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?
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.
