Respuesta :

for this question, lets apply Avagadro's law
when Pressure and temperature are constant, the volume occupied is directly proportional to the number of moles of gases.
[tex] \frac{V}{n} = k[/tex]
where V-volume, n-number of moles and k - constant 
Therefore at 2 instances 
[tex] \frac{V1}{n1} = \frac{V2}{n2} [/tex]
where V1 and n1 are for 1st instance 
and V2 and n2 are for 2nd instance 
therefore 
[tex] \frac{V1}{n1} = \frac{V2}{n2} [/tex]
V1 = 2.4 L
n1 = 3.7 mol
n2 = 3.7 + 1.6 = 5.3 mol
since more He moles are added at the 2nd instance its the sum of the moles.
V2 needs to be calculated [tex] \frac{2.4}{3.7} = \frac{V2}{5.3} [/tex]
V2 = 2.4 x 5.3 / 3.7
     = 3.4 L
Answer is 1st option 3.4 L


mahima6824soni

If 1.6 mol He is added to the balloon, the new volume is 3.4 L.

What is Avogadro's law?

The volume occupied is directly proportional to the number of moles of gases when pressure and temperature are constant.

[tex]V = \dfrac{n}{k}[/tex]

where V is volume,

n = number of moles

and k = constant

Here

[tex]\dfrac{V_1}{n_1} = \dfrac{V_2}{n_2}[/tex]

where V1 and n1 are for 1st instance

and V2 and n2 are for 2nd instance

V1 is 2.4 L

n1 is 3.7 mol

n2 is 3.7 + 1.6 = 5.3 mol

Putting the values

[tex]\dfrac{2.4l}{3.7} = \dfrac{V_2}{5.3\;mol} = 3.4\;L[/tex]

Thus, the new volume is 3.4 L.

Learn more about volume

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