Exam 1 Solutions

21 September 1998

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  1. Dry air is 78% N2, 21% O2, and 1% Ar; so its average molecular weight is given by MW=0.78(28)+0.21(32)+0.01(40)=28.96 g/mol.
    1. Find c*, the most probable speed, of an average air molecule at 20°C.

      c* = [2RT/M]½
      c* = [2(8.315 J/mol K)(293 K) / 28.96×10-3 kg/mol]½
      c* = 410.19 m/s

    2. The speed of sound in dry air is 343.37 m/s at 20°C. How much less likely is it to find an air molecule moving at the speed of sound than at c*?

      P(v)/P(c*) = (v/c*)2 eM(c*2-v2)/2RT
      M(c*2-v2)=0.02896 kg/mol(410.192-343.372)(m/s)2 = 1458 J/mol
      M(c*2-v2)/2RT=1458 J/mol/[2(8.315 J/mol K)(293 K)]=0.2993
      P(v)/P(c*) = (343.37/410.19)2 e0.2993 = 0.945
      or v is 5.5% less likely than c*.

  2. A mole of dry air (as an ideal gas) originally at 1 atm and 20°C expands reversibly and adiabatically to double its volume. If CP(air)=1.007 J/g K at that temperature, what is the final temperature of the gas?

    VTCV/R = constant for rev. adiab. exp.
    CV=CP-R = 1.007 J/g K(28.96 g/mol) - 8.315 J/mol K = 20.85 J/mol K
    Tf = Ti(Vi/Vf)R/CV = 293 K (½)8.315/20.85 = 222 K

  3. Usually DtrsH completely dominates DPV even for evaporations where DV is large. But for He, DvapHq=82.9 J/mol at its normal boiling point of 4.22 K, the two terms are at least in the same ballpark. (Note carefully that is J/mol and not kJ/mol.) At that temperature, the densities of the liquid and gas are r(liq)=124.90 g/L and r(gas)=16.89 g/L. You have all the data now that you need to evaluate q, w, DU, and DH for the condensation of He gas to He liquid at its normal boiling point.

    Condensation is isothermal and isobaric and the reverse of evaporation. So qP = DcondH = - DvapHq = - 82.9 J/mol.

    w = - PextDVm where for normal condensation, Pext=1 atm.

    r = MW/Vm or Vm = MW/r
    DVm = Vm(liq) - Vm(gas) = MW × {[1/r(liq)] - [1/r(gas)]}
    DVm = 4.003 g/mol [(1/124.90 g/L) - (1/16.89 g/L)] = - 0.2050 L/mol
    w = - 1 atm × (-0.2050 L/mol) (8.315 J/0.08206 atm L) = + 20.77 J/mol


    DU = q + w = - 82.9 + 20.8 J/mol = - 62.1 J/mol

  4. At low pressures, we are justified in truncating the virial equation of state

    PV/RT = 1 + B'P + . . .

    at the B'P term. It's a simple matter to determine B' by experiment on near-ideal gases, but B' is a function of temperature. We can estimate the temperature dependence of B' easily using the Joule-Thompson equation

    µCP = T(dV/dT)P - V

    Find the simple expression relating (dB'/dT)P to µ, CP, and T.

    V = (RT/P)(1 + B'P) = RT[(1/P) + B']
    (dV/dT)P = RT(dB'/dT)P + R[(1/P) + B'] = RT(dB'/dT)P + V/T
    µCP = T(dV/dT)P - V = RT2(dB'/dT)P + V - V
    (dB'/dT)P = µCP/RT2

  5. Bunsen burners combust methane in air, but they'd be much hotter if we used pure oxygen in stoichiometric amounts! Estimate the temperature of a Bunsen "torch" if all the heat of combustion is consumed in heating up (only) the products. That's called the "adiabatic flame temperature," and it's always a fairly gross overestimate! We'll make the (ho ho ho) assumption that the heat capacities are constant over thousands of degrees. And we'll not bother with the standard temperature liquid water heat of vaporization by cleverly evaluating DcombHq to H2O(gas) instead of H2O(liquid).

    Molecule CH4(g) O2(g) CO2(g) H2O(g)
    DfHq, kJ/mol - 74.81 0.00 - 393.51 - 241.82
    CP, J/mol K NA NA 37.1 33.6

    CH4(g) + 2 O2(g)  arrow right CO2(g) + 2 H2O(g)
    DcombHq = DfHq[CO2(g)] + 2 DfHq[H2(g)] - DfHq[CH4(g)]
    DcombHq = -393.51 + 2(-241.82) - (-74.81) = - 802.34 kJ/mol

    DheatH = - DcombHq = DT × {Cp[CO2(g) + 2CP[H2O(g)]}
    DT = + 802,340 J/mol / [37.1 + (2)33.6] J/mol K = 7,693 K
    Tflame = Tq + DT = 298 + 7,693 K @ 8,000 K (50% hotter than sun!)


Last modified 21 September 1998