# Fire piston temperature

Sometime this week I will be demonstrating an example of Boyle's law and Gay-Lussac's law for a small class project. I will be using a fire piston to ignite different objects: https://en.wikipedia.org/wiki/Fire_piston

I've been doing some calculations, and I want to know if I missed something.

I would start with the piston at 5 cm3, at 1atm (not sure how our elevation affects it, my watch has a barometer), and a room temperature of 300K.

Wikipedia and my teacher tell me that P1V1=P2V2.

1 atm * 5 cm3 = 5 atm * 1 cm3

So I predict that the pressure would be 5 atm. Wikipedia (and my teacher) tell me that, according to Gay-Lussac's law:

P1/T1 = P2/T2

So...

1atm/300K = 5atm/1500K

1500K = 1227C

That seems a bit excessive to be produced just by smacking down a piston, is it realistic or did I miss something?

Thanks.

There is a problem with your calculation. $P_1V_1 = P_2V_2$ is only true at constant temperature, which is clearly not the case here.

Instead you need to look at this as an adiabatic process, in which case

$$TV^{\gamma-1}=\rm{const}$$

In this expression, $\gamma=\frac{C_p}{c_v}$ is the ratio of specific heat at constant pressure and constant volume; for air it is about 1.4.

This means that if you can compress the cylinder from 5 cc to 1 cc without any heat loss through the cylinder, the temperature inside would be given by

$$T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{0.4} \approx 1.9 T_1 = 571~K$$

That's not as hot as you calculated - but hot enough, according to the wiki article you linked, to set fire to dry tinder.

• You're welcome. Good luck with the class project. I hope you manage to get some things to catch fire... :-) – Floris Jun 2 '17 at 2:26

Gay-Lussac's law only works if the volume of the gas is kept constant. In fact, it is a special case of the more general ideal gas law,

$$pV=N k_{B} T$$

where $N$ is the number of molecules in the gas and $k_{B}=1,38 \cdot 10^{-23}\$ J/K is a constant known as the Boltzmann constant. Solving the law for $p/T$ you get

$$p/T=Nk_{B}/V$$

so that if $V$ is kept constant,

$$p_{1}/T_{1}=Nk_{B}/V=p_{2}/T_{2}$$

On the other hand, Boyle's law only works if you keep the temperature constant. In this case, again from the ideal gas law, you get

$$p_{1}V_{1}=N k_{B} T=p_{2}V_{2}$$



I understand that in your experiment you will be changing the volume of the gas by acting on the piston. Generally, what happens when you do so is that both the temperature and the pressure of the gas change, so that in principle you cannot use neither of those laws (on the other hand, the ideal gas law still remains valid). How they change actually depends on the amount of heat that the gas is able to exchange with the environment. For example, if you push down the piston very slow, any increase in temperature is compensated by the environment: the environment itself re-cools down the gas to room temperature as you decrease the volume. Thus you can indeed demonstrate Boyle's law, provided that you don't push down the piston too fast. On the other hand, as Gay-Lussac's law, as I said, only makes sense when the volume is kept constant, it cannot be demonstrated by pushing down the piston. What you should do to demonstrate it is, for example, heating up the enclosure of the gas while keeping its volume fixed. In this case, the pressure will increase proportionally to the increase in temperature.

P.S.: I understand that a fire piston is used precisely in the opposite regime, namely by pushing down the piston very fast. In this regime, the temperature of the gas quickly increases, but the pressure increases too! So in this regime you cannot use neither Gay-Lussac's nor Boyle's law, nor the third equation that you would deduce from the ideal gas law, namely

$$V_{1}/T_{1}=V_{2}/T_{2}$$

which assumes that $p$ is constant.