Homework Problems Chapter 10 Homework Problems 22 24

Homework Problems Chapter 10 Homework Problems: 22, 24, 34, 40, 44, 48, 50, 56, 62, 66, 70, 81, 82, 88, 94, 106, 128

CHAPTER 10 Gases

Properties of Gases were the first state of matter extensively studied. This is because small changes in temperature and pressure lead to measurable changes in volume, and because gases, to a good first approximation, behave in a simple manner.

Units For Pressure = Force/Area The MKS unit for pressure is the Pascal 1 Pascal = 1 Pa = 1 N/m 2 1 Newton = 1 N = 1 kg. m/s 2 Other common units include 1 bar = 100, 000 Pa = 105 N/m 2 (exact) 1 atmosphere = 1 atm = 101, 325 Pa = 1. 01325 bar (exact) 1 atm = 760. Torr = 760. mm Hg (exact)

Barometer A barometer is a device for measuring atmospheric pressure. In a barometer pressure = force/area = hdg h = height (m) d = density (kg/m 3) g = gravitational constant (9. 807 m/s 2) The pressure unit atmosphere (atm) represent the approximate value for the pressure of the Earth’s atmosphere at sea level. The unit torr (mm Hg) comes from the most commonly used barometer for experimentally measuring pressure, the mercury barometer. 1 torr = 1 mm Hg (exact) d(Hg) = 13. 53 g/cm 3

Manometer A manometer is a device for measuring differences in pressure. In a mercury manometer h (in mm Hg) = pgas - patm

Boyle’s Law Consider an experiment where volume is measured as a function of pressure under conditions of constant temperature and amount of gas. Experimentally, the following is observed p. V = constant This observation is called Boyle’s law. pressure volume p. V (torr) (L. torr) 160. 0 100. 0 16000. 320. 0 50. 0 16000. 640. 0 25. 0 16000. 800. 0 20. 0 16000. 1200. 0 16000.

Charles’ Law Now consider an experiment where volume is measured as a function of temperature under conditions of constant pressure and amount of gas. Experimentally, a plot of V vs T is found to be linear. When different amounts of gas or pressures are used different lines are obtained, but all lines appear to intersect at approximately the same temperature, T - 273. C. If we define a new temperature scale, the Kelvin scale, with T(K) = T( C) + 273. 15 then V = (constant) T This observation is called Charles’ law.

Relationships From Boyle’s Law and Charles’ Law Useful relationships can be derived from Boyle’s law and Charles’ law. Boyle’s law. If n, T = constant, then p. V = constant. Therefore pi. Vi = pf. Vf Charles’ law. If n, p = constant, then V ~ T (or V = c. T, where c is a constant). From this, it follows that V/T = constant, or Vi/Ti = Vf/Tf. Example: A sample of gas at T = 300. K and p = 1. 00 atm occupies a volume V = 586. m. L. What volume will the gas occupy if the pressure is changed to 5. 00 atm while keeping temperature constant?

Example: A sample of gas at T = 300. K and p = 1. 00 atm occupies a volume V = 586. m. L. What volume will the gas occupy if the pressure is changed to 5. 00 atm while keeping temperature constant? Since n, T are constant, it follows that pi. Vi = pf. Vf Vf = Vi (pi/pf) = (586. m. L) (1. 00 atm/5. 00 atm) = 117. m. L

Avogadro’s hypothesis Based on the observed combining ratios for gas reactions, Amadeo Avogadro (1811) proposed the following hypothesis (Avogadro’s hypothesis) Equal volumes of any two gases at the same pressure and temperature contain the same number of molecules. Avogadro’s hypothesis was generally ignored until revived in the 1850 s by Stanislao Cannizzaro.

The Ideal Gas Law The above observations (Boyle’s law, Charles’ law, and Avogardo’s hypothesis) can be summarized in a single equation, called the ideal gas law. p. V = n. RT p = pressure n = moles of gas V = volume R = gas constant T = temperature All of the previous observations can be recovered from the ideal gas law. For example, if n and T are held constant, then p. V = n. RT = constant (Boyle’s law)

Comments 1) The ideal gas law is approximate. 2) The ideal gas law is an example of an equation of state, that is, a relationship among state variables (p, V, n, T). 3) The state variables can be divided into two types a) Extensive variables - Depend on the size of the system b) Intensive variables - Independent of the size of the system p. A = p. B VA = V B TA = T B n. A = n. B p = p. A = p. B V = VA + VB = 2 VA = 2 VB T = T A = TB n = n. A + n. B = 2 n. A = 2 n. B Vm = molar volume = V/n

