Ideal Gas Equation Contradiction [duplicate]

Question

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• When I heat up a balloon, does the air inside increase in pressure as well as volume? 2 answers

You have a regular balloon, and you decide to heat it up. The gas particles inside the balloon will gain kinetic energy, and so, the gaseous particles will collide with the internal wall of the balloon with a higher frequency and more energy. This is what we call pressure, which will, as a result, increase. This is more commonly generalised to be Gay-Lussac's law, which formally states that the pressure of a gas is directly proportional to its absolute temperature.

Now, as pressure has increased, meaning that the gaseous particles are colliding with the internal wall of the balloon with a higher frequency and more energy, the balloon will begin to inflate, or in other words, the volume of the balloon will increase.

This seems to contradict Boyle's law, which states that the pressure of a gas is inversely proportional to its volume. In the scenario I described above, it seems clear to me that it was the increase in frequency of the gaseous particles colliding with the internal wall of the balloon, that too with more energy, which altogether is known as an increase in pressure, that caused the balloon to inflate.

I perfectly understand Boyle's law when the idea of a pump is used, where you can use a slider to change the volume of a gas and clearly see its inversely proportional relationship with the pressure of the gas, however, can't seem to understand what is incorrect in my logic, from which I seemingly have contradicted Boyle's law.

I am a high-school student who just learnt about these laws, and out of curiosity decided to ask this question in hopes of clearing my doubts and fundamentally gaining a better understanding. Any help will be greatly appreciated, thanks in advance.

Answer 1

Your arguments do not contradict each other. You are trying to separate a process into two processes, which are coupled of course, but in first approximation this is a good way at looking it.

Let's state our model system again: A balloon with the pressure $p_1$, the volume $V_1$ is heated from temperature $T_1$. As a result we see an inflated balloon. Let's call those variables with $2$ as an index. Of course we treat everything as ideal as possible.

Now we look at the coupled process as two separate processes:

1. Isochoric heating, i.e. $V=\mathrm{const}$: \begin{align} &&p_1V&=n\mathcal{R}T_1\\ &&p_2V&=n\mathcal{R}T_2\\ \therefore &&\frac{p_1}{p_2}&=\frac{T_1}{T_2}\\ \therefore &&p&\propto T\\ \end{align} That is Gay-Lussac's law.
2. Isothermal expansion, ie. $T=\mathrm{const}$ \begin{align} &&p_1V_1&=n\mathcal{R}T\\ &&p_2V_2&=n\mathcal{R}T\\ \therefore &&\frac{p_1}{p_2}&=\frac{V_1}{V_2}\\ \therefore &&p&\propto \frac{1}{V}\\ \end{align} That is Boyle's law.

You can also look at the whole process as the pressure stays the same while the balloon is heated. That then is an isobaric process, i.e. $p=\mathrm{const}$: \begin{align} &&pV_1&=n\mathcal{R}T_1\\ &&pV_2&=n\mathcal{R}T_2\\ \therefore &&\frac{V_1}{V_2}&=\frac{T_1}{T_2}\\ \therefore &&V&\propto T\\ \end{align} That is known as Charles' law

Of course in reality there will be a mixture happening as the rubber of the balloon will expand, but the pressure that is possible is dependent on the material of it. In combination you will arrive at: \begin{align} &&p_1V_1&=n\mathcal{R}T_1\\ &&p_2V_2&=n\mathcal{R}T_2\\ \therefore &&\frac{p_1V_1}{T_1}&=\frac{p_2V_2}{T_2}\\ \therefore &&\frac{pV}{T}&\propto \mathrm{const}\\ \end{align}

Answer 2

There's no contradiction because Gay-Lussac's law only applies if the volume is held constant and Boyle's law only applies if the temperature is held constant. In the balloon example, both the volume and temperature change so neither law applies.