1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mere approximation problem or something else? Thermodynamics

  1. Apr 19, 2015 #1
    1. The problem statement, all variables and given/known data
    The question is :
    Calculate the work done in joules when 1.0 mole of N2H4 decomposed against a pressure of 1.0 atm at 300 K for the equation:
    3N2H4
    (l)→4NH3(g)+2N2(g)
    2. Relevant equations
    None

    3. The attempt at a solution
    I did it like as:
    Assuming 100% dissociation of N2H4
    4/3 moles of NH3 and 2/3 moles of N2
    will be formed. That means a total of 2 moles of gas will be formed , which will have a volume of 2×22.4L at STP.
    (SINCE AT STP VOLUME OF 1 MOLE OF GAS=22.4L)
    Which will be equal to 44.8 ×10-3 m3
    So work done =PΔV=105×44.8×10-3=4480J (with a negative sign)
    But its answer is -4988.4 J which is much different from mine. What I think is that mere approximations in the values of Molar volume at STP and pressure change from atm to N/m2 cannot cause such a large deviation. So is my method incorrect? Or the given answer?
     
  2. jcsd
  3. Apr 19, 2015 #2

    Quantum Defect

    User Avatar
    Homework Helper
    Gold Member

    What temperature is "STP" -- how does that compare with the 300 K of the original problem?
     
  4. Apr 19, 2015 #3
    Okay....that was a silly mistake....STP is for 273K not 300K . Now I will never forget it. Thank you!!!!
     
  5. Apr 23, 2015 #4
    It seems to me there is something wrong here. If the initial number of moles were 1, and the final number of moles were 2, then at 273K, the change in volume would have been 22.4 liters, not 44.8 liters. So it seems to me that the "correct answer" is a factor of 2 too high. What do you guys think?

    Chet
     
  6. Apr 23, 2015 #5
    But ##N_2H_4## is a liquid at initial conditions.
     
  7. Apr 23, 2015 #6
    Ah. Thanks. That explains it. Duh!!!

    Chet
     
  8. Apr 23, 2015 #7
    Sir I think you knew it but asked the question to check mooncrater.:wink:
     
  9. Apr 23, 2015 #8
    You're very kind, but, no, I didn't notice that (l) in the problem statement. And I definitely didn't know from prior experience that N2H4 is a liquid at ordinary conditions. o:)

    Chet
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Mere approximation problem or something else? Thermodynamics
  1. Thermodynamics Problem (Replies: 4)

  2. Thermodynamics problem (Replies: 4)

  3. A thermodynamic problem (Replies: 16)

Loading...