The discussion revolves around the combustion of hydrogen and oxygen in an expandable container, focusing on the final volume of gases produced after ignition. The context includes the ideal gas law and thermodynamic principles, particularly under conditions of constant pressure and no heat loss.
The discussion is active, with participants providing hints and guidance regarding the calculations needed for the problem. There is an exploration of various interpretations of the combustion process and its effects on volume, as well as the properties of water vapor. Some participants express uncertainty about specific values and concepts, indicating a collaborative effort to clarify these points.
Participants note the complexity of the problem, including the need for specific thermodynamic data and the assumptions made about heat loss and pressure conditions. There is also mention of the stoichiometry involved in the reaction and the resulting state of the products.
Yes. This is all correct.mike l said:thinking along the lines that when the hydrogen is ignited it releases energy which would increase temperature. increased temperature, increased volume because pressure remains the same. knowing that the end product of hydrogen combustion is water would this container then be filled with water vapor?
1.Thats right and one value you need is the molar heat capacity of water vapor in calories/mole Kmike l said:thinking along the lines that when the hydrogen is ignited it releases energy which would increase temperature. increased temperature, increased volume because pressure remains the same. knowing that the end product of hydrogen combustion is water would this container then be filled with water vapor?
The heat of combustion to the hypothetical state of water vapor at 25 C and 1 atm is -242 kj/mole (this already includes the heat of vaporization). What temperature would the release of that amount of heat cause the temperature of 1 mole of water vapor to rise to?krispykreme said:22.4 L H2+22.4 L O2->
2(22.4 L H2)+ 22.4 L O2->
44.8 L H2+22.4 L O2-> (g)2H2O *not STP*
2H2+O2->(g)2H2O
2 mol H2O at STP is solid ice but we will use liquid for the example. 2 mol (l)H2O is 36 grams of water at STP. Once heat of vaporization is achieved, and all the water is evaporated and the surrounding environment remains at STP the volume increases virtually 3000:1. Correct me of I am wrong but 36 mL would now be 108,000 mL of water vapor. This is assuming all water has been brought to the point of vaporization, not before and not beyond. From two gases combining to make water vapor the expansion doesn't seem that great. However, I think the exothermic reaction is much greater than "just enough" to reach heat of vaporization. So at the very least it would be like 50% increase in volume compared to the starting volume of gases at STP, but that is an unrealistic base value considering combustion energies far exceed the energy needed to make steam.
Edit: further research shows 286kj of energy is released from 1 mol H2 combustion. Enthalpy of Vaporization is 40kj+ Molar heat capacity 75j*100 °C which is only a fraction of the energy released during combustion. molar heat capacity of steam is 36 j/mol so there is plenty of energy to exceed the volume of steam at 100°C. My numbers aren't exact but I think it would be a pretty grand difference in pressure or volume considering the containing device and a perfect stoichiometric reaction.