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Determining final temperature, pressure and work done

  1. Jul 11, 2012 #1
    hi guys im swamp with work and have not been able to go through my assignments proper

    1. one cube meter of an ideal gas at 600K and 1000Kpa expands to 5 times its initial volume: by mechanically reversible, isothermal process and by a mechanically reversible, adiabatic process


    2.for each case calculate the final temperature, pressure, and the work done by the gas. take Cp= 21 J/mol.K



    3. For isothermal process:
    T2= 600
    P1V1=P2V2
    thus P2= (1*1000)/5 =200Kpa
    W= -P1V1 ln V2/V1
    = -(1000000000*1) ln (5/1)
    = -1609.438 KJ
     
  2. jcsd
  3. Jul 11, 2012 #2
    Re: hi guys im swamp with work and have not been able to go through my assignments pr

    P2 is ok.

    Check the value of P1V1 in the calculation for the work.
     
  4. Jul 11, 2012 #3
    Re: hi guys im swamp with work and have not been able to go through my assignments pr

    P1= 1000kpa = 1000*10^3 pa =1000000000 pa
    V1= 1m^3

    thus W using pa is equal to -1609.438 KJ
     
  5. Jul 11, 2012 #4
    Re: hi guys im swamp with work and have not been able to go through my assignments pr

    P1= 1000kpa = 1000*10^3 pa =1000000 pa
     
  6. Jul 11, 2012 #5

    TSny

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    Your work looks good except that when a gas expands reversibly, the gas does a positive amount of work.

    It's always important to distinguish between the work done on a system and the work done by the system.
     
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