Effects (1661). Boyle s J-tube experiment Trapped air at the closed end of the J-tube: h (in mm) P 1 atm mm atm Add Hg and measure the volume of air (V): C 1 P C V V or PV C Robert Boyle (UK, ) C: a constant at constant T and fixed amount of gas 400 Fig. 9.3 Boyle s J-tube. (a) Same Hg height on two sides P confined = P air (b) Hg added Difference in Hg heights, h V confined compressed

intercept C on PV-axis) The value of C at 0 o C and for 1

6 401 (a) P vs. V hyperbola (b) P vs. 1/V straight line passing through the origin (slope: C) (c) PV vs. P straight line independent of P (parallel to P-axis, intercept C on PV-axis) The value of C at 0 o C and for 1 mol of gas, ~ good for all gases at very low pressure C = PV = L atm Temperature and Charles s law 402 V constant T (at constant n and P) o V t C 1 V0 t T V V01 V o C T(Kelvin) t(celsius) Jacques Charles (France,)

7 403 Fig. 9.5 Volume of a gas confined at constant P increases as T increases. 404 Fig. 9.6 The volume of a sample of a gas is a function of temperature at constant pressure.

405 Absolute Temperature Scale, K Kelvin, Lord William Thomson (UK, 1824-1907) 0 K : absolute zero temperature 273.16 K : triple point of water 0 K = - 273.

15 K Kelvin scale: absolute, thermodynamic temperature scale absolute zero temperature at which all thermal motion ceases in the classical description of

8 405 Absolute Temperature Scale, K Kelvin, Lord William Thomson (UK, ) 0 K : absolute zero temperature K : triple point of water 0 K = o C 0 o C = K Kelvin scale: absolute, thermodynamic temperature scale absolute zero temperature at which all thermal motion ceases in the classical description of thermodynamics. 9.3 THE IDEAL GAS LAW 405 Boyle s law: V 1/P (at constant T and n) Charles law: V T (at constant P and n) Avogadro s hypothesis: V n (at constant T and P) PV = nrt nt V P Equation of state Limiting law for real gases as P 0 Universal gas constant, R R = J K 1 mol 1 = x 10 2 L atm K 1

2NO 3- (aq) 2NO 2 (g) + Cu 2+ (aq) + 2H 2 O(l) Suppose that 6.80 g copper is consumed in this reaction, and that the NO 2 is collected at a pressure of 0.

9 408 EXAMPLE 9.5 Concentrated nitric acid acts on copper to give nitrogen dioxide and dissolved copper ions according to the balanced chemical equation Cu(s) + 4H + (aq) + 2NO 3- (aq) 2NO 2 (g) + Cu 2+ (aq) + 2H 2 O(l) Suppose that 6.80 g copper is consumed in this reaction, and that the NO 2 is collected at a pressure of atm and a temperature of 45 o C. Calculate the volume of NO 2 produced = =0.214 = = = MIXTURE OF GASES 408 Partial pressure (P i ) of the i th gas in a mixture of gases pressure that the i th gas would exert if it occupied the container alone

10 409 Dalton s Law of Partial Pressures The total pressure of a mixture of gases is the sum of the partial pressures of its component. P PA PB Pi i Mole fraction of the component A, x A na x A, xa xb 1 n n A B P nrt V P n n A A, nrt RT A B V np P A A V n n A B x P A P x P A A 409 EXAMPLE 9.6 When NO 2 is cooled to room temperature, some of it reacts to form a dimer, N 2 O 4, through the reaction 2NO 2 (g) N 2 O 4 (g) Suppose 15.2 g of NO 2 is placed in a 10.0 L flask at high temperature and the Flask is cooled to 25 o C. The total pressure is measured to be atm. What partial pressures and mole fractions of NO 2 and N 2 O 4 are present? = = = + = =0.500 = (0.500 )(10.0 ) ( )(298 ) =0.204 =., =.

9.5 THE KINETIC THEORY OF GASES 410 1. A gas consists of a collection of molecules in continuous random motion. 2. Gas molecules are infinitesimally small (mass) points. 3.

- Collision with walls: consider molecules traveling only in one dimensional x with a velocity of v x.

11 9.5 THE KINETIC THEORY OF GASES A gas consists of a collection of molecules in continuous random motion. 2. Gas molecules are infinitesimally small (mass) points. 3. The molecules move in straight lines until they collide. 4. The molecules do not influence one another except during collisions. - Collision with walls: consider molecules traveling only in one dimensional x with a velocity of v x. 411 The change in momentum of one molecule: 2mv x All the molecules within a distance v x t of the wall and traveling toward it will strike the wall during the interval t. If the wall has area A, all the particles in a volume Av x t will reach the wall if they are traveling toward it.

