Applied Physical Chemistry by C. Heald, A. C. K. Smith (auth.)

By C. Heald, A. C. K. Smith (auth.)

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955 mol. 795 mol. 7 with the gas mixture entering at 298 K. Assuming that the very centre of the flame is adiabatic (as it probably is), this problem is simply a balance between the heat produced by the reaction that goes to heat the reaction products, and the flame temperature. This simple basis is complicated somewhat by the effect of temperature on the heats of reaction and on the heat-capacity values. 10 for reactions (a) and (b) (see example 2. 25 x w- 3 T 2 where Tis the flame temperature.

1 X I0- 3 X 10 3 where 0. 1 x 10- 3 are the density in kg m- 3 and molar mass in kg of ethanol respectively. 33 mol s- 1 . 5 kJ s- 1 . 02 kg s- . It should be pointed out that the following assumptions have been made in this calculation: (i) reaction (a) is the only reaction taking place; (ii) the cooling water must be capable of removing all the heat liberated; (iii) fermenting grape juice may be regarded as an infinitely dilute solution of glucose and ethanol; (iv) the value of M-G- for reaction (a) is the same at 303 K as it is at 298 K.

38 x I0- 4 m3 mol- 1 ; the molar mass of this substance is 154 g. 1. 12 is more difficult since the equation is cubic; graphical solutions are usually suitable in calculations of this type. 50 MN m- 2 and at a temperature of 298 K. 38 x I0- 5 m3 mol- 1 respectively. 1 and this indicates the range of V values over which the graph should be plotted. 0802 m3 ; this last quantity is the volume required. 3 Other equations of state More accurate calculations require correspondingly more accurate, and necessarily more cumbersome, equations of state.

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