4) The ideal gas law is based on two underlying assumptions a) The volume occupied by the gas molecules is small b) Attractive forces between gas molecules are small These assumptions become true in the following limits V (T, n held constant) p 0 (T, n held constant) T (p, n held constant) 5) The units for the gas constant, R, depend on the units used for p, V, and n. R = p. V/n. T can be used to determine the units. R = 8. 314 J/mol. K = (MKS unit) R = 0. 08314 L. bar/mol. K R = 0. 08206 L. atm/mol. K (1 J = 1 kg. m 2/s 2)

Sample Problems 1) A sample of N 2 gas with mass m = 8. 238 g is confined in a container of volume V = 2. 000 L at a temperature T = 30. 0 C. What is the pressure exerted by the gas? Give your answer in atm and torr. 2) A glass bulb has a mass m = 49. 618 g when empty and m = 51. 203 g when filled with an unknown pure gas. The volume of the bulb is V = 1. 283 L. The measurements are made at p = 0. 978 atm and T = 21. 5 C. What is the molecular mass of the gas?

1) A sample of N 2 gas with mass m = 8. 238 g is confined in a container of volume V = 2. 000 L at a temperature T = 30. 0 C. What is the pressure exerted by the gas? Give your answer in atm and torr. p. V = n. RT ; so p = n. RT M(N 2) = 28. 01 g/mol V n = 8. 238 g 1 mol = 0. 2941 mol T = 30. 0 C = 303. 2 K 28. 01 g p = (0. 2941 mol) (0. 08206 L. atm/mol. K) (303. 2 K) = 3. 659 atm 2. 000 L In units of torr p = 3. 659 atm 760 torr = 2781 torr 1 atm

2) A glass bulb has a mass m = 49. 618 g when empty and m = 51. 203 g when filled with an unknown pure gas. The volume of the bulb is V = 1. 283 L. The measurements are made at p = 0. 978 atm and T = 21. 5 C. What is the molecular mass of the gas? M = m/n mgas = mfilled - mempty = 51. 203 g - 49. 618 g = 1. 585 g p. V = n. RT ; so n = p. V RT n= T = 21. 5 C = 294. 6 K (0. 978 atm) (1. 283 L) = 0. 05190 mol (0. 08206 L. atm/mol. K) (294. 6 K) So M = m = 1. 585 g = 30. 5 g/mol n 0. 05190 mol

Density (d) is defined as d = m/V For solids and liquids density is approximately independent of temperature. For gases, density depends strongly on the conditions for which it is measured. Gas density is often reported at STP, standard temperature and pressure, which is T = 0 C = 273. 15 K p = 1. 000 atm Note that the molar volume for an ideal gas at STP is V = RT = (0. 08206 L. atm/mol. K) (273. 15 K) = 22. 41 L/mol n p (1. 000 atm)

Density and Molecular Mass The density of a gas can be related to its molecular mass. Since p. V = n. RT m = n. M n = m/M So p. V = m. RT M M = m. RT = m RT = d. RT p. V V p p So by measuring the density of a gas for known conditions of pressure and temperature, the molecular mass of the gas can be found. We can also rearrange the above equation to solve for other variables such as density, temperature, or pressure.

Reactions Involving Gases Since p. V = n. RT, then V = n. RT/p For gases at a particular pressure and temperature, n ~ V. We can therefore interpret a balanced chemical equation involving gases either in terms of moles of gas or volume of gas. For example 2 CO(g) + O 2(g) 2 CO 2(g) can be interpreted in two ways. CO 2. 1) 2 moles of CO will react with 1 mole of O 2 to form 2 moles of 2) 2 liters of CO will react with 1 liter of O 2 to form 2 liters of CO 2 (assuming the gases obey the ideal gas law, and are at the same temperature and pressure).

