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Adiabatic expansion problems

  1. Dec 6, 2017 #1
    1. The problem statement, all variables and given/known data
    I do not know if I am allowed to have two question in one post, so forgive me if I am breaking a rule. These two are frustrating me as I cannot see where the error in my process is.

    1) A 1.00-mol sample of an ideal diatomic gas, originally at 1.00 atm and 10 ∘C, expands adiabatically to 1.75 times its initial volume. What are the final temperatures for the gas?

    2) An ideal monatomic gas, consisting of 2.8 mol of volume 8.3×10−2 m3 , expands adiabatically. The initial and final temperatures are 95 ∘C and -81 ∘C. What is the final volume of the gas?

    2. Relevant equations

    P_1*V_1 = P_2*V_2

    PV = nRT

    PV^γ = constant

    TV^(γ-1) = constant

    3. The attempt at a solution

    Attempt at #1
    : I know we needed final pressure for this; I had calculated it and was told it was correct. Though the answer was rounded, I kept my final pressure unaltered, since the calculator I am using allows me to assign single letter variables to numerical constants. The final pressure I got was 1 * (1/1.75)^1.4 =~ .457 when rounded.

    I calculated the true initial volume via the constants given; I converted the pressure (1 atm) to 101325 Pascals, and the temperature from Celsius to Kelvin. I then solved for Volume, then I multiplied by 1.75 to get the final volume, which gave me approximately .0398, but I again did not round and assigned it to a letter. I used algebra to get T_f = P_f*V_f / n*R = about 221 when rounded (like the other values, I assigned it a letter for an exact number). Then I subtracted 273.15 from that number and got about -52 degrees Celsius; rounded to two sigfigs as requested by the problem. But that is still wrong; I do not know where my error is.

    Attempt at #2: I felt this was simple, T_1*(V_1)^(1.4-1) = T_2*(V_2)^(1.4-1). Once again I set up my equation by changing Celsius to Kelvin, and we already had V_1, so I felt it was simple algebra.

    I solved for V_2, which came out to be about .42 (4.2 * 10^-1) which I felt was reasonable given there was such a sharp decrease in temperature. Again, I was told I was wrong and I am not sure why.
     
  2. jcsd
  3. Dec 6, 2017 #2

    TSny

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    Homework Helper
    Gold Member

    For problem 1 your answer is not very far off, but you may have too much "round off" error that has accumulated in your many separate calculations. I have not looked at problem 2 yet except to note that now you are dealing with a monatomic gas.

    There is a much simpler way to get the answer using just one of the equations that you listed. You do not need to find any pressures or the values of any volumes.
     
  4. Dec 6, 2017 #3
    For problem 1, the statement says diatomic. But, I agree that the T-V relevant equation is easier to use.
     
  5. Dec 6, 2017 #4
    I am afraid I do not understand. Is "round off" error an error that accumulates from rounding off the separate answers? If so, that should not be a factor. I am using Desmos Graphing Calculator so that I can assign the full answer for each separate calculation to a variable - the values I am calculating with are up to 12-13 digits.

    The true values for final pressure, volume and final temperature in Kelvin are 0.456821706524, 0.0397568566494 and 221.342781882 respectively. I calculate with these and then round at the very end. While there may still be round off errors, wouldn't they be small enough to not matter?
     
  6. Dec 6, 2017 #5

    TSny

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    Homework Helper
    Gold Member

    Yes, if you are calculating each step with that many sig figs, then you would not need to worry about round off error.

    I agree with your value for the final pressure, but not your final volume or temperature.

    As a check, try to work the problem the easy way by just using the same formula as you did in the second problem.
     
  7. Dec 6, 2017 #6
    I figured it out! Thank you! I will work on the second problem; I think I understand that since it is monatomic, the exponential will not be (1.4-1) but ((5/3)-1).
     
    Last edited: Dec 6, 2017
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