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Mixture of hydrogen and oxygen in container

  1. Jan 4, 2014 #1
    1. The problem statement, all variables and given/known data
    In a closed container with a volume of V = [STRIKE]5[/STRIKE] 50 l is a mixture of hydrogen and oxygen with the total mass m = 50 g. The container have pressure p = 300 000 Pa and the temperature t_1 = 20 ° C.
    A) What is the number of moles n_1 (hydrogen) and n_2 (oxygen) in the container?
    B) What heat Q is released after an electrical spark ignited the mixture? The calorific value of the hydrogen is 120 000 000 J / kg.
    C) What is the pressure in the container when the temperature go down to t_2 = 100 ° C (after elektrical spark) and what is m_v weight of condensed water at this temperature?

    ... p_a = 1, 013 * 100 000 Pa; M_mhydrohen = 2,02/1000 kg/mol; M_moxygen = 32/1000 kg/mol; R = 8,31 J/(mol*K)

    2. Relevant equations
    A) pV = nRT
    B) Q = Hm
    C) p/T = const.

    3. The attempt at a solution
    A) pV = nRT = m/M_m * RT; m/V = pM_m/ RT; m/V(hydrogen) = pM_mhydrogen/RT; V_(oxygen) = pM_moxygen/RT
    The system of equations:
    - V_hydrogen +V_oxygen = V (V_oxygen = V - V_hydrogen ... substituted into the second equation)
    - V_hydrogen*(p*M_mhydrogen/RT) + V_oxygen*(p*M_moxygen/RT) = m
    Ok, I solve the system of equation... And I get V_oxygen = 0,01 m^3; V_hydrogen = 0, 04 m^3
    Now back to equation pV = nRT; n = pV/2RT (2 because it H_2 and O_2, it is right?)
    And I get n_oxygen = 0, 626 mol; n_hydrogen = 2, 45 mol
    It is OK?
    B) Q = Hm = H*n*M_mhydrogen*2 (2 because it is H_2, it is right?)
    Q = 1 187 760 J... It is OK?
    C) Hmmm... And this is the biggest problem.
    It is isochoric progress? Than p_1/T_1 = p_2/T_2; p_2 = p_1*T_2/T_1
    p_2 = 300 000 * (273,15 + 100)/(273, 15 + 20) Pa = 381 867 Pa
    Condensed water... Hmm, basic assumption is that the system is in steady state - System of saturated steam and liquid. Saturated steam is not an ideal gas ... The equation p / T = const. ist probably wrong. I really don't know any equation for saturated steam... Can you help me?
    Thank you and sorry for my bad English :-)
     
    Last edited by a moderator: Jan 5, 2014
  2. jcsd
  3. Jan 4, 2014 #2

    SteamKing

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    How many liters are in 1 m^3?
     
  4. Jan 4, 2014 #3
    1 l = 1 dm^3, so 1000 l = 1 m^3...
    yeees, error, 5 l = 0, 005 m^3... Thank you.
    However, the process, is good? And what with C?
     
  5. Jan 4, 2014 #4
    Maybe I'm doing something wrong, but I calculated that the total number of moles of H2 plus O2 in the container is 0.616 gm-moles. However, if x is the number of grams of H2, and 50-x is the number of grams of O2, there was no combination that gives 0.616 moles. Are you sure about that 50 grams? Are you sure it's not 5 gm?

    Chet
     
  6. Jan 4, 2014 #5
    Mistake, big mistake in task...
    In a closed container with a volume of V = 50 l ... Other info is OK.
    So, what now, it is good?
     
  7. Jan 4, 2014 #6
    In that case, I get 1.26 moles of O2 and 4.9 moles of H2.
     
  8. Jan 4, 2014 #7
    Hmmm, so, you get twice larger values (my are 0, 626 moles and 2,45 moles). I divide n by 2, because I consider consider H2 and O2. The resoults 1, 26 moles and 4,9 moles are only for H and O, I think? Or no?
     
  9. Jan 4, 2014 #8
    And what about part B and C? Do you have any idea?
     
  10. Jan 4, 2014 #9
    No. From the ideal gas law, the total number of moles is 6.16, irrespective of the individual molecular weights: (300000)(0.05)/(8.31)(293). So figure out where you went wrong.
     
  11. Jan 4, 2014 #10
    Ok, ok... The mistake was the n dividet two, now I have same resoults.
     
  12. Jan 4, 2014 #11
    There doesn't seem to me to be enough data provided to properly solve part B. But, I can help you with part C.
    After the reaction is complete, how many moles each of H2, O2, and H2O are in the container? If there is liquid water present in the container at the final temperature of 100 C, what is the partial pressure of the water vapor in the gas phase?

    Please answer these questions first, and then we can continue.

