Adiabatic expansion is known to cool gases to a substantial temperature and it's caused by pressure drop. Now if I have a tank full of an ideal gas (nitrogen) at high pressure and temperature, and I inflate a large balloon with the tank until pressure in both equals, the tank becomes less pressurized by a lot, thus a lot of cooling occurs. Why does the temperature have to change if the system experiences no heat loss?

  • $\begingroup$ I've deleted some comments here; please keep in mind that comments are for suggesting improvements to, or requesting clarification on, their parent post. $\endgroup$ – David Z Mar 14 at 5:38

Temperature of a gas system can surely change without supply or removal of heat energy. W (Work done) and Q (Heat energy) are to modes to change thermodynamic state of a system. In other words temperature can be changed by these two modes. According to first equation of thermodynamics,


where W= work done on the system , U= change in internal energy of system, Q= Heat energy transferred to the surroundings by the system (please note that signs will be taken accordingly. For e.g. work done will be negative if work is done by the system)

Now, even if $Q=0$ Temperature ($T$) can change by second mode $W$ . T will fall if W is negative(work done by gas) and T will rise if W is positive(work done on gas). This is what happens in your case. Also work done in an adiabatic process is related to T as follows :


where, n= number of moles, R=ideal gas constant, y=Cp/Cv

  • $\begingroup$ How do one define work. Assuming I connect both the balloon and tank with just a hose and no valve control, what should happen within the hose before the gas reaches the balloon for work to be positive? $\endgroup$ – TechDroid Mar 13 at 11:49

Consider the tank and the balloon together to be a closed system. Boundary work is done by or on a closed system when its boundaries are expanded against or contracted by an external force, respectively. The balloon expands against atmospheric pressure. The system does positive work $W$. If it expands rapidly enough so that there is time no for heat transfer then $Q=0$. Consequently the work done by the system has to be at the expense of its internal energy per the first law $\Delta U=Q-W$.

Finally, since a change in internal energy is a function of temperature only, a decrease in internal energy of an ideal gas results in a decrease in its temperature.

Hope this helps


Temperature may feel closely related to heat transfer. But it shouldn't be thought of like that. Those two terms should be kept separated.

  • A temperature can change without heat transfer.
  • Heat can be transferred without a temperature change.

The reason is that energy can be transferred, not only as heat, but also as work. Work can cause temperature changes, just as well as heat can.

In the case of an adiabatic process, the process may be happening so fast that no significant heat exchange with the surroundings has time to happen. This is the case when you pop a champagne bottle, for instance, where you see a white mist or "smoke", which is ice particles forming due to the suddenly changing temperature. No heat entered or left the released gas, because there was no time. Still the temperature dropped rapidly.

Any (ideal) gas still must fulfill the ideal-gas law: $$pV=nRT$$

This is a separate requirement, unrelated to the issue of no-time-for-heat-exchange. And it just so happens that when a high pressure is released, the most natural response is for both volume and temperature to adjust, in order for this equation to remain true. Why would only volume change with temperature remaining constant? In fact, both variable parameters will change to adjust simultaneously during an expansion where none of them are restricted.

This is clear, since it takes work to expand volume. Where does such work energy come from? Since it isn't coming from a supply of heat, it might come from the internal thermal energy, causing a temperature drop.

  • $\begingroup$ So let say I hook both tank and balloon up with just a hose, what order of magnitude lower of temperature should I be expecting in the balloon? $\endgroup$ – TechDroid Mar 13 at 8:58
  • $\begingroup$ I thought in free expansion the pressure and volume change but not the temperature. And isn't free expansion an example of but needing to do work to change volume? $\endgroup$ – Aaron Stevens Mar 13 at 10:41
  • $\begingroup$ @TechDroid If we assume a very large balloon (or a small tank) that can expand freely, so only atmosphere adds pressure, then the gas that enters the balloon will reach the surrounding, atmospheric pressure. So, we know the final pressure $p_2$ but not the final volume $V_2$ nor temperature $T_2$. We do know all the initial values though. Thus you can use an adiabatic formula: $$p_1V_1^\gamma=p_2V_2^\gamma$$ where $\gamma=c_p/c_v$ includes the heat capacities for the gas (can be looked up in a gas table most likely) $\endgroup$ – Steeven Mar 13 at 10:42
  • $\begingroup$ @AaronStevens True, I have been unclear, I see. Thanks, I will edit the answer to fit. $\endgroup$ – Steeven Mar 13 at 10:43
  • $\begingroup$ How do one define work. What should be happening between the tank and balloon for work to be either positive or negative? $\endgroup$ – TechDroid Mar 13 at 11:54

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