Deriving ideal gas law from Boyle and Charles

Question

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?

Answer 1

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.

Answer 2

$PV$ 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.

Answer 3

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

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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.

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

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Reference: from 10:46 of this video

Answer 4

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 $PV/T = Rn$