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Gas law, how to fine the final T and Final P?

  1. May 13, 2012 #1
    i don’t understand how they got this answer

    can you please explain the steps
    needed to fine the Final T and How the final P was found.
    thanks


    N2H4 + 3O2 ==> 2NO2 + 2H2O


    assume there are 60 moles of O2, what is the final T and P

    V = 20L
    T1 = 290K
    n (N2H4) = 20
    Cp = 57 kJ/C
    Hc = 194.1 x 105 J/kg


    final T:
    20 mol N2H4 * 32g N2H4 / 1mol N2H4 * 1kg / 1000g =0.640 kg N2H4

    0.640kg N2H4 * 194.1 * 10^5 J/ Kg N2H4 * 1kJ/ 1000j =12,420 KJ

    12,420KJ * 1c/57Kj= 218C T f = 290K+218K=508K




    Final P : PV=nRT ==> P=nRT/V

    n=20 moles N2H4 react with 60 mole of O2 sp no left over O2


    N2H4 + 3O2 ==> 2NO2 + 2H2O
    20N2H4====> 40NO2 + 40 H2O =80moles

    R= 0.0821L*atm /1mole *K

    T=508K
    V= 20L

    P=nRT/V

    = 80mol * 0.0821 L*at, / 1mol K * 508K * 1/20L = 167atm
     
  2. jcsd
  3. May 13, 2012 #2
    You want to fine the final T for speeding :D ?

    Anyway, what specifically don't you understand about this? And are you sure that Cp and not Cv is given?
     
  4. May 14, 2012 #3
    ok i figure out the first step.
    all that they did was converted from
    moles to g to kg then multiply the specific heat then convert to J then to KJ then to C and then to K

    now i am confuse as to how they found the P

    i understand that the equation is pv=nrt which becomes p=nrt/v

    but they did this

    N= 20 moles N2H2 react with 60 mol O2 , so there is no left over O2
    what i don't understand, is what do they mean by left over O2?

    so then it continues with
    N2H4 + 3O2 ==> 2NO2 + 2H2O

    N2H4--->40NO2 + 40H2O = 80 moles

    i dont understand how they got the second equation N2H4--->40NO2 + 40H2O = 80 moles


    can't i just add 60 and 20 to get the 80moles?

    thanks
     
  5. May 14, 2012 #4
    The (properly equilibrated) chemical equation for the reaction gives you the proportions in which the reactants react with each other. In your case, for each mole of N2H4 you need 3 moles of O2 (thrice the amount of N2H4). That will give you 2 moles of NO2 (twice the amount of N2H4) and two moles of H20 (twice the amount of N2H4). If you have 20 moles of N2H4 you need 20*3 = 60 moles of O2, which is exactly what you have. Had you had 100 moles of O2, there would be 40 moles of O2 left over. Knowing that 20 moles of N2H4 will react, you know that you'll have 20*2 = 40 moles of each one of the products. Since 40 + 40 = 80, you'll end up with 80 moles of gas after the reaction finishes. Once again, had you had 100 moles of O2, you'd still have 80 moles of product by the end plus 40 moles of O2. In this reactions the total number of moles is conserved (1 + 3 = 2 +2 ), so it's easy. There are several reaction where moles of gas are created or destroyed. Most explosives, for example, involve reactions that produce huge amounts of gas, which will expand, creating massive destruction.

    A chemical equation gives you the proportions of reactants you need and the proportion of the products you will obtain. That's what you need to keep in mind. You should always measure everything with respect to the quantity of limiting reactant (the reactant there's not enough of).
     
    Last edited: May 14, 2012
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