A flexible vessel contains 43L of gas where the pressure is 1.3 atm. What will the volume be when the pressure is 0.61 atm, the temperature remaining constant?

Respuesta :

Answer:

91.64L

Explanation:

Step 1:

Data obtained from the question.

This includes the following:

Initial volume (V1) = 43L

Initial pressure (P1) = 1.3 atm

Final volume (V2) =..?

Final pressure (P2) = 0.61 atm

Step 2:

Determination of the final volume.

To solve for the final volume, we'll apply the Boyle's law equation since the temperature is constant. The final volume is obtained as follow:

P1V1 = P2V2

1.3 x 43 = 0.61 x V2

Divide both side by 0.61

V2 = (1.3 x 43)/0.61

V2 = 91.64L

Therefore, the new volume is 91.64L

The volume will be 91.64 L when the pressure is 0.61 atm

From Boyle's law equation, we have that

P₁V₁ = P₂V₂

Where P₁ is the initial pressure

V₁ is the initial volume

P₂ is the final pressure

and V₂ is the final volume

From the given information

P₁ = 1.3 atm

V₁ = 43 L

P₂ = 0.61 atm

V₂ = ?

Putting the parameters into the formula, we get

1.3 × 43 = 0.61 × V₂

∴ [tex]V_{2} = \frac{1.3 \times 43}{0.61}[/tex]

V₂ = 91.64 L

Hence, when the pressure is 0.61 atm, the volume will be 91.64 L

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