Let's say that an ideal gas does work on a piston, thus increasing the volume of the gas in its insulated cylinder. The pressure of the gas is assumed to be constant; therefore, by the ideal gas law, $PV =nRT$, the temperature of the ideal gas should increase.

Now let's look at this problem from the angle of the First Law of Thermodynamics, $\Delta U = Q-W$. The gas does work on the piston, therefore $W$ is positive. There is no heat added to the system, so $Q=0$. So, $\Delta U = -W$. So, the internal energy of the ideal gas particles has now decreased according to the First Law of Thermodynamics, meaning the ideal gas's temperature has decreased.

According to the Ideal Gas Law, temperature has increased whereas according to the First Law it has decreased.

Please tell me what I am doing wrong here. Your help would be greatly appreciated!

  • 2
    $\begingroup$ You can probably guess that the answer is “No”, so it would make more sense to phrase the question as “Why doesn’t...?”. Otherwise it sounds like you are pushing a non-mainstream personal theory. $\endgroup$
    – G. Smith
    Commented Aug 18, 2020 at 20:18
  • $\begingroup$ See update to my answer applying the first law to the constant pressure process. $\endgroup$
    – Bob D
    Commented Aug 18, 2020 at 21:28

3 Answers 3


You are comparing two different processes. When you applied the ideal gas law it was a constant pressure expansion process and $Q$ doesn’t equal zero.. When you applied the first law it was an adiabatic expansion process where $Q=0$.

That being said, let's compare apples to apples. For the constant pressure process applying the ideal gas equation to one mole of gas

$$P\Delta V=R\Delta T$$

$$\Delta T=\frac{P\Delta T}{R}$$

Now apply the first law for the constant pressure proccess

$$\Delta U=Q-W$$

$$\Delta U=C_{P}\Delta T-P\Delta V$$

For an ideal gas, any process

$$\Delta U=C_{V}\Delta T$$

For derivation of this see: $\Delta U$, $C_p$, $C_v$ for an ideal gas process

$$C_{V}\Delta T=C_{P}\Delta T-P\Delta V$$

$$(C_{P}-C_{V})\Delta T=P\Delta V$$

For an ideal gas (see same link for derivation)



$$R\Delta T=P\Delta V$$

$$\Delta T=\frac{P\Delta V}{R}$$

which is the same as applying the ideal gas law.

Hope this helps


In an adiabatic reversible expansion of an ideal gas, both the temperature and the pressure decrease, not just the temperature. But, even if it were irreversible and the pressure was constant (at a new lower value), the temperature would decrease, and this would again be consistent with the first law of thermodynamics and the ideal gas law.


In the first case it is said that gas expands at constant pressure that means heat is supplied to system .


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