Question

Container A holds 717 mL 717 mL of ideal gas at 2.50 bar 2.50 bar . Container B holds 179 mL 179 mL of ideal gas at 4.30 bar 4.30 bar . If the gases are allowed to mix together, what is the partial pressure of each gas in the total volume?

Answer 1

`p_(fA) = 2.467 atm ((0.717 L)/(0.896 L))=1.974 atm`

`p_(fB) = 4.244 atm ((0.179 L)/(0.896 L))=0.848 atm`

`p_(f)= p_(fA) +p_(fB)= 1.974 +0.848 atm = 2.822 atm`

Step-by-step explanation:

For this case we can use the Boyle law given by:

Gas A

`p_(iA) v_(iA) = p_(fA) v_(fA)`

The final volume for this case would be the addition of the two volumes given

`v_(fA) = 717 + 179 ml = 896 ml *(1L)/(1000 ml)=0.896 L`

We can calculate the final pressure of the gas A:

`p_(fA) = p_(iA) ((v_(iA))/(v_(fA)))`

We need to remember that 1 bar = 1.01325 atm, we convert the initial pressure for the gas A and we got:

`p_(iA) = 2.5 bar *(1 atm)/(1.01325bar)= 2.467 atm`

And finally we got:

`p_(fA) = 2.467 atm ((0.717 L)/(0.896 L))=1.974 atm`

Gas B

`p_(iB) v_(iB) = p_(fB) v_(fB)`

The final volume for this case would be the addition of the two volumes given

`v_(fB) = 717 + 179 ml = 896 ml *(1L)/(1000 ml)=0.896 L`

We can calculate the final pressure of the gas A:

`p_(fB) = p_(iB) ((v_(iB))/(v_(fB)))`

We need to remember that 1 bar = 1.01325 atm, we convert the initial pressure for the gas A and we got:

`p_(iB) = 4.3 bar *(1 atm)/(1.01325bar)= 4.244 atm`

And finally we got:

`p_(fB) = 4.244 atm ((0.179 L)/(0.896 L))=0.848 atm`

And the total pressure for this case at the end would be:

`p_(f)= p_(fA) +p_(fB)= 1.974 +0.848 atm = 2.822 atm`