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An air bubble at the bottom of lake with temperature 7 0C and pressure of 3.2 atm has a radius of 2.5 cm. It rises to the surface where the temperature is 25 oC and atmospheric pressure is 1 atm. What is the volume of the bubble at the surface? Assume perfect gases. V=(4/3)πr3.

Answer :

dsdrajlin

Answer:

[tex]2.0 \times 10^{2} cm^{3}[/tex]

Explanation:

Step 1: Given data

  • Initial pressure (P₁): 3.2 atm
  • Initial temperature (T₁): 7 °C = 305 K
  • Initial radius (r₁): 2.5 cm
  • Final pressure (P₂): 1 atm
  • Final temperature (T₂): 25 °C = 298 K
  • Final volume (V₂): ?

Step 2: Calculate the initial volume

We will use the following expression.

V = 4/3 × π × r³

V = 4/3 × π × (2.5 cm)³

V = 65 cm³

Step 3: Calculate the final volume

Assuming ideal behavior, we can calculate the final volume of the bubble using the combined gas law.

[tex]\frac{P_1 \times V_1 }{T_1} = \frac{P_2 \times V_2 }{T_2}\\V_2 = \frac{P_1 \times V_1 \times T_2 }{T_1 \times P_2} = \frac{3.2atm \times 65cm^{3} \times 298K }{305K \times 1atm} = 2.0 \times 10^{2} cm^{3}[/tex]

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