3
$\begingroup$

Why does the temperature of a tyre rise when it is pumped up rapidly?

The way my textbook explains this is that $V\propto T$, so, an increase in volume causes an increase in temperature.

However, I think that this justification is flawed, because air is a real gas instead, and follows (approximately) the van der Waals' equation of state: $\left(P+\frac{a}{V^2}\right)(V-b)=RT$. I'm not not sure how I can neglect the $a/V^2$ term in it, considering the volume in a tyre is not that high.

Moreover, I actually had thought the reverse statement to be true. Pumping air "rapidly" implies the conditions are adiabatic i.e. heat exchange is negligible. So, $TV^{\gamma-1}=\text{constant}$ holds, implying an inverse relation between volume and temperature.

So, why actually does pumping a tyre increase its temperature? And what is the fault in my reasoning above?


Source: KS Verma; Physical Chemistry for JEE (Advanced): Part 1; Exercise 5.1 Question 6

$\endgroup$
  • $\begingroup$ Hmm, I'm tempted to downvote all answers. Probably more important process is loss of energy as heat, which accumulates in fast process. $\endgroup$ – Mithoron Mar 24 '18 at 18:34
  • $\begingroup$ Isn't $TV^{\gamma-1}=\text{constant}$ and ideal gas property? $\endgroup$ – Eashaan Godbole Mar 29 '18 at 11:20
  • $\begingroup$ @EashaanGodbole Oh you're right. Well it doesn't. But I thought it may hold "approximately" if not exactly. Like the "inverse relation between volume and temperature" may still be true. $\endgroup$ – Gaurang Tandon Mar 29 '18 at 11:39
  • $\begingroup$ Well, around 34 psi and 298K, the molar volume of an ideal gas would be around 10 L/mol. Ideal gas law won't be too far off. (It's usable.) $\endgroup$ – Eashaan Godbole Mar 29 '18 at 11:43
3
$\begingroup$

Once the pressure in the tire has risen a little, the volume of the tire doesn't change much any more. But, you are increasing the number of moles of air inside the tire when you pump air into it. The new air pumped into the tire compresses the gas that was already in there previously and raises its pressure. So the new gas is doing work on the previous gas to compress it adiabatically and approximately reversibly. This causes the internal energy of the gas and its temperature to increase.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Oh, nice! This makes sense. Could you please also add a line or two about why my attempted explanation was leading to the wrong result? $\endgroup$ – Gaurang Tandon Mar 25 '18 at 2:21
  • $\begingroup$ It was leading to the wrong result because your answer didn't take into account the fact that the number of moles of air inside the tire was increasing. $\endgroup$ – Chet Miller Mar 25 '18 at 2:59
  • 1
    $\begingroup$ Plus, the pressure is building up in the tire, so the new air being injected is getting adiabatically compressed by the pump even before it enters the tire. This causes the new air to enter at a higher temperature. $\endgroup$ – Chet Miller Mar 25 '18 at 3:38
  • $\begingroup$ Yep, that makes sense. $\endgroup$ – Gaurang Tandon Mar 25 '18 at 5:49
1
$\begingroup$

We know that $\Delta U=Q-W$, $\Delta U$ being change in internal energy. Further, $\Delta U=nC_V\Delta T$. As process is adiabatic, $Q=0$. Therefore, all work done is gone to change the internal energy which implies its temperature rises rapidly. Also, $TV^{\gamma-1}$ does hold. Consider $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}=\text{constant}$ ($T_1$ and $T_2$ being initial and final temperatures). Clearly, $V_2$ is less than $V_1$, implying $T_2$ is greater than $T_1$ for above equation to hold. Thus it is not reverse but decreasing volume increases temperature in adiabatic process.

| improve this answer | |
$\endgroup$
  • $\begingroup$ " it is not reverse but decreasing volume increases temperature in adiabatic process." Actually that contradicts the given statement, which is that increasing volume increases temperature. $\endgroup$ – Gaurang Tandon Mar 25 '18 at 5:45
  • $\begingroup$ What I am saying is decreasing volume increases temperature.When you do work on gas,you reduce gas volume.By PV^γ=Constant,decreasing volume implies temperature increases for to above eqn to hold true.You must have used pump to inflate tires of bicycles in your childhood.The pump can be seen as piston container system,where when you pump down the volume decreases thereby increasing temp. $\endgroup$ – Panda Mar 25 '18 at 8:01
1
$\begingroup$

Kinetic theory of gases, if your volume is constant, but your pressure is increasing as consequence of adding molecules, they are going to increase the number of hits against the wall, in this case the wire. In turn, the frequency by which each molecule transfer energy to the wall is going to be larger than the previous state. Then, temperature rises.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Give an argument on why -1, don't be mediocre . $\endgroup$ – Another.Chemist Mar 24 '18 at 18:32
  • $\begingroup$ (not my downvote) I don't know why there was any need to mention the KTG theory at the beginning of the answer. I guess it only applies to ideal gases and the rest of the answer seems to make sense without it as well. What specific postulate of it were you trying to refer to? $\endgroup$ – Gaurang Tandon Mar 25 '18 at 2:22
0
$\begingroup$

Suppose the tire's (absolute) pressure is increased by the tire pump from $P_0 \ge P_1$, where $P_1$ = atmospheric pressure, to a final pressure $P$ at the ambient temperature $T_0$ to $T_1$.

Also, the process is adiabatic, as you mentioned it is done rapidly.

$TP^{1/\gamma-1} = \text{constant}$

Therefore, $T_0P_0^{1/\gamma-1} = T_1P_1^{1/\gamma-1}$

Putting $\gamma=7/5$ for air,

$T_0P_0^{-2/7} = T_1P_1^{-2/7}$

$T_1/T_0 = P_1^{2/7}/P_0^{2/7}$

Since, $P_1>P_0$, $T_1>T_0$.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.