Dalton’s Law of Partial Pressures To this point we have discussed pure gases. For mixtures of ideal gases we can use Dalton’s law of partial pressures. For N gases p = ptotal = p 1 + p 2 + p 3 + … + p. N p 1 = partial pressure of gas 1; p 2 = partial pressure of gas 2; etc. Now since p. V = n. RT, p = n. RT/V, and so p 1 = n 1 RT/V ; p 2 = n 2 RT/V ; … ptotal = p 1 + p 2 + … + p. N ptotal = n 1 RT + n 2 RT + … + n. NRT V V V If we factor out (RT/V) then

ptotal = RT [ n 1 + n 2 + … + n. N ] = ntotal. RT V V Also, p 1 = n 1 RT = n 1 ntotal. RT = X 1 ptotal V ntotal V V where X 1 = n 1 is the mole fraction of gas 1 in the mixture ntotal Note that X 1 + X 2 + X 3 + … + XN = 1

Sample Problem A gas mixture is prepared by mixing together 10. 00 g of nitrogen (N 2, M = 28. 01 g/mol) and 10. 00 g of argon (Ar, M = 39. 95 g/mol). The gas is confined in a container with volume V = 20. 00 L. The total pressure of the gas mixture is ptotal = 0. 8488 atm. Find the following: 1) Partial pressure of N 2 and Ar in the gas mixture. 2) Temperature of the gas mixture.

A gas mixture is prepared by mixing together 10. 00 g of nitrogen (N 2, M = 28. 01 g/mol) and 10. 00 g of argon (Ar, M = 39. 95 g/mol). The gas is confined in a container with volume V = 20. 00 L. The total pressure of the gas mixture is ptotal = 0. 8488 atm. Find the following: 1) Partial pressure of N 2 and Ar in the gas mixture. 2) Temperature of the gas mixture. moles N 2 = 10. 00 g N 2 1 mol N 2 = 0. 3570 mol N 2 28. 01 g N 2 moles Ar = 10. 00 g Ar 1 mol Ar = 0. 2503 mol Ar 39. 95 g Ar Total number of moles of gas = 0. 3570 mol + 0. 2503 mol = 0. 6073 mol X(N 2) = 0. 3570 mol = 0. 5878 0. 6073 mol X(Ar) = 0. 2503 mol = 0. 4122 0. 6073 mol

From Dalton’s law, pi = Xi ptotal, so p(N 2)= (0. 5878)(0. 8488 atm) = 0. 4989 atm p(Ar) = (0. 4122)(0. 8488 atm) = 0. 3499 atm Finally, since p. V = n. RT, it follows that T = p. V/n. R so T = (0. 8488 atm)(20. 00 L) (0. 6073 mol)(0. 08206 L. atm/mol. K) = 340. 6 K

Kinetic Theory The above description of gas behavior is based on experimental observation. However, by use of kinetic theory, a description of the behavior of gases on a molecular level, the above equations can be derived. Assumptions: 1) Gases are composed of molecules in random motion. 2) The volume occupied by the molecules is small. 3) The interaction forces between molecules are weak. 4) Collisions between molecules, or between a molecule and the walls of the container, are elastic (total kinetic energy remains constant).

Average Speed and Kinetic Energy Based on the above assumptions the following equations can be derived for urms, the root mean square average speed and (KE)ave, the average kinetic energy of a gas molecule urms = [3 RT/M]1/2 (KE)ave = 3 RT/2 (per mole of gas) Note the following: 1) For a particular value of temperature all gases have the same value for average kinetic energy. 2) urms ~ T 1/2 urms ~ (1/M)1/2 So gas speds increase as temperature increases, and at a particular temperature light gases move faster (on average) than heavy gases.

Distribution of Molecular Speeds In a gas mixture there will be some molecules that are moving faster or slower than the average speed. The distribution of molecular speeds in a gas can be derived. This distribution, called the Maxwell. Boltzmann distribution, has the following appearance.

Temperature Dependence of the Maxwell-Boltzmann Distribution As temperature increases the peak in the Maxwell-Boltzmann distribution shifts to higher speeds, and the height of the distribution decreases.

Example: What is the average speed of an N 2 (MW = 28. 01 g/mol) molecule at T = 300. K and at T = 1000. K?

Example: What is the average speed of an N 2 (MW = 28. 01 g/mol) molecule at T = 300. K and at T = 1000. K? urms = [3 RT/M]1/2 At T = 300. K urms = [3 (8. 314 J/mol. K) (300. K)/ (28. 01 x 10 -3 kg/mol)]1/2 = 517. m/s (1160 mph) At T = 1000. K urms = [3 (8. 314 J/mol. K) (1000. K)/ (28. 01 x 10 -3 kg/mol)]1/2 = 944. m/s Notice we use R in MKS units (J/mol. K), and M in units of kg/mol.

