Question

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.

Answer 1

New volume of the gas will be 9.33 L.

Step-by-step explanation:

Let's assume the gas behaves ideally.

So we can write,
`(P_(1)V_(1))/(P_(2)V_(2))=(n_(1)RT_(1))/(n_(2)RT_(2))`

where
`P_(1)` and
`P_(2)` represents initial and final pressure of gas respectively.

`V_(1)` and
`V_(2)` represents initial and final volume of gas respectively

`n_(1)` and
`n_(2)` represents initial and final number of moles of gas respectively.

`T_(1)` and
`T_(2)` represents initial and final temperature (in kelvin scale) of gas respectively.

R is gas constant.

Here
`P_(1)=P_(2)` ,
`T_(1)=T_(2)` ,
`V_(1)=5.00L` ,
`n_(1)=0.965mol` and
`n_(2)=1.80mol`

So
`V_(2)=(V_(1)n_(2))/(n_(1))` =
`((5.00L)* (1.80mol))/((0.965mol))` = 9.33 L

Hence new volume of the gas will be 9.33 L.