FLORIDA INTERNATIONAL UNIVERSITY



CHM 3400 – Problem Set 1

Due date: Wednesday, January 16th

Do all of the following problems. Show your work. (NOTE: Conversion factors between different pressure units are given in Table 0.1, p 5 of Atkins. Values for R in various units are given in Table 1.1, p 16 of Atkins. Note that 1 dm3 = 1 L.)

1) The density of oxygen gas (O2, M = 32.00 g/mol) in a new oxygen gas cylinder, measured at T = 20.0 (C, is 0.1632 g/mL. Find the following:

a) The number density of oxygen molecules in the gas cylinder (in units of molecules/cm3).

b) The calculated pressure of gas in the cylinder, using the ideal gas law (Atkins, eq 1.2). Give your answer in units of atm, to four significant figures.

c) The calculated pressure of gas in the cylinder, using the van der Waals equation (Atkins, eq 1.23a). Give your answer in units of atm, to four significant figures. The van der Waals coefficients for O2 are a = 1.364 L2(atm/mol2, b = 0.0319 L/mol.

d) Which calculated value for pressure (ideal or van der Waals) do you expect to be closer to the true value for gas pressure for the conditions in the problem? Why?

2) At 100.0 (C and 16.0 kPa the mass density of phosphorus vapor is 0.6388 kg/m3. What is the molecular formula of phosphorus under these conditions?

3) The atmosphere of Mars is composed almost entirely of carbon dioxide (CO2, M = 44.01 g/mol). In this problem we will assume the atmosphere is pure CO2. At the surface of Mars typical values for pressure and temperature are p = 4.5 torr, T = 220. K. Find the following for these conditions of pressure and temperature:

a) crms, the rms average speed of a CO2 molcule in the atmosphere at surface level (Atkins, eq 1.15). Give your answer in units of m/s.

b) (, the mean free path of a CO2 molecule in the atmosphere at surface level (Atkins, eq 1.19). Give your answer in units of nm.

c) z, the collision frequency for a CO2 molecule in the atmosphere at surface level (Atkins, eq 1.19). Give your answer in units of collisions/s.

The value for ( for CO2 is given in Table 1.3, p 28 of Atkins. Note that if all of the quantities appearing in eq 1.15 and eq 1.19 are given in MKS units the values calculated using those equations will also have MKS units. Also, c, (, and z are related to one another by eq 1.18 of Atkins.

4) Consider a mixture of argon (Ar, M = 39.95 g/mol) and helium (He, M = 4.00 g/mol). The density of the mixture, measured at T = 22.8 (C and p = 791. torr, is ( = 0.662 g/L. What is XAr, the mol fraction of argon in the gas mixture? You may assume that argon and helium behave ideally for the conditions in the problem.

5) A “generic” phase diagram for a typical pure chemical substance is given in Fig 5.12, p 115 of Atkins.

a) What is the significance of the critical point in the phase diagram?

b) The van der Waals equation is the simplest equation of state that can be used to describe critical behavior for pure substances. Use the values for the van der Waals a and b coefficients for carbon dioxide (Table 1.5, p 34 of Atkins) to estimate the values for the critical constants for CO2 (pC, VC, and TC, see eq 1.24). Compare your values to the experimental values (pC = 72.85 atm, VC = 94.0 cm3/mol, and TC = 304.2 K), and briefly discuss the agreement between the experimental values and those predicted using the van der Waals coefficients.

6) The Maxwell distribution of molecular speeds in an ideal gas is given by eq 1.16 of Atkins. Starting with this equation find an expression for cmp, the most probable speed of a molecule in a gas. HINT: The most probable speed corresponds to the peak in a plot of number of molecules vs speed (see Atkins, Fig 1.8). Using the condition for an extreme point in a function you can find an expression for cmp.

Solutions.

1) Before we begin, we will convert the density of the gas (in g/mL) to the molar density (in mol/mL and in mol/L). If we let ( = m/V be the density of the gas, and M be the molecular mass of the gas, then

Vm = V = M = (32.00 g/mol) = 196.1 mL/mol = 0.1961 L/mol

n ( (0.1632 g/mL)

a) Using the result above, the number density of molecules in the gas is

N = 1 mol 6.022 x 1023 molecule = 3.071 x 1021 molecule/cm3

V 196.1cm3 mole

b) From the ideal gas law

p = nRT = RT = (0.08206 L(atm/mol(K) (293.15 K) = 122.7 atm

V Vm (0.1961 L/mol)

c) From the van der Waals equation ( a = 1.364 L2(atm/mol2 ; b = 0.0319 L/mol )

p = RT - a

(Vm - b) Vm2

= (0.08206 L(bar/mol(K) (293.15 K) - (1.364 L2(atm/mol2)

[ (0.1961 L/mol) - (0.0319 L/mol) ] (0.1961 L/mol)2

= 146.50 atm - 35.47 atm = 111.0 atm

The percent difference between the value calculated using the ideal gas law and that found using the van der Waals equation of state is

% difference = | 122.7 – 111.0 | . 100. % = 10.5 %

111.0

which is a significant difference.

d) We would expect the value calculated using the van der Waals equation to be closer to the true value for pressure than that found using the ideal gas law, because the van der Waals gas contains corrections for the approximations made in the ideal gas law.

