Solutions to Exam 1     20 September 1999

Answer any 4 questions!
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  1. The atmosphere on Jupiter is only 160 K at the cloud tops. Its composition is roughly 92.5% H2, 7.2% He, 0.15% CH4, and 0.15% NH3 in mole percent. (There's many more molecules at ignorable concentrations like the H2S that makes the yellows in Jupiter's bands.)

    At a depth in Jupiter's atmosphere where the pressure is 1 bar, what is the mean free path for the molecules given that s = 0.266 nm2 is the composition-weighted average cross-section for H2 and He?

    l = (½)½kT / sp

    s = 0.266×10-18 m2
    p = 105 Pa = 105 N m-2 = 105 J m-1 m-2 = 105 J m-3

    l = 0.707 × 1.381×10-23 J K-1 × 160 K / [ 2.66×10-19 m2 × 105 J m-3 ]
    l = 5.87×10-8 m = 58.7 pm = 587 Å

  2. Derive the expression for DT for reversible adiabatic expansion (or compression, it's the same expression) processes in terms of Ti and both Vi and Vf (where, of course, i means initial and f means final).

    The standard expression is TV1/c = constant where c = CV / R.
    Therefore TfVf1/c = TiVi1/c.

    DT = Tf - Ti = Ti(Vi / Vf)1/c - Ti = Ti [ (Vi / Vf)1/c - 1 ]

  3. You pump up a bicycle tire by compressing (adiabatically) air at 1 bar and 25°C into a volume of 1.3×10-3 m3 at 5 bar. If the air is ideal, and its CP = 29.17 J mol-1 K-1, what will the temperature of the air inside the tire become? Now you know why hand pumps get so bloody hot. (Ignore the temperature dependence of CP.)

    Use a P,T form of the adiabatic expression or you'll drive yourself crazy.

    If we want a p,T adiabatic expression, we can get it by substitution of the ideal gas expression Vm = RT/p into TV1/c fixed.

    TV1/c = T(RT/p)1/c = R1/c T1+(1/c) p-1/c = fixed

    Since R is fixed regardless, we can fold it into whatever constant "fixed" stands for. Likewise we can raise the entire expression to the cth power and the constant similarly. This means that the expression we seek is

    Tc+1 / p = T(CV / R) + 1 / p = TCP / R / p = fixed
    (since for an ideal gas, CV+R = CP)

    But it's temperature in which we're interested, so we should rearrange the expression by raising it to the R/CP power to get

    T p=R/CP is fixed, or Tf pf-R/CP = Ti pi-R/CP or Tf = Ti (pf / pi)R/CP

    Tf = 298 K ( 5 bar / 1 bar)8.315 J mol-1 K-1 / 29.17 J mol-1 K-1
    Tf = 298 K 5 0.2850 = 471 K well above boiling!

    In fact, there are devices made which use sudden adiabatic compression to generate heat high enough to light kindling. Compressional matches. J

  4. Find the general total differential dT given that T is a function of p and V. Express the answer in terms of the expansion coefficient, a, and the isothermal compressibility, kT. Then substitute the ideal gas expressions for a and kT, and verify by direct integration that the partials yield Charles' laws for T,V and T,P. (That way you know you got it right.)

    dT = (dT/dp)V dp + (dT/dV)p dV

    (dT/dV)p = 1 / (dV/dT)p = 1 / (aV)
    (dT/dp)V = - (dT/dV)p (dV/dp)T = - (1 / [aV]) (-kTV) = kT / a

    dT = (kT/a) dp + (1/aV) dV

    a = (1/V) (dV/dT)p =° (1/V) (d[RT/p]/dT)p = (1/V) R/p = R/pV = 1/T
    kT = - (1/V) (dV/dp)T =° - (1/V) (d[RT/p]/dp)T = - (1/V) (- RT/p2) = + (RT/pV)/p = 1/p

    dT =° (T/p) dp + (T/V) dV = T d(ln p) + T d(ln V)
    (1/T) dT = d(ln T) = d(ln p) + d(ln V)

    Integrating either partial alone gives

    ln(Tf / Ti) = ln(pf / pi) or Tf / T i = pf / pi or T / p fixed at fixed V.

    ln(Tf / Ti) = ln(Vf / Vi) or Tf / Ti = Vf / Vi or T / V fixed at fixed p.

  5. How many liters of methane at STP does it take to boil a liter of water? The water starts at 25°C and has CP = 75.29 J mol-1 K-1 and a DvapH = 40.66 kJ/mol. Ignore temperature variations in these parameters (and those below).

    Molecule CH4(g) O2(g) CO2(g) H2O(liq)
    DfHq, kJ/mol - 74.81 0.00 - 393.51 - 285.83

    nH2O = 1 kg × (1 mol H2O / 0.018 kg) = 55.5 mol

    DH = nCPDT + nDvapH
    DH = 55.5 mol [ 0.07529 kJ mol-1 K-1 × (75 K) + 40.66 kJ/mol ]
    DH = 55.5 mol (46.31 kJ/mol) = 2570 kJ

    CH4 + 2 O2 arrow right CO2 + 2 H2O
    DrH° = DfH°[CO2] + 2 DfH°[H2O] - DfH°[CH4]
    DrH° = (-393.51) + 2(-285.83) - (-74.81) = - 890.36 kJ/mol

    Thus, we need nCH4 = 2570/890.36 = 2.886 mol

    V = nRT/p
    V = 2.886 mol × 0.08206 atm L mol-1 K-1 × 273 K / (1/1.01325) atm
    V = 65.5 L CH4 to boil 1 L H2O

  6. Benzene is a rotten ideal gas. Its van der Waals parameters are a = 18.24 atm L2/mol and b = 0.1154 L/mol. Find its compression factor, Z, at 100°C. (It only boils at 80°C!)

    (p + a/V2)(V - b) = RT if we interpret V as Vm.
    Z = pV/RT at 1 atm and 373 K requires knowledge of V.

    V = b + RT/(p + a/V2) solve by successive approximations.

    The first guess should be
    V° = RT/p = 0.08206 atm L mol-1 K-1 × 373 K / 1 atm = 30.61 L

    V = 0.1154 L + 30.61 L atm / (1 + 18.24/V2) atm

    Guessed VCalculated V
    30.6130.14
    30.1430.12
    30.1230.12

    Z = pV/RT = 1 atm (30.12 L) / 30.61 atm L = 0.984

Last modified 20 September 1999