1. Not finding help here? Sign up for a free 30min 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!

Thermodynamics : molar specific heats for gases

  1. Dec 5, 2014 #1
    1. The problem statement, all variables and given/known data

    An audience of 2750 fills a concert hall of volume 35000 m3. If there were no ventilation, by how much would the temperature of the air rise over a period of 2.0 h due to the metabolism of the people (70 W/person)?

    2. Relevant equations

    Q= nCvΔT
    Cv= (3/2) R

    3. The attempt at a solution

    Since the v keeps constant, we use Cv to calculate the change of T of the air. We want to get the value of ΔT and we know n and Cv.

    The total heat release by audience: 2750 x 70 x 2.0 x 60 x 60 = 1386000000 J

    PV=nRT, so n = PV/RT = (1.013 x 10^5 x 35000) / (8.314 x (20 + 273.15)) = 1454713 mole

    Cv = 1.5 x 8.314 =12.471

    ΔT = Q/ (n x Cv) = 76.3 K... which doesn't make sense.
     
  2. jcsd
  3. Dec 5, 2014 #2

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    What about your answer doesn't make sense?
     
  4. Dec 5, 2014 #3
    Oh sorry! I forgot to add that the answer is 43.1 K.
     
  5. Dec 5, 2014 #4

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    (3/2)R is the constant volume heat capacity of a monatomic ideal gas. Give you any ideas?
     
  6. Dec 6, 2014 #5
    Yes I am kind of confused about which volume heat capacity I should use, since there are both monatomic and diatomic gas in air.
    For diatomic gas, the heat capacity is 5/2 R, but I don't know if I should use that.
     
  7. Dec 6, 2014 #6

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Which gases are monatomic? Nitrogen (79% of air)? Oxygen (21%)? Looks like everything else in the air is basically at trace levels only.
     
  8. Dec 6, 2014 #7
    So Cv should be 5/2 R instead of 3/2 R!
    but I plug in Cv = 5/2 R I get 45.8 K as the answer, still not correct.. I don't know what is going wrong here with three degrees difference.
     
  9. Dec 6, 2014 #8
    anyone helps? : P
     
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: Thermodynamics : molar specific heats for gases
  1. Molar Specific Heat (Replies: 1)

  2. Molar specific heat (Replies: 1)

  3. Molar Specific Heat (Replies: 5)

Loading...