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!

Pumping Bicycle tire: Find volume, work done, temperature

  1. Jul 6, 2015 #1
    1. The problem statement, all variables and given/known data
    Daniel Schroeder, introduction to thermal physics problem 1.36 (page 26):

    "In the course of pumping up a bicycle tire, a liter of air at atmospheric pressure is compressed adiabatically to a pressure of 7 atm. (Air is mostly diatomic nitrogen and oxygen.)

    a) What is the final volume of this air after compression?

    b) How much work is done in compression the air?

    c) If the temperature of the air is initially at 300K, what is the temperature after compression?

    Also, can anyone tell me how to properly write things as fractions in LaTeX? I'm finding that \frac is annoying to use.

    2. Relevant equations
    Ideal gas law, given by [tex] PV = NkT. [/tex]

    Assuming quasistatic compression, so for work done we can use [tex] W= \int_{V_i}^{V_f} P dV. [/tex]

    For adiabatic + quasistatic processes, we know that [tex] VT^{f/2} = constant. [/tex]

    It also follows that [tex] PV^{(2+f)/f} = constant. [/tex]

    To make things simpler to write, let [tex] \gamma := (2+f)/f. [/tex]

    First law of thermodynamics might help, too. That's [tex] \Delta U = Q + W. [/tex]

    Equipartition theorem: For a system at temperature T, each quadratic degree of freedom has energy [itex] {(1/2)} (fkT). [/itex]

    3. The attempt at a solution
    a) doesn't trouble me, but b) does, and for c) there seem to be two methods that disagree with one another. I'll put all my working for all three of them anyway.

    a) To find final volume, we recall that for an adiabatic, quasistatic process (which, according to Schroeder, is a fairly accurate approximation of real-life processes), we have [itex] PV^{(2+f)/f} = constant. [/itex] Since we know initial pressure, final pressure, and initial volume we should be able to find final volume fairly quickly, it's just rearrangement. Since the product of pressure and volume is constant it follows that initial pressure times initial volume raised to the power of gamma must equal final pressure times final volume raised to the power of gamma.

    That is, [tex] P_i V_i^{\gamma} = P_f V_f^{\gamma} . [/tex]

    Using logarithms and exponentiation of both sides gives that [tex] V_f = ((P_i V_i^{\gamma})/P_f)^{1/\gamma} .[/tex]

    Substituting numerical values after unit conversion (1 atmosphere is 10^5 pascals, 1 liter is 0.001 cubic metres, degrees of freedom f = 5 for a diatomic gas at room temperature) gives that [tex] V_f = 2.49*10^{-4} m^{3} [/tex] or about 0.25 liters.

    b) We must find the work done in compressing the air. We will use [tex] W= \int_{V_i}^{V_f} P dV. [/tex]

    Now, since it's not an isothermal process we can't just use P = NkT/V in the integral. We need to use that [itex] PV^{\gamma} = constant. [/itex] So, it would follow that for an arbitrary P and corresponding V, we have that [tex] PV^{\gamma} = P_i V_i ^{\gamma}. [/tex] We should also be able to use P_f and V_f instead, it would give us the same result since the product of them is constant.

    But according to a solution I found this is wrong, and you should instead use [itex] PV_i^{\gamma} = P_i V^{\gamma}, [/itex] then rearrange for P and put this into the integral. What?! That seems completely wrong since the values of P_i and V_i aren't corresponding to the right values of P and V. Blindly trusting the [itex] PV_i^{\gamma} = P_i V^{\gamma} [/itex] rearrangement and plugging it into the integral gives work done as about 40 Joules. But I don't care about getting the right answer until I understand why [itex] PV_i^{\gamma} = P_i V^{\gamma} [/itex] is correct.

    c) We must find the final temperature due to the changes in volume and pressure, given that initial temperature T_i = 300 kelvin.

    The high school method which I completely forgot about, but do understand, is one method of doing it. That is, we know that Boltzmann's constant k and number of molecules N are constant. So, [tex] P_i V_i = NkT_i [/tex] and [tex] P_f V_f = NkT_f, [/tex] so it follows that [tex] (P_i V_i )/T_i = (P_f V_f) / T_f. [/tex] Substituting numbers in and rearranging gives an answer of 522 kelvin, which seems ridiculous. It doesn't seem right that pumping a bicycle tire increases its temperature by over 200 degrees.

