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A driver blows 0.75 -L air bubble 10 m under water. As it rises to the surface, the pressure goes from 2.25 atm to 1.03 atm. What will be the volume of air in the bubble at the surface ?

Respuesta :

colegatecg

Answer:

1.64 L

Explanation:

P₁V₁ = P₂V₂

P₁ = 2.25 atm

V₁ = 0.75 L

P₂ = 1.03 atm

V₂ = ?

(2.25 atm)(0.75 L) = (1.03 atm)V₂

[(2.25 atm)(0.75 L)]/(1.03 atm) = V₂

V₂ = 1.64 L

Cricetus

The volume of air in the bubble will be "1.64 L".

Pressure and Volume,

Whenever the volume of a confined gas is reduced, this same pressure increases, and that when the volume is increased, this same pressure decreases.

Indeed, when the volume rises by a particular factor, the pressure reduces according to the similar proportion, as well as vice versa.

According to the question,

Pressure,

  • [tex]P_1 = 2.25[/tex] atm
  • [tex]P_2 = 1.03[/tex] atm

Volume,

  • [tex]V_1 = 0.75[/tex] L
  • [tex]V_2 = \?[/tex] ?

We know the relation,

→ [tex]P_1 V_1 = P_2 V_2[/tex]

or,

→     [tex]V_2 = \frac{P_1 V_1}{P_2}[/tex]

By putting the values,

           [tex]= \frac{2.25\times 0.75}{1.03}[/tex]

           [tex]= 1.64[/tex] L

Thus the above answer is right.  

Find out more information about volume here:

https://brainly.com/question/25736513

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