Acceptable Answer
Answer
The irreducible representations generated by this T-systems are A;, A;, and 2E". The energies and the wave functions for the first two orbitals may be obtained in a straightforward manner. On the other hand, the derivation of the four E" hasis functions is more complicated. A convenient set is
Initially, the final temperature of the system must be deter^ mined.
+I +Z
+
c -e) (2)-"Z(c + e ) = (6)-'12(b + 2d f ) G4 = (2)-lI2(b - /) = (6)F'Wa
$9 =
+
Qreleared = Qabsorbed by the bmtk Qreleesed
= 2-50g C ~ H S N ~ O ~ 3.43 X 103 kJ 1 mol C7HSN306 X 227.1 g C7HsNsOs 1mol C7H5N306 = 37.76 kJ
Let TI be the final temperature
and +r and that between By symmetry, the interaction between +S and $3 both lead to the following determinant:
T h e six resultant orbital energies and wave functions are summarized in the following table2. Now using the balanced chemical equstion, we may determine the amount of gas in the bottle. 1 mol C7H~N306 227.1 g C7HsN306 = 0.0110 mol C7H6N3O6 28 mol COz Amount of Con = 0.0110 rnol C~HsNa0sX 4 mol C7HsNsOs = 0.0770 mol Con 6 mol Nz Amount of Np = 0.0110 ma1 C7H6N30fiX mol C7HSN306
Amount of TNT = 2.50 g C7HsN306X
= 0.0165 mol Nz But at 28.3'C, the vapor pressure of water is 3.85 kPa. Therefore the amount of water vapor present must be determined i.e.
n = 1.11 X
mol
Thus the total amount of gas in the bottle is: 0.0770 mol COn(g) 0.0165 mol Ndg) 0.00111 mol HzO(g) 0.09461mol of gas
Pressure and the Exploding Beverage Container Robert R. Perkins Memorial Universify Sir Wilfred Grenfell College Corner Brook, Newfoundland,Canada. AZH 6P9
T h e public has recently become aware of t h e problem of exploding pop bottles. T h e following question is a n extention of this concept t o illustrate t h e balancing of a chemical equation, enthalpy, stoichiumetry a n d vapor pressure calculations, and t h e use of t h e Ideal Gas Equation. T h e question is aimed at t h e first-year level student.
Question
We may now determine the total uressure inside the bottle nRT 0.09461 mol X 8.31 kPa dm3 X 301.3 K p=-= V mol K 0.725 dm3 1atm P = 326.7 kPa X 101.3 kPa (a) P = 3.23 atm
.
Before the reaction began we had Amount of Oz = 0.0110 ma1 C~HsN306X
completely with all t h e T N T . T h e initial temperature is 20.0°C, and t h e oroducts of t h e reaction are nitrogen a n d and liquid water. T h e enthalpy for comcarbondioxide bustion of T N T is -3.43 X lo3 kJ mol-I and t h e specific heat of glass is 6.73 J g-'.deg-'. Calculate: (a) the final pressure inside the bottle (h) the difference in pressure from the start of the reaction. (You may ignore the heat absorbed by the products.)
21 mol0z 4 mol C7H8N306
= 0.0578 rnol Oz
A wine bottle, of mass 675 g and volume 725 cm3, contains 2.50 g of T N T (C7H5N306) and enough oxygen gas t o react
.
nRT = 0.0578 mol X 8.31 kPa .dm3. 293 K poz= mol. K X 0.725. dm3 V 1 atm 101.3 kPa
PO2= 194.1 kPa X Po* = 1.92 atm The difference in pressure is 3.23 stm - 1.92 atm ( b ) or 1.31 atm
Volume 58
Number 4
April 1981
363
