# Question #db8f6

Aug 1, 2015

Here's how you can approach this problem.

#### Explanation:

The first important thing to recognize here is that you can use Boyle's Law to predict what will happen to the pressure of the mixture after the containers are switched.

According to Boyle's Law, pressure and volume have an inverse relationship when temperature and number of moles (amount of gas) are kept constant.

This means that a decrease in the volume of the container will result in an increase in the total pressure of the mixture.

Mathematically, this is written like this

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$, where

${P}_{1}$, ${V}_{1}$ = the pressure and volume of the mixture in the first container;
${P}_{2}$, ${V}_{2}$ - the pressure and volume of the mixture in the second container.

You know that the second container is half the size of the first one, so

${V}_{2} = \frac{1}{2} \cdot {V}_{1}$

This means that the pressure of the mixture in the second container will be

${P}_{2} = {V}_{1} / {V}_{2} \cdot {P}_{1}$

${P}_{2} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{V}_{1}}}}}{\frac{1}{2} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{V}_{1}}}}} \cdot {P}_{1} = 2 {P}_{1}$

SImply put, the new pressure will be twice the initial pressure

${P}_{2} = 2 {P}_{1}$

The initial pressure will be the sum of the partial pressures of the two gases - think Raoult's Law.

${P}_{1} = \left({P}_{\text{argon" + P_"nitrogen}}\right)$

Therefore,

${P}_{2} = \textcolor{g r e e n}{2 \cdot \left({P}_{\text{argon" + P_"nitrogen}}\right)}$