Question

General Electric Co. has developed an advanced fuel-cell battery that will supply the primary power needed during orbit by space capsules. The peak load delivered by the battery will be just under 2kW. The facility will be assembled from ion-exchange-membrane cells, in which the membranes are sheets of a tough, undisclosed plastic. The fuels supplied to the cells will be hydrogen and oxygen. Though fed in the gaseous state during battery operation, these two elements will be sorted as liquid, to conserve space.The equipment is inherently reliable because it is simple, and it achieves 60% thermal efficiency under normal load. Also, the battery will yield a by-product that is highly valuable in space-flight---- a pint of pure drinkable water for each kilowatt-hour of operation. If the O2 and the H2 are stored as saturated liquids at their normal boiling point, what is the enthalpy change per g mol of O2 between the liquid storage tank and the state of which the O2enters the fuel cell (1.5 atm, 27 C)

Answer 1

Enthalpy change is the amount of heat evolved or absorbed in a reaction. It is the difference of enthalpy of products & that of reactants.

Change in enthalpy per g mol of O₂  between liquid storage tank & the gaseous state in which O₂ enters fuel cell is given by : ${\color{Red} }\Delta H=\Delta U+P\Delta V$ , where the symbols have their usual meanings.

Now, O₂ is a diatomic gasous molecue. So, the value of $\large \Delta U= 5/2 \Delta nRT$ . Also, we know $P\Delta V=\Delta n_{g}RT$ . Therefore, $\large \Delta H=5/2\Delta nRT +\Delta nRT$ $\large = 7/2 \Delta nRT$ . Since, the state of change of O₂ is from gaseous state to liquid state, $\large \Delta n=2-1= 1$ $\large \left ( O_{2\rightleftharpoons 2O} \right )$ .

Therefore, $\large \Delta H= 7/2 \times 8.314\times 300= 8729.7 Jmol^{-1}$ . Since the instrument is said to have 60 % thermal efficiency under normal work load, $\large \Delta H=8729.7 \times 0.60= 5237.82 Jmol^{-1}$