Hydrogen oxygen ignited volume expansion

[mike l]

Homework Statement

there is an expandable container with 1m3 of hydrogen and 1/2 m3of oxygen at stp. the gas is ignited.what is th final volume if there is no heàt lost and the pressure outside the container is 1atm? assume complete combustion

Homework Equations

other than the ideal gas equation not sure what other equations to- use

The Attempt at a Solution

 

[DrClaude]

And how could you apply the ideal gas law?

 

[mike l]

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?

 

[Chestermiller]

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?

Yes. This is all correct.

Here are some hints:

If there is no heat lost and the expansion occurs against constant pressure of 1 atm., from the first law, what is the change in enthalpy of the system?

If there had been no change in temperature, what would the change in enthalpy have been, and how much heat would have been lost?

Chet

 

[mike l]

thanks chet will get back to you-m

 

[Point Conception]

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 K
2. How many moles of H₂O produced ? ( 22.4moles/liter STP) How many liters in m³
3. What is enthalpy in calories released in total reaction ?
4. From above you should get K° for T
5.R should be ..08206 L atm/mole K
6. V = nRT/P

I got xx.x m³

 

[mike l]

hate to'sound ignorant (but i am) 1) i am not finding any molar heat capacity of water VAPOR. would not the heat capacity of vapor depend on temperature which would change once the hydrogen is combusted? 2)there are aproximately 67 moles in the 1 1/2 m3. 44.6 of hydrogen and 22.3 for oxygen. not understanding "number of moles H2O" moles being gaseous does that mean water vapor? Am i making this more complicated than it has to be? I really appreciate the help ,I am trying to get my brain in the right perspective and not just do the math.

 

[Chestermiller]

So how many moles of water vapor are produced when 44.6 moles of hydrogen react with 22.3 moles of oxygen. What is the heat of this reaction (with the product being water vapor)? This is the same as the heat of formation of water vapor at standard conditions. Are you saying that you can't find any data on Cp for water vapor? (Did you consider getting the value per unit mass, and then multiply by the molecular weight?).

Chet

 

[krispykreme]

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.

 

[Chestermiller]

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.

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?