Five liters of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what new volume will result? The temperature and pressure remain the same.

Respuesta :

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.