CHM 5414 Thermodynamics

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Quiz #3 Solutions

29 October 1996

Molecule CS2(liq) O2(g) N2(g) CO2(g) SO2(g)
Delta Hf°, kJ/mol 89.0 0 0 -393.5 -296.8
CP, J/mol°K 79.7 21.911 29.1 37.1 39.9
  1. When I was a lad, I served a term apprenticed to a chemist firm. It was there that I first heard that carbon disulfide, CS2(liq), burned with a "cool flame." I didn't believe it then, but after finding its flame temperature, given the data below, you might suggest a reason for why its flame should be "cool" in spite of the magnitude of Tflame. Of course, it is burning in air, X(N2)=0.8; take that into account in the flame temperature.

Flames burn while expanding their hot gases against atmospheric pressure. The assumption we must make (but it's a rotten one) is that that expansion is adiabatic; that is, all of the heat of combustion is available to heat up the products. So enthalpy and CP are the proper tools.

Since X(O2)=0.2 in air, the ratio of moles of N2 to O2 is 4:1.

CS2(liq) + 3 O2(g) + 12 N2(g) arrow right CO2(g) + 2 SO2(g) + 12 N2(g)


We've made the additional assumption that the flame temperature is so low that N2 cannot be persuaded to oxidize. That assumption turns out to be all right.

Delta T = - Delta H° / Summation nu CP(products+diluents)


Summation nu CP(products+diluents) = CP[CO2(g)] + 2 CP[SO2(g)] + 12 CP[N2(g)]
= 37.1 + 2(39.9) + 12(29.1) J/mol°K = 466.1 J/mol°K.

Delta H° = Delta Hf°[CO2(g)] + 2 Delta Hf°[SO2(g)] - Delta Hf°[CS2(liq)]
= (-393.5) + 2(-296.8) - (+89.0) kJ/mol = 1076.1 kJ/mol = - 1.0761x106 J/mol

Thus, Delta T = + 1.0761x106 J/mol / 466.1 J/mol°K = + 2309°K

And Tflame = Delta T + 298°K = 2607°K

Which is ridiculous, but that's because the adiabatic assumption is poor. The same failure makes the 11,500°K oxyacetylene torch flame temperature (even without N2 diluent) absurd. Nevertheless, if oxyacetylene is, in fact, a high temperature flame, then CS2(liq) a factor of 4 less must be classified as a cool one. Also CS2(liq), being a liquid, doesn't mix nearly as well with O2 as does HCCH(g). Slower kinetics leads to an even worse adiabatic assumption.


  • The van der Waals equation of state for a real gas incorporates both molecular attractions which reduce the effective pressure, P + a(n/V)2, and molar volumes which reduce the free volume, V - nb, combined as

    [ P + a(n/V)2 ]*[ V - nb ] = nRT


    Find an algebraic expression for the molar entropy of isothermal expansion of a van der Waals gas from V1 to V2.

    • HINT: One of the Maxwell Relations comes in really handy here.

    (So does the correct statement of van der Waals expression! Fortunately, the typos on the quiz had no effect on the answer.)

    The first thing to realize is that for one mole (n=1), the expression simplifies to:

    [ P + aV-2 ] [ V - b ] = RT

    or
    P = [RT/( V - b )] - aV-2

    which makes the loss of pressure obvious.

    But the key to the problem is to realize that we need:

    Delta S = integral V1V2 [ Partial S/ Partial V]T dV (amazing! subscripts can have subscripts in HTML)

    And that leads us to look for a Maxwell Relation with SdT (narrows it to dA and dG), but that variation with respect to V makes us look for ?dV, and that nails it as dA,
    dA = - SdT - PdV,
    from whose cross-second derivatives we draw the Maxwell Relation:

    [ Partial S/ Partial V]T = [ Partial P/ Partial T]V

    Differentiating that expression for P above leaves [ Partial P/ Partial T]V = R/(V-b). The derivative of the 2nd term, -aV-2, vanishes at constant V.

    And the integration becomes easy,
    Delta S = R integral V1V2 dV/(V-b) = R ln[ (V2-b) / (V1-b) ]
    which makes sense; it says the expansion can only occur into the volume not already occupied!
    Last modified 30 October 1996