Assume we have a cylinder of given volume filled with gas of given temperature and pressure. The cylinder is enclosed from the top by a piston of provided mass. Now, we place a small mass on the piston. Is it possible to find how much does the temperature of gas change? We assume, for the simplicity, that no heat is exchanged with the environment and that the gas is ideal.

  • 1
    $\begingroup$ You need to assume something about the characteristics of the gas. For example is it justifiable to think about it as an Ideal Gas? $\endgroup$
    – Benjamin
    Mar 27, 2016 at 17:56
  • $\begingroup$ I made an edit, everything should be justified now. $\endgroup$
    – user89768
    Mar 27, 2016 at 17:57
  • $\begingroup$ Since there is no heat exchanged with the environment, the answer would be obviously, "Yes. The temperature of the gas will be changed." However there are three parameters, P, V and T that are all free to change satisfying the Ideal Gas Law. But, I am trying to see how we can tackle the problem mathematically. $\endgroup$
    – Benjamin
    Mar 27, 2016 at 18:06
  • $\begingroup$ This problem, as it stands, is not solvable as there is less information on the given system. We need three more pieces of information: exact value of the added mass, the cross section area of the piston and the amount of subsidence it experiences due to mass. $\endgroup$
    – Benjamin
    Mar 27, 2016 at 18:46
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    $\begingroup$ You would call it adiabatic. Since the weight does work on the gas that is energy. First assume T does not change and get the new volume and calculate work. The solve for T based on the thermal capacity of the gas. But then you will have new final V so you have to iterate. If you remove the weight then the cylinder would do work on the atmosphere so that pressure is potential energy but I am not sure how to deal with that. $\endgroup$
    – paparazzo
    Mar 27, 2016 at 19:14

1 Answer 1


Let $x$ be the amount the piston moves down due to the mass $m$:

enter image description here

The answer to your question is "no, there is insufficient information provided to find the change in temperature". However, if you know the cylinder's "shape" (ie, the $h:r$ ratio above) and the gas' heat capacity (https://en.wikipedia.org/wiki/Heat_capacity), the answer is yes:

  • To support the extra mass $m$, the force on the piston must increase by $g m$. Since force is pressure times area, we have

$\Delta P (\pi r^2) = m g\to \Delta P = \frac{g m}{\pi r^2}$

Note that $\Delta P$ is independent of $x$.

  • The change in volume is simply:

$\Delta V=-\pi r^2 x$

  • By Joule's Second Law, the total energy of an ideal gas depends only on its temperature:

$E_{\text{total}}=c n R T$

where $c$ is the gas' heat capacity.

Since the gas gains the $m g x$ of potential gravitational energy that the mass loses (and $c$, $n$, and $R$ remain constant):

$\Delta E_{total} = c n R \Delta T = m g x \to \Delta T = \frac{m g x}{c n R}$

  • By the Ideal Gas Law, we know that $\frac{P V}{n R T}$ remains constant. Since $n$ and $R$ are already constant, this means $\frac{P V}{T}$ remains constant. Thus, using $P$, $V$, and $T$ as the original pressure, volume and temperature, we have:

$\frac{P V}{T}=\frac{(P+\Delta P) (v+\Delta V)}{T+\Delta T}$

Using the values of $\Delta P$, $\Delta V$, and $\Delta T$ above, we can solve for $x$:

$ x\to \frac{c g m n R T V}{\pi r^2 \left(c n R T \left(g m+\pi P r^2\right)+g m P V\right)} $

Plugging this into our formula for $\Delta T$ yields:

$ \Delta T = \frac{g^2 m^2 T V}{\pi r^2 \left(c n R T \left(g m+\pi P r^2\right)+g m P V\right)} $

We can further simplify, since $n R T\to P V$:

$\Delta T = \frac{g^2 m^2 T}{\pi P r^2 \left((c+1) g m+\pi c P r^2\right)}$

(of course, we could've also made this simplification earlier when computing $\Delta T$)

While this answer is technically correct, it appears to have an odd dependency on $r$, the radius of the cylinder, and no dependency on $V$, the initial volume.

However, if we define, $k=\frac{h}{r}$ (which measures the "shape" of the cylinder in some sense), we have $h=k r$ and thus:

$V=\pi r^2 h=\pi r^2 (k r) = \pi r^3 k \to r = \sqrt[3]{\frac{V}{\pi k}}$

Making this final substitution, we have:

$ \Delta T = \frac{g^2 m^2 T}{\sqrt[3]{\pi } P \left(\frac{V}{k}\right)^{2/3} \left((c+1) g m+\sqrt[3]{\pi } c P \left(\frac{V}{k}\right)^{2/3}\right)} $

Disclaimer: This is a purely mathematical answer. You should check that it makes sense in real life.


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