From Boyle's law:

$$P_1V_1 = P_2V_2 \quad\implies\quad V_2 = \frac{P_1V_1}{P_2} \label{eqn:boyle}\tag{1}$$

From Charles' law: $$\frac{V_1}{T_1} = \frac{V_2}{T_2} \quad\implies\quad V_2 = \frac{V_1T_2}{T_1} \label{eqn:charles}\tag{2}$$

Equating \eqref{eqn:boyle} and \eqref{eqn:charles} gives:

$$\frac{P_1V_1}{P_2} = \frac{V_1T_2}{T_1},$$

which upon solving will give:

$$P_1T_1 = P_2T_2,$$

which is exact opposite of Gay-Lussac's law. I can't find the flaw in my logic. Help!

  • 6
    $\begingroup$ (1) is true for isothermal processes, (2) for isobaric, and it would be prudent not to expect anything good when you mix the two. $\endgroup$ Jun 3, 2019 at 12:19
  • 4
    $\begingroup$ There is no flaw. Or rather, if there is a flaw it is to think that your equation holds for any T1/T2 and P1/P2 pair when in fact (as Ivan explains) you have implicitly imposed constant T and P conditions during the derivation. $\endgroup$
    – Buck Thorn
    Jun 3, 2019 at 12:41
  • 1
    $\begingroup$ @Ivan Neretin But my textbook says you can prove the Gay Lussac's law by Boyle's law and and Charles' law. If mixing those two wouldn't get anything meaningful how would you prove Gay Lussac's law through Charles' law and Boyle's law ? $\endgroup$
    – Ali
    Jun 3, 2019 at 13:20

2 Answers 2


As was pointed out in the comments, the problem arises when you assume that the two starting points are equivalent. @IvanNeretin pointed out that Boyle's Law is only useful for situations where the temperature doesn't change (isothermal) and Charles' Law is only useful in situations where the pressure doesn't change (isobaric).

This means that even though we can mathematically set the equality, it doesn't have any physical significance.

This may be a bit clearer if we consider that both Boyle's Law and Charles' Law can be seen as simplified versions of the ideal gas law.

$PV = nRT$

I'm adding the subscripts B (Boyle's Law) and C (Charle's Law) to highlight the differences.

The only way for Boyle's Law to hold true is if the $nRT$ portion of the equation is identical in both cases.

$P_{1B}V_{1B} = nRT$ and $P_{2B}V_{2B} = nRT$

$P_{1B}V_{1B} = nRT = P_{2B}V_{2B}$

For this to be true, both the number of moles in the system n and the temperature must be held constant between the two systems.

We can do the same analysis for Charles' Law, in this case holding the pressure (and the number of moles) between the two systems constant.

$PV_{1C} = nRT_{1C}$ and $PV_{2C} = nRT_{2C}$

$\dfrac{V_{1C}}{T_{1C}} = \dfrac{nR}{P} =\dfrac{V_{2C}}{T_{2C}}$

So although we can establish two equations that are equal to $V_2$

$V_{2B} = \dfrac{P_{1B}V_{1B}}{P_{2B}} = \dfrac{nRT}{P_{2B}}$


$V_{2C} = \dfrac{V_{1C}T_{2C}}{T_{1C}} = \dfrac{nRT_{2C}}{P}$

It should be clear that the $V_{2B} \neq V_{2C}$, and therefore the premise of the rest of your derivation isn't valid.


The derivation is not that hard. If my memory is correct this how we have done that:

From Boyle's Law:

$$P_1V_1 = P_2V_2 \label{eqn:1}\tag{1}$$

From Charles' Law:

$$\frac{V_1}{T_1} = \frac{V_2}{T_2} \label{eqn:2}\tag{2}$$

Multiplying \eqref{eqn:1} and \eqref{eqn:2} gives:

$$\frac{P_1V_1^2}{T_1} = \frac{P_2V_2^2}{T_2} \label{eqn:3}\tag{3}$$

If $V_1=V_2$ (as Gay-Lussac's law states), \eqref{eqn:3} can be simplified to give:

$$\frac{P_1}{T_1} = \frac{P_2}{T_2} \tag{4}$$

  • $\begingroup$ How can you multiply 1 and 2? $\endgroup$
    – Wolgwang
    Jun 16, 2021 at 15:43
  • $\begingroup$ The two law apply for the same amount of gas. $\endgroup$ Jun 16, 2021 at 18:26
  • $\begingroup$ But Temperature is constant for (1) and pressure is constant for (2). $\endgroup$
    – Wolgwang
    Jun 16, 2021 at 18:27
  • $\begingroup$ They still cancel out if that condition applies. The key is same amounts. $\endgroup$ Jun 16, 2021 at 20:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.