At constant of 200 cm³ of $\ce{N2}$ at 720 mmHg and 400 cm³ of $\ce{O2}$ at 750 mmHg are put together into one litre flask. The final pressure of the mixture is:

My question is how to use ideal gas law for mixture of two gases. For one gas, it is easy but for two it becomes difficult for me. I am assuming that these gases won't react or else it would have been given in the question. Can somebody tell me how to apply ideal gas law for two gases and what happens when they are kept together?

  • $\begingroup$ Imagine they are one gas. Since they don't react (which is true), this is a perfectly reasonable assumption. Then it must become easy again. $\endgroup$ Jun 22, 2018 at 11:40
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    $\begingroup$ On second thought, this might be a bad idea. Look at it this way: find the resulting pressure of one gas as if the other one were not there. Then find the pressure of another, this time disregarding the first one. Then add the two pressures together. $\endgroup$ Jun 22, 2018 at 11:45
  • $\begingroup$ So we have to just calculate the partial pressure of each gas $\endgroup$ Jun 22, 2018 at 12:08
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    $\begingroup$ Yes, exactly.$\!$ $\endgroup$ Jun 22, 2018 at 12:12
  • $\begingroup$ "At constant of 200 cm³ …" – Is there anything missing, like "at constant temperature a volume of 200 cm³"? $\endgroup$
    – Loong
    Jun 5, 2021 at 15:37

1 Answer 1


Note that since $1\ \mathrm{cm^3} = 1\ \mathrm{mL}$, I will talk in mL.

I doubt that you are supposed to use the ideal gas law for this problem since no temperature is specified, and moles are not talked about. Instead, let's look at a few instances of Boyle's Law ($P_1V_1=P_2V_2$ -- for comparing the pressures and volumes of gasses).

The gasses are, indeed, unreactive, and, according to Kinetic-Molecular Theory, have nearly identical properties where gas laws are concerned. Let's find the partial pressure of each gas with Boyle's Law and sum them with Dalton's Law of Partial Pressures ($P_\mathrm{tot} = P_1 + P_2 +\dots$).

Note that to make the units cancel, I will use $1\ \mathrm L = 1000\ \mathrm{mL}$.

$$720\ \mathrm{mmHg} \times 200\ \mathrm{mL} = x\ \mathrm{mmHg} \times 1000\ \mathrm{mL}$$

$$\frac{720\ \mathrm{mmHg} \times 200\ \mathrm{mL}}{1000\ \mathrm{mL}} = x\ \mathrm{mmHg} = 144\ \mathrm{mmHg}$$

$$750\ \mathrm{mmHg} \times 400\ \mathrm{mL} = x\ \mathrm{mmHg} \times 1000\ \mathrm{mL}$$

$$\frac{750\ \mathrm{mmHg} \times 400\ \mathrm{mL}}{1000\ \mathrm{mL}} = x\ \mathrm{mmHg} = 300\ \mathrm{mmHg}$$

$$144\ \mathrm{mmHg} + 300\ \mathrm{mmHg} = P_\mathrm{tot} = 444\ \mathrm{mmHg}$$

444 mmHg is the final pressure in the container.

  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. Please note that the proper term for "(number of) moles" is amount of substance. The former would be the same as referring to the mass as "(number of) kilograms". $\endgroup$ Jun 22, 2018 at 14:53

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