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Edge Of Pain

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## Homework Statement

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.

## Homework 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]

## 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.