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Fynjy0 [20]
2 years ago
10

If the balloon had a volume of 3 L at a depth of 50 m, what was the original volume of the balloon it we assume the pressure at

the surface of the water is 14.7 psi? Express your answer using one significant figure and include the appropriate units.
Chemistry
1 answer:
disa [49]2 years ago
8 0

If the balloon had a volume of 3 L at a depth of 50 m, the original volume of the balloon will be

<h3 /><h3>What is Significant figures ?</h3>

Significant figures are the number of digits in a value, often a measurement, that contribute to the degree of accuracy of the value.

The pressure at a depth of 40 m is the hydrostatic pressure of the water plus the atmospheric pressure.

The hydrostatic pressure P of a liquid is given by the formula ;

P = ρgh

where,

  • ρ = the density of the liquid (Density of water = 1000 kg m⁻³)
  • g = the acceleration due to gravity (9.81 ms⁻²)
  • h = the depth of the liquid (Given : 50 m)

P =  1000 kg m⁻³ x 9.81 ms⁻² x 50 m

   = 1000 kg m⁻¹ x 9.81 s⁻² x 50

   = 4.9 x 10⁵ Pa

Lets's Convert Pa in to atm

    =   4.9 x 10⁵ Pa x 1 / 103. 325 x 10³ Pa

    = 4.7 atm

Mow, lets calculate P(atm) ;

P(atm) = 14.7 psi x 1 atm / 14.7 psi

          = 1 atm

Total Pressure at 50 m is ;

P(total) = 4.7 atm + 1 atm

            = 5.7 atm

Now we can apply Boyle's Law to calculate the volume of the balloon at the surface.

P₁V₁ = P₂V₂

V₂ = P₁V₁ / P₂

According to the question ;

  • P₁ = 5.7 atm  
  • P₂ = 1 atm
  • V₁ = 3 L
  • V₂ = ?

V₂ = 3 L x 5.7 atm  / 1 atm

     = 17.1 L

     = 20 L ( 1 significant figure)

Learn more about Boyles law here ;

brainly.com/question/1437490

#SPJ1

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