Answer :
Answer:
[tex] p_{fA} = 2.467 atm (\frac{0.717 L}{0.896 L})=1.974 atm [/tex]
[tex] p_{fB} = 4.244 atm (\frac{0.179 L}{0.896 L})=0.848 atm [/tex]
[tex] p_{f}= p_{fA} +p_{fB}= 1.974 +0.848 atm = 2.822 atm[/tex]
Explanation:
For this case we can use the Boyle law given by:
Gas A
[tex] p_{iA} v_{iA} = p_{fA} v_{fA}[/tex]
The final volume for this case would be the addition of the two volumes given
[tex] v_{fA} = 717 + 179 ml = 896 ml *\frac{1L}{1000 ml}=0.896 L[/tex]
We can calculate the final pressure of the gas A:
[tex] p_{fA} = p_{iA} (\frac{v_{iA}}{v_{fA}}) [/tex]
We need to remember that 1 bar = 1.01325 atm, we convert the initial pressure for the gas A and we got:
[tex] p_{iA} = 2.5 bar *\frac{1 atm}{1.01325bar}= 2.467 atm[/tex]
And finally we got:
[tex] p_{fA} = 2.467 atm (\frac{0.717 L}{0.896 L})=1.974 atm [/tex]
Gas B
[tex] p_{iB} v_{iB} = p_{fB} v_{fB}[/tex]
The final volume for this case would be the addition of the two volumes given
[tex] v_{fB} = 717 + 179 ml = 896 ml *\frac{1L}{1000 ml}=0.896 L[/tex]
We can calculate the final pressure of the gas A:
[tex] p_{fB} = p_{iB} (\frac{v_{iB}}{v_{fB}}) [/tex]
We need to remember that 1 bar = 1.01325 atm, we convert the initial pressure for the gas A and we got:
[tex] p_{iB} = 4.3 bar *\frac{1 atm}{1.01325bar}= 4.244 atm[/tex]
And finally we got:
[tex] p_{fB} = 4.244 atm (\frac{0.179 L}{0.896 L})=0.848 atm [/tex]
And the total pressure for this case at the end would be:
[tex] p_{f}= p_{fA} +p_{fB}= 1.974 +0.848 atm = 2.822 atm[/tex]