My textbook states

Notice that since $PV = \text{constant}$ and $\frac{V}{T} = \text{constant}$ for a given quantity of gas, then $\frac{PV}{T}$ should also be a constant.

I tried to prove this, but no success:
$$PV = a$$ $$\frac{V}{T} = b$$

  $$\frac{PV^2}{T} = ab$$ $$PT = \frac{a}{b}$$

But I am not able to cook up $\frac{PV}{T}$... Any help?


4 Answers 4


This formulation of Boyle's law $$PV=\text{const}$$ is very misunderstandable. Actually the constant on the right side is only meant to be independent of $P$ and $V$. But it may still depend on other parameters, like $T$ (temperature) and $N$ (number of molecules). So a better way to write this law is $$PV=a(T,N) \tag{1}$$ where $a(T,N)$ is some unknown function of $T$ and $N$.

Likewise this formulation of Charles's law $$\frac{V}{T}=\text{const}$$ is misunderstandable in the same way. A better way to write it is $$\frac{V}{T}=b(P,N) \tag{2}$$ where $b(P,N)$ is some unknown function of $P$ and $N$.

Now we can divide equation (1) by $T$ and multiply equation (2) by $P$ to get $$\frac{PV}{T} = \frac{a(T,N)}{T} = P\ b(P,N).$$ The only way for this to hold true while varying $P$ and $T$ is that $\frac{a(T,N)}{T}$ is independent of $T$, and $P\ b(P,N)$ is independent of $P$. Hence it only depends on $N$, and we can call this function $c(N)$.

So finally we arrived at the combined gas law $$\frac{PV}{T}=c(N) \tag{3}$$ where $c(N)$ is some unknown function of $N$ only.

  • 1
    $\begingroup$ The missing step between $(2)$ and $(3)$ is dividing $(1)$ by $T$ and multiplying $(2)$ by $P$ to get $\frac{PV}{T} = \frac{a(T,N)}{T} = P*b(P,N)$. The only way for this to hold true while varying $P$ and $T$ is if they cancel both out, meaning $a(T,N) = T*c(N)$ and $b(P,N) = \frac{c(N)}{P}$ $\endgroup$ Sep 12, 2020 at 18:04
  • $\begingroup$ @BlueRaja-DannyPflughoeft You're right. Thanks for filling the gap in the logic. $\endgroup$ Sep 12, 2020 at 18:21

$PV$ is constant for fixed $T$, and $V/T$ is constant for fixed $P$. Hence

$PV=f(T)$ and $V/T=g(P)$.

From these we can write

$V=f(T)/P=T\times g(P)$.

This implies that

$f(T)=kT$ and $g(P)=k/P$ for some constant $k$.

Hence $PV/T = k$ (constant, actually $nR$) is the required answer.

  • $\begingroup$ Nice! Pretty sure you meant $g(P) = k/P$ in the last but one line... $\endgroup$
    – across
    Sep 12, 2020 at 18:12
  • $\begingroup$ Of course I did ;) Thanks - fixed now. $\endgroup$
    – Peter
    Sep 14, 2020 at 0:45

You can't derive it like that because the proportionality relations hold only when the third parameter is kept constant.

However, you can derive the ideal gas law by noting that for high temperature, we get a limit as shown below:

$$ \lim_{ p \to 0 } p \overline{V} = f(T)$$

So, the limit of the product as pressure drops to zero is a unique function $ f(T)$ for all gases independent of the substance used. We can use this to define the linear kelvin scale. Using the triple point of water and absolute zero as our reference,

$$ f(T) = \frac{f(T_{trip-point})}{273.16K} T$$

Where $f(T_{trip-point})$ is the value of the limit at the triple point, using this and our first equation, we can write,

$$ \lim_{ p \to 0} p \overline{V} = \frac{f(T_{trip-point})}{273.16K} T$$

and now, the universal gas constant is defined as follows:

$$ R = \frac{f(T_{trip-point})}{273.16K}$$

Which leads us to:

$$ \lim_{ p \to 0} p \overline{V} = RT$$

Now, we call an ideal gas is one which obeys the above relation even when the limit is not there.

$$ p \overline{V} = RT$$

Reference: from 10:46 of this video

  • $\begingroup$ pretty sure you meant ... even when the limit is there. in the last line. thank you for pointing me to that mit course XD $\endgroup$
    – across
    Sep 12, 2020 at 9:41
  • $\begingroup$ Ahhh I fixed it now :D Btw are you Indian? If so check out my profile , you may find something there :D $\endgroup$ Sep 12, 2020 at 9:45

Boyle’s Law: V ∝ (1/P) (constant T, n)

Charles’s Law: V ∝ T (constant P, n)

Avogadro’s Hypothesis: V ∝ n (constant T, P)

The combination of the three laws for ideal gases yields to

V ∝ nT/P

you can pass from proportionality to equality by introducing a constant R

$V =R nT/P$

and so you have that $PV/T = Rn$

  • $\begingroup$ If you can just combine the proportionalities like that, then why didn't op's method work? $\endgroup$ Sep 12, 2020 at 10:09

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