# What happens when ideal gasses are mixed?

I'm sure their temperatures will be equal, but what about their volumes and pressures?How does one approach ideal gas mixing problems?

• Just being curious, why is this question downvoted? Commented Aug 9, 2019 at 18:07
• what makes you say their temperatures will be equal. I assume you mean final temperature of the mix. What is the volume of the chamber where mixing occurs and is it both rigid and insulated? Commented Aug 9, 2019 at 18:17
• @BobD if they mix together, then shouldn't the mixture of these gasses have a single temperature? Commented Aug 9, 2019 at 19:02
• @user3711671 Yes, but I wanted to make sure you weren't saying that if the initial temperatures of the individual gases will are the same then the final temperature of the mixture will be necessarily be the same as the initial temperature. Commented Aug 9, 2019 at 19:18
• When you mix ideal gases, their molecules intermingle, and the individual gases lose their separate integrities. So we can't talk about their separate volumes. Both constituents fill the entire volume of the container. But is possible to assign a 'partial pressure" to each gas that is proportional to the mole fraction of the gas. The sum of the partial pressures add up to the total pressure of the mixture. In addition, each gas behaves as if it is present alone in the container, such that its partial pressure satisfies pV=nRT, where p is the partial pressure and V is the total pressure. Commented Aug 10, 2019 at 0:37

For example: 2 monoatomic gases, at $$( T, V, P)$$ and $$(2T, V, 2P)$$ are mixed ( by just connecting a pipe between the containers)(equal moles), then final temperature => $$nC_v T + nC_v (2T) = (2n)C_v T_f$$ ( both monoatomic, same $$C_v$$) This is the energy conservation equation. Hence get $$T_f =3T/2$$ . Volume will be $$2V$$. Use ideal gas eq to get $$P_f$$.