### Ideal Gas Law Combining Multiple Gases (Ex3)

Sometimes you need to use the ideal gas law when multiple cylinders are combined into a single tank.

Cylinder A contains gas A, it has a volume of 48.2 L, at a certain temperature, at a pressure of 8.35 atm. Another cylinder, B containing gas B has a gas at pressure 9.5 atm at the same temperature as vessel A. The two cylinders are combined, and the gases mix and the resulting pressure is 8.71 atm, with no change in temperature during mixing. What is the volume of cylinder B? Use the ideal gas law assumption.

The trick with this question is that you can write 3 equations, it's the PV=nRT equation for A, for B, and for the combined system. Then you manipulate the equations to solve for the volume of B. I'll round my equations, but you should keep all decimals in your calculator when possible!

Starting with vessel A, and we'll call the unknown temperature T1:

${n_A} = \frac{{{P_A}{V_A}}}{{R{T_1}}}$

${n_A} = \frac{{(48.2L)(8.35atm)}}{{\left( {0.08206\frac{{Latm}}{{molK}}} \right){T_1}}}$

${n_A} = \frac{{4904.58}}{{{T_1}}}$

Repeat for vessel B. But we don't know the volume of B so continue leaving it as an unknown.

${n_B} = \frac{{{P_B}{V_B}}}{{R{T_1}}}$

${n_B} = \frac{{(9.5){V_B}}}{{0.08206{T_1}}}$

${n_B} = \frac{{115.77{V_B}}}{{{T_1}}}$

And we can use the ideal law for the combined system. Now we use the resultant pressure, total moles, and the sum of both volumes. Our example says that the temperature doesn't change after mixing.

${P_{}}({V_A} + {V_B}) = \left( {{n_A} + {n_B}} \right)R{T_1}$

What is $${{n_A} + {n_B}}$$? Use the equations we developed previously.

${n_A} + {n_B} = \frac{{4904.58}}{{{T_1}}} + \frac{{115.77{V_B}}}{{{T_1}}}$

$\left( {{n_A} + {n_B}} \right){T_1} = 4904.58 + 115.77{V_B}$

Notice how we have $$\left( {{n_A} + {n_B}} \right){T_1}$$ in our ideal gas law equation for when both gases are combined? As well, substitute in the pressure of 8.71 atm, volume of A as 48.2 L and R.

$8.71\left( {48.2 + {V_B}} \right) = \left( {4904.58 + 115.77{V_B}_{}} \right)\left( {0.08206} \right)$

There we go! The way we manipulated our equations, we don't even need to solve for what T1 even was. In the equation above, all we have is one equation and one variable, so just isolate and solve for the volume of vessel B.

${V_B} = 21.96L$

Please see our other worked-out examples of the ideal gas law below:

Ideal Gas Law with Molecules (Ex1)

Ideal Gas Law Mole Fractions (Ex2)

Ideal Gas Law Partial Pressures and Mole Fractions (Ex4)

Ideal Gas Law Mixing Two Gas Vessels Together (Ex5) 