Solution manual for physical chemistry 2nd edition by ball

Solution Manual for Physical Chemistry 2nd Edition by Ball Full file at https://TestbankDirect.eu/Solution-Manual-for-Physical-Chemistry-2nd-Edition-by-Bal Chapter Gases and the Zeroth Law of Thermodynamics 1.2 A system is any part of the universe under observation The “surroundings” includes everything else in the universe Consider a solution calorimeter in which two aqueous solutions are mixed and temperature changes are recorded In this case the solutes (reactants and products) would be considered the “system” The water, calorimeter, and rest of the lab and world would be the “surroundings”   1.4 1000 mL cm   12,560 cm  1.256  10 cm (a) 12.56 L  1L mL (b) 45ºC + 273.15 = 318 K (c) 1.055 atm  1.01325 bar 100,000 Pa   1.069  105 Pa atm bar (d) 1233 mmHg  torr atm 1.01325 bar    1.644 bar mmHg 760 torr atm cm  125 cm (e) 125 mL  mL (f) 4.2 K – 273.15 = –269.0°C (g) 25750 Pa  bar  0.2575 bar 100,000 Pa   1.6 patm = pmouth + hg, where hg correspond to the pressure exerted by the liquid patm - pmounth = hg = (1.0x103 kg/m3)(0.23m)(9.80m/s2) = 2254 N/m2 = 2.3 x103 Pa 1.8 In terms of the zeroth law of thermodynamics, heat will flow from the (hot) burner or flame on the stove into the (cold) water, which gets hotter Then heat will move from the hot water into the (colder) egg 1.10 For this sample of gas under these conditions, F(T) = 2.97 L  0.0553 atm = 0.164 Latm If the pressure were increased to 1.00 atm: 0.164 Latm = (1.00 atm)  V; therefore V = 0.164 L   1.12 V  R pV nT , which rearranges to R  nT p Therefore: R    (2.66 bar)(27.5 L) L  bar  0.0830 (1.887 mol)(466.9 K) mol  K   Full file at https://TestbankDirect.eu/Solution-Manual-for-Physical-Chemistry-2nd-Edition-by-Bal Solution Manual for Physical Chemistry 2nd Edition by Ball Full file at https://TestbankDirect.eu/Solution-Manual-for-Physical-Chemistry-2nd-Edition-by-Bal Instructor’s Manual 1.14 V1=67 L, p1=1.04 atm x atm atm ; x = 6.34 atm Therefore p2 = 1.04 atm + 6.34 atm = 7.38 atm  64.0 m 10.1 m p1V1 = p2V2; so V2 = 1.16 1.82  1013 g SO  nRT  p V (1.04 atm)(67 L)  L (7.38 atm) moleSO  2.84  1011 molesSO 64.07 g L  atm )(273.15  17.0K ) mol  K   10 10 atm 21  10 L (2.84  1011 molesSO )(0.0821 L  atm L  atm / K    0.0456 mol  K R mol  R 1.18 0.0821 1.20 Calculations using STP and SATP use different numerical values of R because the sets of conditions are defined using different units It’s still the same R, but it’s expressed in different units of pressure, atm for STP and bar for SATP   1.22 The partial pressure of N2 = 0.80  14.7 lb lb  11.8 2 in in The partial pressure of O2 = 0.20  14.7 lb lb  2.9 2 in in 1.24 4.50torr(44.01g / mole)  0.01706 g / L  1.71 10 5 g / cm3 m pM m pV   RT;   L  torr V RT M (62.36 )(273.15  87K) mol  K   1.26 Using the ideal gas law, the number of moles of CO2 = (.965 atm)(1.56 L) = 0.0621 moles L·atm (0.0821 )(273.15K+22.0K) K·mol mole C6 H12 O6 180.16 g × = 5.60 g C6 H12 O6 0.0622 moles CO2 × mole C6 H12 O6 moles CO2 Full file at https://TestbankDirect.eu/Solution-Manual-for-Physical-Chemistry-2nd-Edition-by-Bal Solution Manual for Physical Chemistry 2nd Edition by Ball Full file at https://TestbankDirect.eu/Solution-Manual-for-Physical-Chemistry-2nd-Edition-by-Bal     Chapter 1.28 Following the normal rules of derivation: (a) y  3w z xy z y2 z3  (b) (c) 32 y w w2 w z xyz 3y z 3y z xy   (d) Using the answer from part a, we get  (e)  (f) w w w 32 y Using the answer from part e, we get 3y z w2   1.30 p  n  (a)     V  T, P R  T  T  V (b)     p  V,n n  R - pV  n  (c)     T  P,V RT RT  p  (d)    V  n  T ,V 1.32 (a) V(T, p)  nRT p  V   V  (b) dV    dp  dT    T  p  p  T (c)    Latm   Latm     1mole 0.0821  1 mole 0.0821 350K      nRT   nR  Kmole   Kmole    0.10atm  dp   dT    dV   10.0K    2     p 08 atm p 1.08atm              = -1.70 L    V   nRT   V  2nRT   1.34 (a)  ;    p p3  p  n ,T  p  n ,T nR   p   p   0 (b)    ; V  T  n , V  T  n , V 1.36   (a) Z=1 for an ideal gas (b) If the gas truly follows the ideal gas law, Z will always be regardless of the pressure, temperature, or molar volume   Full file at https://TestbankDirect.eu/Solution-Manual-for-Physical-Chemistry-2nd-Edition-by-Bal Solution Manual for Physical Chemistry 2nd Edition by Ball Full file at https://TestbankDirect.eu/Solution-Manual-for-Physical-Chemistry-2nd-Edition-by-Bal Instructor’s Manual 1.38 Using equation 1.23, TB  a , and using data from Table 1.6, we have: bR L2 atm mol for CO2: TB   1026 K L atm   0.04267 L/mol  0.08205  mol K   3.592 L2 atm mol for O2: TB   521 K L atm   0.03183 L/mol  0.08205  mol K   1.360 L2 atm mol for N2: TB   433 K   L atm   0.03913 L/mol  0.08205  mol K   1.390   1.40 1.42 C In order for the term to be unitless, C should have units of V2 (volume)2/(moles)2, or L2/mol2 The C’ term is C’p2, and in order for this term to have the L·mol (The unit same units as p (which would be Latm/mol), C’ would need units of atm bar could also be substituted for atm if bar units are used for pressure.) The C term is Gases that have lower Boyle temperatures will be most ideal (at least at high temperatures) Therefore, they should be ordered as He, H2, Ne, N2, O2, Ar, CH4, and CO2   1.44 (a) He: b= m3 0.0237L m   2.37  10 5 mole mole 1000 L m3 1mole  29 m   3.94 10 2.37 10 atom mole 6.02 10 23 atoms 5 V m3 r  3.94  10  29 ; r  2.11  10 10 m or 2.11 Å atom (b) H2O: b= 0.03049L ; using a similar method to part a, r = 2.30 Å mole (c) C2H6: b= 0.0638L ; using a similar method to part a, r = 2.94 Å mole Full file at https://TestbankDirect.eu/Solution-Manual-for-Physical-Chemistry-2nd-Edition-by-Bal Solution Manual for Physical Chemistry 2nd Edition by Ball Full file at https://TestbankDirect.eu/Solution-Manual-for-Physical-Chemistry-2nd-Edition-by-Bal     Chapter 1.46 Let us assume standard conditions of temperature and pressure, so T = 273.15 K and p = 1.00 atm Also, let us assume a molar volume of 22.412 L = 2.2412  104 cm3 Second virial coefficient terms can be calculated using values in Table 1.6 Therefore, we have for hydrogen: pV B 15.7cm / mol  1  1  1.00070 , which is a 0.070% increase in the V RT 2.2412 10 cm / mol compressibility For H2O, we have: pV B 213.3cm3 / mol  1  1  0.9905 , which is a 0.95% decrease in the V RT 2.2412 10 cm / mol compressibility with respect to an ideal gas 1.48 By comparing the two expressions from the text B C a 1 b  Z  1 b        and Z      V V RT  V  V   it seems straightforward to suggest that, at the first approximation, C = b2 Additional terms involving V may occur in later terms of the first expression, necessitating additional corrections to this approximation for C   1.50 The Redlich-Kwong equation of state: p  RT a  Vb T V ( V  b)  RT a a  p       2 T V ( V  b)   V  T ,n ( V  b) T V ( V  b) an 1.52 The van der Waals equation of state: (p  )(V  nb)  nRT ; As V approaches ∞, the V an2/V2 term goes to and the nb term becomes negligible The equation then reduces to pV=nRT 1.54 The Redlich-Kwong equation of state: p  RT a  Vb T V ( V  b) As T approaches infinity, the second term on the right side goes to The molar volume of a gas at high temperature will generally be high so the correction factor, b, is negligible The equation reduces to p V  RT     Full file at https://TestbankDirect.eu/Solution-Manual-for-Physical-Chemistry-2nd-Edition-by-Bal Solution Manual for Physical Chemistry 2nd Edition by Ball Full file at https://TestbankDirect.eu/Solution-Manual-for-Physical-Chemistry-2nd-Edition-by-Bal Instructor’s Manual RT 1.56 Using the ideal gas law, p  ,p  V (0.0821 L  atm )(273.15K ) K  mol  1.00atm 22.41L Using the Dieterici equation of state: p  (0.0821 L  atm e )(273.15K ) K  mol Latm    (10.91atmL2 ) /  ( 22.41L )( 0.0821 )( 273.15 K )  K mol   (22.41L  0.0401L)  0.981atm ; It varies by ~2%   1.58 In terms of p, V, and T, we can also write the following two expressions using the cyclic rule:  T   p       V   