Diffusion and Effusion Diffusion refers to the process of mixing of gases due to their random motion. Effusion refers to the escape of a gas to vacuum through a small hole. For both processes the rate of the process is proportional to u, the average speed of the gas molecules Therefore rate ~ (1/M)1/2

Comparison of Rates of Diffusion and Effusion Since for both effusion and diffusion rate ~ (1/M)1/2 It follows that the relative rate of either diffusion or effusion for two gases at the same conditions of temperature and pressure is (rate 2)/(rate 1) = (M 1/M 2)1/2 where M 1 = molecular mass of gas 1 M 2 = molecular mass of gas 2

Real Gases As previously discussed, the ideal gas law is an approximate description of the behavior of real gases. One way of showing this is by measuring the volume occupied by one mole of a real gas at STP (standard temperature and pressure, taken as T = 0 C, p = 1. 000 atm). For an ideal gas at STP the molar volume is 22. 414 L/mol.

Compressibility Factor (Z) The compressibility factor (Z) is defined as Z = p. V n. RT For an ideal gas Z = 1 for all pressures and temperatures. Deviations from this value for real gases can be taken as a measure of nonideal behavior.

Van der Waals Equation Various nonideal gas laws have been proposed to account for deviations from ideal behavior. A simple and useful equation is the van der Waals equation. We can obtain the equation as follows: From the ideal gas law p = n. RT V We can correct for excluded volume (volume occupied by other gas molecules) by modifying this equation p= n. RT (V – nb) where nb = “excluded volume”

Now consider the effect of attractive forces on the pressure For molecules in the middle of the container the attractive forces appear in all directions and tend to cancel. However, when a molecule approaches the walls of the container the attractive forces tend to slow the molecule down and thus lower the pressure.

We correct for this decrease in pressure by subtracting a term from the pressure whose value depends on the strength of the attractive forces and the molar density (n/V) of the gas. p= n. RT - an 2 (V - nb) V 2 a, b are constants the van der Waals equation. a coefficient - depends of strength of intermolecular attractive forces b coefficient - depends on size of molecules Values for a and b are different for different gases, and are found by fitting to experimental data.

Note that in general b increases as the size of the molecule increases, and a increases as the strength of intermolecular forces increases.

Example: What is the pressure of 1. 000 mole of ammonia (NH 3) at T = 300. 0 K and V = 10. 000 L according to the ideal gas law and according to the van der Waals equation (a = 4. 17 L 2. atm/mol 2; b = 0. 0371 L/mol)? Ideal gas law p = n. RT V van der Waals equation p = n. RT - an 2 (V - nb) V 2

Example: What is the pressure of 1. 000 mole of ammonia (NH 3) at T = 300. 0 K and V = 10. 000 L according to the ideal gas law and the van der Waals equation (a = 4. 169 L 2. atm/mol 2; b = 0. 0371 L/mol)? Ideal p = n. RT = (1. 000 mol) (0. 08206 L. atm/mol. K) (300. 0 K) V 10. 000 L = 2. 462 atm van der Waals p = (1. 000) (0. 08206 L. atm/mol. K) (300. 0 K) [ 10. 000 L - (1. 000 mol) (0. 0371 L/mol) ] - (4. 17 L 2. atm/mol 2) (1. 000 mol)2 (10. 000 L)2 = 2. 4710 atm - 0. 0417 atm = 2. 429 atm (or ~ 1. 3% lower)

So V(ideal) = 2. 462 atm V(van der Waals) = 2. 429 atm Which volume is correct? The result from the van der Waals equation is likely to be closer to the correct (experimental) value because the van der Waals equation usually does a better job of predicting pressure, volume, etc. than the ideal gas law. However, since all gas laws are approximate, it is incorrect to say that one result is wrong and the other result is right. It is better to say that the van der Waals equation does a better job of predicting these quantities than the ideal gas law.

End of Chapter 10 “. . . equal volumes of gases, at the same temperature and pressure, contain the same number of molecules. ” - Amadeo Avogadro “ Two attacks on Boyle’s work were immediately published, one by Thomas Hobbes…and the other by…Franciscus Linus. Hobbs based his criticism on the physical impossibility of a vacuum (“A vacuum is nothing, and what is nothing cannot exist”). Linus claimed that the mercury column (in the barometer) was held up by an invisible thread, which fastened itself to the upper end of the tube. The theory seemed quite reasonable, he said, for anyone could easily feel the pull of the thread by covering the end of the barometer tube with his finger. ” - John Moore, Physical Chemistry