My physics professor during a lecture presumably mathematically derived Boyle's law from the two Gay-Lussac laws for ideal gasses. What he said is also printed in his own textbook.

He states that, given the variables $P$, $V$, $T$,

assuming only:

  • $T \propto V$ when $P$ is constant (1st Gay-Lussac law)

  • $T \propto P$ when $V$ is constant (2nd Gay-Lussac law)


  • $P \propto{1\over V}$ when $T$ is constant (Boyle's law)

  • $PV \propto T$ for any $P, V, T$

If this is correct, it seems very tidy and compact so why do most of more rigorous textbooks take Boyle's law as an experimental assumption instead? This way only three experimental laws (these two along with Avogadro's) are needed to justify the importance of the ideal gas model.

  • 1
    $\begingroup$ Aren't all three laws (Boyle, Gay-Lussac & Charles) derived from observations? $\endgroup$ – Kyle Kanos Feb 11 '16 at 15:09
  • $\begingroup$ That's what I thought, but apparently the first is only a confirmation of the other two that imply it, it isn't theoretically indipendent and can be derived from them $\endgroup$ – Sebastian Giles Feb 11 '16 at 15:28
  • $\begingroup$ I'd find that hard to believe since Boyles law was published in the mid 1600s and the other two in the early 1800s. It seems to me that all three can be derived from the other two as the ideal has law relates all three variables. $\endgroup$ – Kyle Kanos Feb 11 '16 at 15:33
  • $\begingroup$ Shouldn't there be an agreement about which ones are the fundamental laws and which one is the derived one? Ignoring any historical factor, maybe by selecting the two with the most reliable experiments backing them, and leaving the third as a consequence. $\endgroup$ – Sebastian Giles Feb 11 '16 at 15:50
  • $\begingroup$ Wouldn't the ideal gas law be the fundamental relation? And the three experiments follow from that one relation? $\endgroup$ – Kyle Kanos Feb 11 '16 at 15:53

I think it can be done mathematically. Let T = T(P,V). Then, if T is proportional to P at constant V, then $$\frac{T}{P}=F(V)$$. Therefore, $T=PF(V)$. Now, if T is proportional to V at constant P, then $$\frac{T}{V}=P\frac{F(V)}{V}=kP$$In the above equation, $F(V)/V$ must be a constant in order for the right hand side to be independent of V. So T=kPV.


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