I was just working on a special question but I ignored the effect of temperature on it and now it becomes very important to me.

What is the relation between Pressure and Temperature?

Suppose we have a balloon or something that we can fill it with air {air pressure is 1 a.t.m }, if we increase the temperature, what will happen for the pressure? Is there a formula for measuring it?

For answering that question, please consider the elasticity of the balloon.

  • $\begingroup$ Have you heard of the ideal gas law? $\endgroup$ – Kyle Kanos Jan 27 '15 at 18:27
  • $\begingroup$ Also note that the pressure in these relations are absolute pressure, not gauge. For example, if the absolute pressure inside a balloon at your house is 1 atm, the balloon is not inflated. If the gauge pressure is 1 atm, the absolute will be 2 atm. $\endgroup$ – Bill N Jan 27 '15 at 18:33
  • $\begingroup$ of course I heard it , but isn't it different for rubbers & elastics???? $\endgroup$ – Michaele Jan 27 '15 at 19:08
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    $\begingroup$ I didn't derive this formally (and thus check properly), which is why I write this as a comment rather than as an answer. Young-Laplace gives $p = 2\gamma/r$ (assuming the balloon is tight) and the ideal law $pV = NkT$. Taking $\gamma\propto A$, and combining the equations we have $p \propto T^{1/4}$. $\endgroup$ – alarge Jan 28 '15 at 2:24
  • $\begingroup$ I couldn't understand , Can you tell me the real formula??? $\endgroup$ – Michaele Jan 28 '15 at 16:00

A well-known result from statistical mechanics is the ideal gas law,

\begin{equation} PV = nRT\end{equation}

which comes in a variety of forms. Here, $n$ denotes the amount of gas, $R$ is a constant, $T$ is the temperature, $V$ the volume and $P$ the pressure.

If you increase the temperature, either the volume, the pressure or both must increase proportionally. If the balloon cannot expand, the volume cannot increase; thus, the pressure will increase (with $\frac{nR}{V}$ per degree). If there is a certain degree of elasticity, the volume may increase somewhat; however, not following the ideal gas law. As an astronomer, I have not worked with elasticities much, so an applied physicist can probably help you further.


An ideal gas is a theoretical gas composed of many randomly moving point particles that do not interact except when they collide elastically. It all depends on your case. I mean if the pressure and temperature are low, you can use Ideal Gas law to calculate the relation between pressure and temperature.

enter image description here


enter image description hereis the pressure of the gas

V is the volume of the gas

n is the amount of substance of gas (also known as number of moles)

R is the ideal, or universal, gas constant, equal to the product of the Boltzmann constant and the Avogadro constant.

T is the temperature of the gas

And we know :

enter image description here

where :

m is mass (grams)

M is molar mass (grams per mole)


enter image description here

You should check the case you are facing with and then decide to use this or not to use it. but something really important is that ideal gas law doesn't answer for elastic cases.


Make sure you use T in Kelvins, and have the other units compatible with each other.

You should also look up " pressure altitude " and " temperature altitude ", and " Lapse Rate " to see if these apply to your problem.

As you increase altitude, the confining atmospheric pressure and temperature decreases, so the Balloon increases in size compared to lower altitudes.


Quick derivation

The Young-Laplace law states that $$p-p_0 = \frac{2\gamma}{R}$$ whereas the equation of state of the ideal gas goes as $$p = \frac{Nk_BT}{V}$$ Solving for $R$ and assuming that we are dealing with a spherical balloon ($V = \frac{4}{3}\pi R^3$, $A = 4\pi R^2$), and that the elasticity is described by a Hookean force (with equilibrium at zero size), $\gamma = \alpha A$, $$\left(\frac{Nk_BT}{\frac{4}{3}\pi p}\right)^{1/3} = R = \frac{p-p_0}{8\pi\alpha}$$

To make the algebra simpler, I assume that $p_0 = 0$, so that we have $p\propto T^{1/4}$.

Slightly more rigorous derivation

For simplicity I'm going to assume that the pressure outside is zero. Adding non-zero pressure is trivial, though, but makes the equations a bit uglier.

Suppose we have a sphere filled with $N$ molecules of ideal gas, so that the partition function can be written as $$\mathcal{Z} = \iint \mathrm{d}^{3N}p\ \mathrm{d}^{3N}r\ \ e^{-\beta(\mathcal{H}+\gamma A)}$$

So, we are left with $$\mathcal{Z} = C V^N e^{-\beta\gamma A}$$

Now, minimizing the free energy with respect to $R$, $$N\frac{A}{V} = \beta \partial_R(\gamma A)$$

Taking the rubber to be Hookean, $\gamma = \alpha A$, we finally have the size of the balloon: $$R = \left(\frac{3N}{64\pi^2\alpha\beta}\right)^{1/4}$$

Now it is easy to compute the pressure, $$p = -\left(\frac{\partial \mathcal{F}}{\partial V}\right)_A = \frac{N\frac{A}{V}}{\beta A} = \frac{N}{\beta V}$$ No surprise here; this is just the equation of state of the ideal gas. Plugging in the size ($V\leftarrow\frac{4}{3}\pi R^3$), we have $p \propto \beta^{-1/4} \propto T^{1/4}$.

I also wrote a simple Monte Carlo simulation (which could easily be extended to cover more general cases where the gas is non-ideal, say), and my numerical results agree with what I derived above.


Temperature and pressure are directly proportional to each other. This means that as the temperature decreases, the pressure also decreases, and as the temperature increases, the pressure increases. One way to think of this is if you increase the speed of the molecules –by increasing their temperature- the force of the molecules hitting their container increases and this increases the pressure. This relationship is called Gay-Lussac’s Law and makes up part of the ideal gas law.


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