Answer:
New volume of the gas will be 9.33 L.
Explanation:
Let's assume the gas behaves ideally.
So we can write, [tex]\frac{P_{1}V_{1}}{P_{2}V_{2}}=\frac{n_{1}RT_{1}}{n_{2}RT_{2}}[/tex]
where [tex]P_{1}[/tex] and [tex]P_{2}[/tex] represents initial and final pressure of gas respectively.
[tex]V_{1}[/tex] and [tex]V_{2}[/tex] represents initial and final volume of gas respectively
[tex]n_{1}[/tex] and [tex]n_{2}[/tex] represents initial and final number of moles of gas respectively.
[tex]T_{1}[/tex] and [tex]T_{2}[/tex] represents initial and final temperature (in kelvin scale) of gas respectively.
R is gas constant.
Here [tex]P_{1}=P_{2}[/tex] , [tex]T_{1}=T_{2}[/tex] , [tex]V_{1}=5.00L[/tex] , [tex]n_{1}=0.965mol[/tex] and [tex]n_{2}=1.80mol[/tex]
So [tex]V_{2}=\frac{V_{1}n_{2}}{n_{1}}[/tex] = [tex]\frac{(5.00L)\times (1.80mol)}{(0.965mol)}[/tex] = 9.33 L
Hence new volume of the gas will be 9.33 L.
