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A scuba diver exhales 2.65 L of air while swimming at a depth of 36.30 m where the sum of atmospheric pressure and water pressure is 4.51 atm. By the time this exhaled air rises to the surface, where the pressure is 1.00 atm, what is its volume?

Respuesta :

MechEngineer

Answer:

11.9515 liter

Explanation:

Assume constant temperature and ideal gas condition where the product of the volume and pressure is constant.

[tex]P_1V_1 = P_2V_2[/tex]

where P, V are the pressure and volume at 2 different positions, one at the depth of 36.3 m and the other at the surface

[tex]4.51 * 2.65 = 1*V_2[/tex]

[tex]V_2 = 4.51*2.65 / 1 = 11.9515 L[/tex]

So the volume at the surface is 11.9515 liter

Lanuel

The final volume of the air assuming constant temperature is 11.9515 Liters.

Given the following data:

  • Initial volume = 2 Liters.
  • Height = 36.30 meters
  • Final pressure = 1.00 atm.
  • Initial pressure = 4.51 atm.

To calculate the final volume assuming constant temperature, we would apply Boyle's law:

The formula for Boyle's law.

Mathematically, Boyle's law is given by this formula;

[tex]PV = k\\\\P_1V_1 = P_2V_2[/tex]

Where:

  • [tex]P_1[/tex] is the original (initial) pressure.
  • [tex]P_2[/tex] is the final pressure.
  • [tex]V_1[/tex] is the original (initial) volume.
  • [tex]V_2[/tex] is the final volume.

Substituting the given parameters into the formula, we have:

[tex]4.51 \times 2.65 =1 \times V_2\\\\V_2= 4.51 \times 2.65[/tex]

Final volume = 11.9515 Liters.

Learn more about Boyle's law here: https://brainly.com/question/24938688

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