V   T V  p V   and  There are other constructions possible that    p  T  p   T   T  p       V  T  V  p would be reciprocals of these relationships or the one given in Figure 1.11 1.60 Since the expansion coefficient is defined as  V    ,  will have units of V  T  p volume   , so it will have units of K-1 Similarly, the volume temperature temperature  V  isothermal compressibility is defined as    , so  will have units of V  p  T volume   , or atm-1 or bar-1 volume pressure pressure 1.62  1   V     nRT   nRT          V      For an ideal gas,   V  P  T V  p  V  p  p  V p p This is equal to bar-1 at STP and SATP nRT  nRT  V , this last Since   p  V p T 1V  for an ideal gas The expression  is evaluated as expression becomes V p p p 1.64 For an ideal gas,     V  V  p    nRT     V p  p T T T  V  T   nRT  T nR For an ideal gas, the ideal gas law can be        p p V  T  p pV T  p  pV p nR V  , so we substitute to get that this last expression is rearranged to give p T Full file at https://TestbankDirect.eu/Solution-Manual-for-Physical-Chemistry-2nd-Edition-by-Bal Solution Manual for Physical Chemistry 2nd Edition by Ball Full file at https://TestbankDirect.eu/Solution-Manual-for-Physical-Chemistry-2nd-Edition-by-Bal     Chapter T V , which  Thus, the two sides of the equation ultimately yield the same pV T p expression and so are equal RT Therefore, the expression for density becomes, substituting p M pM  The derivative of this expression with respect for the molar volume, d  RT / p RT pM  d  to temperature is  Using the definition of V , this can be rewritten as   RT  T  p.n 1.66 For an ideal gas, V  M  d     VT  T  p.n 1.68  g  m   m   Mgh  mol  s  pe  ; If we convert g to kg and recognize that a J= J RT   K    K  mole   kg  m      kg  m Mgh  mol  s   , All of the units cancel and the exponent is unitless RT s2  J   K   K  mol  Mgh  RT   1.70 If we assume that the average molecular weight of air is 28.967 g/mole, p  e  Mgh RT =e-X m J kg     X X   0.028967  9.8  432m    8.314 273.15  39.0K   0.0473; e  1.05atm K  mole mole s      4 3  5   10 7  7.14   1.72 N=7 score    4 3    7 7   1.74 The probability that the particle is in the higher state = e  E / RT (a) At 200K, probability = e (b) At 500K, probability = e     J    1000 J /   8.314   200 K   K mole    J    1000 J /   8.314  500 K   K mole    (c) At 1000K, probability = e =0.548 =0.786 J    1000 J /   8.314  100 K   K mole      =0.887 Full file at https://TestbankDirect.eu/Solution-Manual-for-Physical-Chemistry-2nd-Edition-by-Bal Solution Manual for Physical Chemistry 2nd Edition by Ball Full file at https://TestbankDirect.eu/Solution-Manual-for-Physical-Chemistry-2nd-Edition-by-Bal Instructor’s Manual probability For the three temperatures  probability these ratios equal: 1.21, 3.67, and 7.85; as the temperature goes up To calculate ratios we can use the equation: 1.76 RT; Erot  RT 3 (b) H2O is a non-linear molecule.  Etrans  RT; Erot  RT 2 (c) Kr is an atom.  Etrans  RT; Erot  3 (d) C6H6 is a non-linear molecule.  Etrans  RT; Erot  RT   2 (a) (CN)2 is a linear molecule.  Etrans  Full file at https://TestbankDirect.eu/Solution-Manual-for-Physical-Chemistry-2nd-Edition-by-Bal ... https://TestbankDirect.eu /Solution- Manual- for- Physical- Chemistry- 2nd- Edition- by- Bal Solution Manual for Physical Chemistry 2nd Edition by Ball Full file at https://TestbankDirect.eu /Solution- Manual- for- Physical- Chemistry- 2nd- Edition- by- Bal... https://TestbankDirect.eu /Solution- Manual- for- Physical- Chemistry- 2nd- Edition- by- Bal Solution Manual for Physical Chemistry 2nd Edition by Ball Full file at https://TestbankDirect.eu /Solution- Manual- for- Physical- Chemistry- 2nd- Edition- by- Bal... https://TestbankDirect.eu /Solution- Manual- for- Physical- Chemistry- 2nd- Edition- by- Bal Solution Manual for Physical Chemistry 2nd Edition by Ball Full file at https://TestbankDirect.eu /Solution- Manual- for- Physical- Chemistry- 2nd- Edition- by- Bal