How do I prove

$${{\displaystyle p_{i}=p_{\text{total}}y_{i}}}$$

where $y_i$ is the mole fraction of the $i^{th}$ component in the total mixture of $n$ components ?

Here is what I know

Mathematically, the pressure of a mixture of non-reactive gases can be defined as the summation:

${\displaystyle p_{\text{total}}=\sum _{i=1}^{n}p_{i}}$


${\displaystyle p_{\text{total}}=p_{1}+p_{2}+p_{3}+...p_{n}}$

And mole fraction $y_i= \dfrac{n_i}{n_{\text {total}}}$

I don't see how to relate , please help.


After having some discussion with @JohnRennie

Let $P_i=n_i \dfrac{RT}{V}$ then $P_i=\dfrac{n_i}{n_{\text{total}}}\dfrac{n_{\text{total}}RT}V$

$P_i=x_i P_{\text{total}}$

Where $x_i$ is mole fraction of $i^{\text{th}}$ particle.

  • $\begingroup$ Ahh, this is what he must have meant. Good answer. +1 $\endgroup$ – Pritt Balagopal Apr 23 '17 at 14:50

Recall the ideal gas law equation


Lets say we have $m$ components in the gas mixture, with number of moles of each being $n_1$,$n_2$,$n_3$,...,$n_m$. Let $n$ denote sum of all moles. Let partial pressure of each be $p_1$,$p_2$,$p_3$,...,$p_m$, and $p$ denote the total pressure.

From the gas law, we get:






I think that's a satisfactorily enough proof.

  • $\begingroup$ Sorry but I have to prove that formula highlighted one with using concept of Dalton's law and mole fraction. $\endgroup$ – Fawad Apr 23 '17 at 13:52
  • $\begingroup$ That is the definition of partial pressure. John Dalton defined a term called partial pressure in that way. You can't prove definitions. $\endgroup$ – Pritt Balagopal Apr 23 '17 at 13:54

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