2) If we assume that the vapor obeys the ideal gas law, then

n = p = 16.0 x 103 N/m2 = 5.156 mol/m3

V RT (8.3145 J/mol.K) (373.2 K)

The molecular mass is

M = (mass density) = 0.6388 x 103 g/m3 = 123.9 g/mol

(molar density) 5.156 mol/m3

If we assume the vapor molecules have the formula Pj, then (since M(P) = 30.97 g/mol)

j = 123.9 g/mol = 4.00 ( 4 The formula for the vapor molecules is P4.

30.97 g/mol

3) a) crms = (3RT/M)1/2 = [3(8.3145 J/mol(K)(220. K)/(44.01 x 10-3 kg/mol) ]1/2 = 353. m/s

b) NA = 6.022 x 1023 molecule/mol

( = 0.52 nm2 = 0.52 x 10-18 m2

p = 4.5 torr (1 atm/760 torr) (1.01325 x 105 Pa/atm) = 600.0 Pa

( = RT = (8.3145 J/mol(K) (220. K)

21/2NA(p 21/2 (6.022 x 1023 molecule/mol) (0.52 x 10-18 m2) (600.0Pa)

= 6.88 x 10-6 m = 6880. nm

c) z = c/( = (353. m/s) = 5.13 x 107 collisions/s

(6.88 x 10-6 m)

4) Assume you have 1.000 L of gas. Based on the density, the mass of gas you have is m = 0.662 g. The moles of gas may be found using the ideal gas law

n = pV = (791. torr) (1. atm/760. torr) (1.000 L) = 0.04286 mol gas

RT (0.08206 L(atm/mol(K) (295.95 K)

Let ntotal = nHe + nAr = 0.04286 mol be the total number of moles of gas. Then

nHe = ntotal – nAr

m = nHeMHe + mAr MAr = (ntotal – nAr)MHe + nArMAr = ntotalMHe + nAr(MAr – MHe)

or, solving for nAr

nAr = m – ntotalMHe = [ (0.662 g – (0.04286 mol) (4.00 g/mol) ] = 0.01365 mol Ar

(MAr – MHe) [ (39.95 g/mol) – (4.00 g/mol) ]

The mol fraction of argon in the gas is then

XAr = nAr = 0.01365 mol = 0.318

ntotal 0.04286 mol

5) a) The significance of the critical point is most easily seen by considering the critical temperature, TC. For a substance initially in the gas phase, the substance can be converted from gas to condensed phase (solid or liquid) by an isothermal reversible compression if T < TC. If T > TC an isothermal reversible compression will never result in a phase transition (that is, the properties of the substance will change continuously from those of a gas at low pressure to those of a liquid at high pressure, but without a phase transition ever taking place). pC and VC represent the values for pressure and temperature at the critical temperature on the liquid-gas phase boundary.

b) For CO2, a = 3.610 L2.atm/mol2, and b = 0.0429 L/mol. So (eq 1.24)

VC = 3 (0.0429 L/mol) = 0.1287 L/mol = 128.7 cm3/mol (experimental value is 94.0 cm3/mol)

TC = 8a = 8 (3.610 L2.atm/mol2)

27Rb 27 (0.08206 L.atm/mol.K)(0.0429 L/mol)

= 303.8 K (experimental value is 304.2 K)

pC = a = (3.610 L2.atm/mol2) = 72.65 atm (experimental value is 72.85 atm)

27b2 27 (0.0429 L/mol)2

The values for the critical temperature and pressure calculated using the van der Waals coefficients are within a few tenths of a percent of the experimental value. The value for VC calculated from the van der Waals coefficients is 37% higher than the experimental value. The van der Waals equation typically does a good job in predicting TC and pC, but usually overestimates the value for VC.

6) The Maxwell distribution for speeds in an ideal gas is given by the expression

f(s) ds = 4( (M/2(RT)3/2 s2 exp(-Ms2/2RT) ds

= A s2 exp(-Bs2) ds where A = 4( (M/2(RT)3/2

B = M/2RT

At an extreme point in f(s) the first derivative will be equal to zero. So

df(s)/ds = 0 = d/ds { A s2 exp(-Bs2) } = A {2s exp(-Bs2) + s2 exp(-Bs2) (- 2Bs) }

= 2As exp(-Bs2) { 1 – Bs2 }

The above will be equal to zero when the term in brackets is equal to zero. Therefore

1 – Bs2 = 0

1 = Bs2

s2 = 1/B, or s = cmp = (1/B)1/2 = (2RT/M)1/2

is the most probable speed for a molecule in an ideal gas.

Note that we have, strictly speaking, only found an extreme point for the function. To show that it is a maximum we could use the second derivative test (maximum if d2f(s)/ds < 0). However, based on our discussion of the Maxwell distribution in class we know the extreme point we have found is a maximum.

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