$411 The number of molecules in the volume Av x tis that fraction of the total volume V, multiplied by the total$ the volume Av x t: = 2 The total momentum change = number of collisions individual molecule change = = Force = rate

the volume Av x t: = 2 The total momentum change = number of collisions individual molecule change = = Force = rate

12 411 The number of molecules in the volume Av x tis that fraction of the total volume V, multiplied by the total number of molecules: = = The average number of collisions with the wall during the interval t is half the number in the volume Av x t: = 2 The total momentum change = number of collisions individual molecule change = = Force = rate of change of momentum = (total momentum change)/t 412 = = = = = for the average value of mean-square speed = + + = 3 = =

= 1 3 = 1 3 = 412 - Kinetic energy of N A molecules, = 1 2 = 3 2 1 3 = 3 2 - average kinetic energy per

Maxwell-Boltzmann distribution of speed = () with =4 =4 2 / 2 / / / 414 Boltzmann constant: k R/ N 1.

13 = 1 3 = 1 3 = Kinetic energy of N A molecules, = 1 2 = = average kinetic energy per molecule, = k B = R/N A - root-mean-square speed = = = M = molar mass = N A m u rms s of gases at 25 o C Maxwell-Boltzmann distribution of speed = () with =4 =4 2 / 2 / / / 414 Boltzmann constant: k R/ N J K B A 23 1 James Clerk Maxwell Ludwig Eduard Boltzmann (Scotland, ) (Austria, )

14 414 Fig A device for measuring the distribution of molecular speeds. 415

15 (1) Most probable speed: 416 = 2 = 2 () =0 (2) Average speed: = = = (3) Mean square speed: 8 = = 3 (4) Root-mean-square speed: = 3 = = 3 = vmp < < =... v vrms 9.6 REAL GASES: INTERMOLECULAR FORCES 417 Compression (or Compressibility) factor, Z V V PV Z V RT P RT m m m ideal m / - For an ideal gas, Z = 1. - Real gases, deviation from Z = 1 as P. Z < 1 for attractive force Z > 1 for repulsion

418 z = PV/nRT The Van der Waals Equation of State 418 Corrections to the ideal equation of state - Attraction at long distance: Reduction in collision frequency Reduction

16 418 z = PV/nRT The Van der Waals Equation of State 418 Corrections to the ideal equation of state - Attraction at long distance: Reduction in collision frequency Reduction in intensity of collision n P Pa V 2 ideal 2 - Repulsion at short distance : n/ V n/ V No overlap of molecules excluded volume effect Reduction in free volume n V V bn ideal

17 419 Van der Waals equation: 2 n Pa 2 V nbnrt V a: atm L 2 mol -2 b: L mol -1 R: L atm mol -1 K -1 - = = = 1 1 Repulsive forces (through b) increase z above 1. Attractive forces (through a) reduce z. - Constant b is the volume excluded by 1 mol of molecules and should be close to V m, the volume per mole in the liquid state. 419

Below are the van der Waals constants (a and b) for three gases: A, B and C. Gas a / atm L mol 2-2 b / L mol -1 A 3.592 0.04267 B 6.714 0.05636 C 5.284 0.

18 Below are the van der Waals constants (a and b) for three gases: A, B and C. Gas a / atm L mol 2-2 b / L mol -1 A B C Identify the gases CO 2, NO 2 and SO 2 as A, B or C, giving brief reasons. CO 2 A - smallest value of constant a (nonpolar molecule) NO 2 C - high value of constant a (polar molecule); similar value of constant b to CO 2 (similar size) SO 2 B - high value of constant a (polar molecule); highest value of constant b (biggest molecule) (a) Using the ideal gas law, calculate the pressure exerted by 50.0 g carbon dioxide in a 1.00 L vessel at 25 o C. (b) Using the van der Waals equation, calculate the pressure under the above condition. For CO 2, a = atm L 2 mol -2 and b = L mol -1. (c) Do attractive or repulsive forces dominate in CO 2? (a) (b)

19 Intermolecular Forces 421 Lennard-Jones Potential: 12 6 VLJ ( R) 4 R R where is the depth and is the distance at which V(R) passes through zero. 421

Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten. Chapter 10. Gases.

Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 10 Characteristics of Unlike liquids and solids, they Expand to fill their containers.