    Chet
     
  13. Jan 4, 2014 #12
    Ok, so I have some solution...
    First - how many moles has each of H2, O2, and H2O?
    m=m, (total mass of container in part A = total mass of container in part C)
    I have to exprees m,... It is sum of m_H, + m_O, + m_H2O, in part C.
    1) hydrogen: m_H = n * M_H = N,*M_H/N_A ... However, we can write N,=N-2N,, (N, is number of partiples H2 in part c, N is nimber of partiples H2 in part a; 2N, is number of partiples for water)
    m_H = (N-2N,,)*M_H/N_A= (n_H - 2N,,/N_A)*M_H = n_H*M_H -2N,,*M_H/N_A (n_H is number of moles of hydrogen from part a)
    2) oxygen: m_O = (N-N,,)*M_O/N_A = (n_O - N,,/N_A)*M_O = n_O*M_O -N,,*M_O/N_A ( n_H is number of moles of oxyhen from part a; ... 2N,, of H_2 and N,, of O_2 for H2O)
    3) water: m_H2O = 3N,,*M_H20/N_A

    m = m_H + m_O + m_H2O = (n_H*M_H -2N,,*M_H/N_A)+(n_O*M_O -N,,*M_O/N_A)+(3N,,*M_H20/N_A)
    m - n_H*M_H - n_O*M_O = -2N,,*M_H/N_A - N,,*M_O/N_A + 3N,,*M_H2O/N_A
    N_A(m - n_H*M_H - n_O*M_O) = N,,(-2M_H - M_O + 3M_H2O)
    N,, = N_A(m - n_H*M_H - n_O*M_O)/(-2M_H - M_O + 3M_H2O)
    N,, = 6,022*10^23(0,05 - 4,9*2,02*10^-3 - 1,26*32*10^-3)/(-2*2,02*10^-3 - 32*10^-3 + 3(2*2,02*10^-3+32*10^-3) = 6,022*10^23 * -2,18*10^-4 / 0,07208 = -1,82*10^21 is the number of moles in hydrogen, hydrogen and oxygen we can easily calculate, but .... Eeeeemh. Crazy resoult. Hmmm... Something is wrong, don't know what.
    If i use my twice smaller values: n_oxygen = 0, 626 mol; n_hydrogen = 2, 45 mol
    Water: 2, 1*10^23 ... 0, 35 moles
    Oxygen: (0,626-0,35)moles = 0,276 moles
    Hydrogen: (2,45-2*0,35)moles = 1,75 moles
    ... Hmmm. I don't understand why, but i think the idea is good. May I am too sleepy and thus stupit :-D
    Do you have some resoult? Is my idea good or do you see a mistake? Thank you and good night.
     
  14. Jan 4, 2014 #13
    What you did is a lot of work, but I can't make sense out of it because of your complicated notation. Anyway, here is my simple reasoning: To start with, you have 1.26 moles of O2 and 4.9 moles of hydrogen. The total number of moles does not stay constant since 2 moles of H2 reacts with 1 mole of O2 to produce 2 moles of H20. The limiting reactant is oxygen, so it all gets used up. This consumes (1.26)(2)=2.52 moles of H2 to produce 2.52 moles of H2O. So, in the final state, you have 0 moles of O2, 4.9 - 2.52 = 2.38 moles of H2, and 2.52 moles of H20. The final number of moles in the container is 4.9.

    Next, please answer my question about the partial pressure of the water vapor in the final state of the system.
     
  15. Jan 4, 2014 #14
    Eeeeh. So simly? Thank you. I will answer your next question for a few hours, I go sleep for a while... However, why do you think that all oxygen have to be consumed? You don't know, when on the container is equilibrium state. Equilibrium state can be before consumed all oxygen, no?
     
  16. Jan 4, 2014 #15
    This reaction is well-known to go virtually to completion. It has an extremely high equilibrium constant. So the amount of O2 remaining is going to be insignificant.

    Chet
     
  17. Jan 5, 2014 #16
    The partial pressure of water vapor = pressure of water vapor in the V of the system and with temperature T system... Is it right? However, why particial pressure? It isn't in task. So, I will try it...
    p_(H2O)V = n_(H2O)RT
    p(H2O) = n_(H2O)RT/V = 2,52 * 8,31 * (100+273,15)/0,05 Pa = 156 284 Pa
     
  18. Jan 6, 2014 #17

    Borek

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    It must be a partial pressure, as not all gases reacted (hydrogen was in excess).

    If there is a condensed water present, what MUST be water vapor pressure at 100 deg C?

    Or, from slightly different angle - what typically happens to water at 100 deg C?
     
  19. Jan 6, 2014 #18
    No. This is not correct for the reasons that Borek stated. In addition, the gas does not fill the entire container, so that the volume of the gas is not 0.05 m^3. There is some liquid water in the container that takes up part of the volume now. So you can't substitute 0.05 into the ideal gas law once the liquid water is present. We will get to all that in due course. So again, to reiterate Borek's question, what is the vapor pressure of water at 100C?

    Chet
     
  20. Jan 6, 2014 #19

    Borek

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    Staff: Mentor

    You can safely ignore the liquid volume, it is orders of magnitude lower than the volume of the gas.
     
  21. Jan 6, 2014 #20
    This is true for this particular problem. But, the OP doesn't know that yet. I think it would be helpful for him, in terms of his learning experience, to not automatically assume this yet. The equations will certainly reveal this.

    Chet
     
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