I was reading Fundamentals of Physics(Halliday, 10th ed.), chapter 19, when I saw a statement "Figure 19-15a shows our usual insulated cylinder, now containing an ideal gas and resting on an insulating stand. By removing mass from the piston, we can allow the gas to expand adiabatically. As the volume increases, both the pressure and the temperature drop." Here the figure is shown But I don't understand why temperature can't stay constant as volume increases and pressure decreases? Why both the pressure and the temperature decrease?

  • $\begingroup$ Doesn’t pV=nRT solve the problem? $\endgroup$ – QuIcKmAtHs Feb 2 '18 at 11:18
  • $\begingroup$ when V increases, P and T must change, too. But I wanted to see how they change. V increases, so T can increase, but as P decreases, T also decreases. So the decrease in p should be more effective than in V. $\endgroup$ – DRLOVER Feb 2 '18 at 11:43
  • $\begingroup$ Of course you must make use of some knowledge of the laws of thermodynamics $\endgroup$ – QuIcKmAtHs Feb 2 '18 at 11:58
  • $\begingroup$ Interesting is that if you manually lift the piston, (pull it up), then as the volume increases, the pressure would fall, but the temperature should remain constant. $\endgroup$ – Charles Bretana Feb 10 '18 at 15:44

The first law of thermodynamic states that the change in internal energy of a system $\Delta U$ is equal to the heat input to the system $Q$ minus the work done by the system $W$.

In your example the process is adiabatic so $Q=0$ and $\Delta U=-W$.

As the system does work expanding $W$ must be positive thus the internal energy of the system decreases - the temperature of the system decreases.

You have the same number of molecules moving slower than before in a container which has a larger surface area thus the pressure is lower.


In a adiabatic process, there is no energy supplied to the system. So, when the gas expands adiabatically it is doing work, and the energy to do work can only come from it's internal energy which is why the temperature decreases.

If temperature remains constant, then it is an isothermal process. Since the temperature here is constant,the energy required to do the work must necessarily come from an external agency. Hence, an isothermal process cannot also be an adiabatic process


A microscopic explanation may also be useful: when molecules are bouncing against the receding piston, the speed with which they bounce back will be lower. This means lower kinetic energy, a lower temperature.


The Combined Gas Law which is a combination of three other gas laws, states that

"The ratio between the pressure-volume product and the temperature of a system remains constant." (Wikipedia).

It is expressed mathematically as ${PV\over T} = k, \qquad\tag 1$ where $P$, $V$, $T$ and $k$ are the pressure, volume, temperature and a constant (with units of energy divided by temperature) for the particular body of the gas.

For the same body of gas that changes between state 1 and 2: ${P_1V_1\over T_1} = {P_2V_2\over T_2}\tag2$

In your question, the process is adiabatic (no energy or mass transfer), so as volume increases to $V_2$, pressure will drop to $P_2$ in inverse proportion (amount of gas is constant) according to a sub-law of the Combined Gas Law called Boyle's Law, and the temperature drops to $T_2$, which we can be calculated using another sub law (Gay-Lussac's Law). This is an example of adiabatic cooling.

  • $\begingroup$ No, this is just not right. For example, the temperature is different for monoatomic and diatomic gases. $\endgroup$ – Pieter Feb 9 '18 at 7:01
  • $\begingroup$ @Pieter. Irrelevant. It is the same gas. Note my use of the word 'particular' in my answer. $\endgroup$ – Dlamini Feb 9 '18 at 7:06
  • $\begingroup$ Equation (2) by itself does not provide enough information to calculate both pressure and temperature from a change in volume. $\endgroup$ – Pieter Feb 9 '18 at 19:53
  • $\begingroup$ @Pieter On the face of it you would seem to be correct. However, the sub laws that make up the Combined Gas Law clearly state that both pressure and temperature would drop. I have edited the text to enunciate this, and add the references. $\endgroup$ – Dlamini Feb 10 '18 at 14:22

protected by Qmechanic Feb 9 '18 at 6:21

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