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Homework Help: Increase in Internal Energy of a gas during compression

  1. May 24, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider a pump that is required to compress air in a factory. The cylinder in the pump has an inner
    diameter of 2.00 cm and length 60.0 cm. Air is drawn into the pump at atmospheric pressure and
    18°C and the pump adiabatically compresses the air to a pressure of 17 atmospheres.

    Calculate the volume and temperature of the compressed air.

    I calculated the volume to be 24.9153 cm3, and the temperature to be 653.8 Kelvin. Volume was found by using the P1V1gamma = P2V2gamma formula, presuming it was a diatomic gas (air), and for temperature just used P1V1/T1 = P2V2/T2 (or could have used the T1V1^gamma-1)

    But this question stumps me:

    What is the increase in internal energy of the gas during the compression?

    2. Relevant equations

    Eint = (5/2) * n * R * T

    Eint = (5/2) * N * Kb * T

    ΔE = Q + W

    W = PΔV

    3. The attempt at a solution

    My first thought was to use the top equation with the different temperatures and minus them from each other, but realised I didn't have the amount of moles.
    My second thought was to use the last equation and sub it into the second to last equation (I know Q=0 since it's adiabatic), but was confused about what pressure to use? Would I use 101300 Pa, or 1722100 Pa (which is 17 atm), or neither?
  2. jcsd
  3. May 24, 2013 #2


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    Staff Emeritus
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    Gold Member

    Welcome to PF Vegie,

    It's an ideal gas, so you know that nR = PV/T, which means that you know nR, and you can use your first equation for internal energy. Another way to say this is that at 1 atm and 18 C, a cylinder of that volume is going to hold a specific number of moles of gas, and this is fully determined.

    Speaking of the cylinder volume, yours doesn't look right to me. I would double check it.
  4. May 24, 2013 #3
    Thanks for your reply.

    I was unsure about the volume, but I'm not too sure how else I could calculate it (Unless I used P1V1 = P2V2 to find V2 and then using P1V1 / T1 = P2V2 / T2 to find T2?)

    Since the whole topic that this question is on sort of focused around adiabatic expansion, I presumed that they want the PiVigamma = PfVfgamma equation used.

    Earlier, I tried using the Ideal Gas Law and rearranging to find 'n', but the amount of moles changed when I put both my results for P2 and V2 in.

    So, in your opinion, do you think I should use P1V1 = P2V2 to find V2, and then find the amount of moles, and then use that for my internal energy question?
  5. May 25, 2013 #4
    What did you get for the initial volume (this is a straight geometry problem)?
    From the initial volume, the initial pressure, and the initial temperature, you should be able to use the ideal gas law to calculate the number of moles (which doesn't change). What do you get for the number of moles?
  6. May 25, 2013 #5

    Andrew Mason

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    Homework Helper

    Yes, use PVγ = K to find the change in volume. Then use Tf/Ti = PfVf/PiVi to find the final temperature and, from that, the change in internal energy.

  7. May 25, 2013 #6
    My initial volume was calculated to be 1.8852x10-4 m3. Using this volume to calculate the number of moles, I get 7.9x10-3 moles.

    That's what I've done, but I'm having trouble with the change in internal energy part. How do I calculate the change in internal energy?
  8. May 25, 2013 #7
    You can calculate the change in internal energy in two different ways, and the results from both of these should be the same:

    Method 1: ΔE = W
    Method 2: ΔE = 2.5 nR ΔT

    What do you get?
  9. May 25, 2013 #8
    Method 1: ΔE = W

    W = PΔV → 101300*(1.8852x10-4 - 24.915x10-6)
    W = 16.57 J ∴ ΔE = 16.57 J

    Method 2: ΔE = 2.5 nR ΔT

    ΔE = 2.5 * 7.9x10-3 * 8.314 * (653.8 - 291)
    ΔE = 59.57 J
  10. May 25, 2013 #9
    You calculated the work incorrectly. The work done by the surroundings on the system is not pΔV. It is W = -∫pdV.

  11. May 26, 2013 #10
    Ahhh I see. I'm not too good with the equation editor, so I'll try and explain this as best as I can.

    Using your equation, I had an intergrand of 1/(V^gamma) dv. My value for gamma was 7/5.

    This integrated to -5/2 * 1/V^(2/5).

    Subbing in my values for V2 and V1, and multiplying by P1 V1^gamma (which was out the front as a constant), I got 59.3.

    This value for 59.3 is very close to the 59.57 I got from the Method 2 equation you suggested. Since these match up, I'll take that these are correct.

    Thanks so much for your help :)
  12. May 26, 2013 #11
    You're welcomel. Nice job well done.

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