    The other method, the one I thought to use gives an apparently wrong answer. I used the first law of thermodynamics. We invoke [tex] \Delta U = Q + W. [/tex]

    Now, it's an adiabatic process so the compression was fast enough to not allow any heat to escape during. So we can set Q to 0. Now we have [tex] \Delta U = W. [/tex]

    For change in internal energy, recall the equipartition theorem: For a system at temperature T, each quadratic degree of freedom has energy [itex] {(1/2)} (fkT). [/itex] In this case we have a system with N molecules so we just multiply this by N to get the total internal thermal energy. The change in energy is provided entirely by the change in temperature, so we have [tex] \Delta U = (1/2) (Nkf) \Delta T. [/tex]

    We now have that [itex] (1/2) (Nkf) \Delta T = W. [/itex]

    Problem is that we don't know what N is. No matter, we know from the ideal gas law that [itex] P_i V_i = NkT_i [/itex] and [itex] P_f V_f = NkT_f [/itex]. So we can rearrange for Nk. We get [itex] (P_i V_i) / T_i = Nk [/itex]. We now substitute this in for Nk in [itex] (1/2) (Nkf) \Delta T = W, [/itex] to get [tex] ((P_i V_i) / T_i) * 0.5 *f * (T_f - T_i) = W. [/tex] Rearranging gives T_f = 348 kelvin, which seems far more reasonable. But both methods seem to be correct, so why do they disagree?

    Thank you for taking the time to read my wall of text.
     
  2. jcsd
  3. Jul 6, 2015 #2
    Hint, the auto-ignition temperature of diesel fuel is 210 deg C (483 K), and diesel engines usually require a compression ratio of at least 14 for autoignition to occur reliably.
     
  4. Jul 6, 2015 #3
    I'm sorry but I'm not sure how this relates to the problem unless I'm missing something.
     
  5. Jul 6, 2015 #4
    Your problem about how compression ratio is related to heating.

    The diesel engine provides an experimental example where a certain compression ratio is required to reach a certain temperature. It provides a reality check which can help you determine which of your two methods is incorrect.
     
  6. Jul 6, 2015 #5
    I'm just going through this textbook "Introduction to Thermal Physics" by Daniel Schroeder and I've not come across compression ratios or diesel engines yet. Also, no heating actually happened. Work was done, increasing the temperature, but no "heat" was added or removed.
     
  7. Jul 6, 2015 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    On the algebra, I agree with you... ##PV^{\gamma}## should be constant, giving a temperature ratio of about 7/4.
    From personal experience, which is far from adiabatic, 348K and 40J are too low.
    @Chestermiller , any comment?
     
  8. Jul 6, 2015 #7
    I agree with you and with Edge of Pain. He correctly assessed that the equation given in the "accepted solution" is incorrect. Regarding the 522 versus the 348, the 522 is correct. The first law equation he used to get the final temperature is just the integrated version of the differential equation that led to the 522 value. I wonder what value he got for the work when he used the correct equation between P and V? I wonder what value that would have given for the final temperature. The final temperatures calculated the two different ways have to agree.

    Chet
     
  9. Jul 7, 2015 #8
    OK, so I should use a rearrangement of [itex] PV^{\gamma} = P_i V_i ^{\gamma} [/itex] in part b)? If so that makes me happy because it fits with what I understand. But I got gamma as 7/5 rather than 7/4 because diatomic at room temperature => f = 5.

    I will now find the work done using the right equation for part b) and see if the work done I find leads to the two methods in part c) agreeing. I'll be back soon - I'll just edit this post with new answers if nobody replies within that time.

    EDIT:

    OK, with the correct equation for b), I get about 186J of work, and it leads to the two methods in part c) agreeing with each other, at 522K!

    Thank you everyone, I've learned to go with what seems intuitively correct rather than abandoning my own thoughts after seeing a solution.

    There doesn't seem to be an option for editing the OP. I would like to edit the OP with the correction at the top of the post, as I think it will be helpful for other users who see this. Can I do that?
     
    Last edited: Jul 7, 2015
  10. Jul 7, 2015 #9
    The 7/4 value is the approximate temperature ratio, not γ.

    Regarding the editing of posts, you have a certain amount of time after submitting a post to edit it. After that time expires, you can no longer edit. In this thread, I see no compelling reason for any of the posts to be edited. The existing sequence accurately reflects the history of the transition to a solution.

    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: Pumping Bicycle tire: Find volume, work done, temperature
  1. Bicycle pump (Replies: 6)

  2. Bicycle pump! (Replies: 3)

Loading...