Question

Container A holds 752 mL of ideal gas at 2.10 atm. Container B holds 184 mL of ideal gas at 4.80 atm. If the gases are allowed to mix together, what is the resulting pressure?

Answer 1

There are at least two different ways to calculate the final pressure,
One is calculate how many moles of A and the moles of B that you have with the equation
PV=nRT
n=PV/RT
R=0.082 atm L/ mol K
T= any value different than 0
Add the two values (moles of A+ moles of B= total moles of gas)
Add the two volumes (mL of A+ mL of B= total mL of gas = total mLx10^(-3) L gas
convert to L to be able to use R
Finally calculate the final pressure with the same formula
PV=nRT => P= nRT/V
n will be now the total moles of gas and V the total volume of gas

The other way is to do the formula calculation and eliminate R and T
\[P _{final}= \frac{ P _{A} \times V _{A} + P _{B} \times V _{B} }{ V _{A}+ V _{B} }\]
Explanation
\[n _{A}=\frac{ P _{A} \times V _{A}}{ R \times T } \]
\[n _{B}=\frac{ P _{B} \times V _{B}}{ R \times T } \]
\[n _{total}= n _{A}+ n _{B}\]
\[n _{total}= \frac{ P _{A} \times V _{A}}{ R \times T } + \frac{ P _{B} \times V _{B}}{ R \times T } \]

\[n _{total}=\frac{ 1 }{R \times T } \left( P _{A} \times V _{A} +P _{B} \times V _{B} \right)\]

\[P _{final}= n _{total} \times R \times T \div V _{total}\]
\[V _{total}= V _{A}+V _{B}\]
\[P _{final}= \frac{ 1 }{R \times T } \left( P _{A} \times V _{A} +P _{B} \times V _{B} \right) \times R \times T \div V _{total}\]
cancel R x T
final formula to calculate the final pressure
\[P _{final}= \frac{ P _{A} \times V _{A} + P _{B} \times V _{B} }{ V _{A}+ V _{B} }\]