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A scuba diver 40.0 feet below the surface of the ocean inhales 55.5 ml of compressed air from a tank at a pressure of 2.95 atm and a temperature of 12.5 °C. What is the pressure of the air in the lungs if the gas expands to 175 ml at a body temperature of 37.0 °C?

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Considering the combined gas law, the pressure of the air in the lungs if the gas expands to 175 ml at a body temperature of 37.0 °C is 1.016 atm.

Combined gas law

The combined gas law combines the three gas laws: Boyle's Law, Charles' Law, and Gay-Lussac's Law.

The combined gas law expresses the relationship between the pressure, volume, and absolute temperature of a fixed amount of gas. It states that the ratio of the product of pressure and volume and the absolute temperature of a gas is equal to a constant:

(P×V)÷T=k

Also, the combined gas law relates "before and after" conditions of a gas:

(P₁×V₁)÷T₁= (P₂×V₂)÷T₂

Pressure of the air in the lungs

In this case, you know:

  • P₁= 2.95 atm
  • V₁= 55.5 mL
  • T₁= 12.5 °C= 285.5 K
  • P₂= ?
  • V₂= 175 mL
  • T₂= 37°C= 310 K

Replacing in the combined gas law:

(2.95 atm× 55.5 mL)÷285.5 K= (P₂×175 mL)÷ 310 K

Solving:

[(2.95 atm× 55.5 mL)÷285.5 K] × (310 K÷175 mL)= P₂

1.016 atm= P₂

Finally, the pressure of the air in the lungs if the gas expands to 175 ml at a body temperature of 37.0 °C is 1.016 atm.

User Avijit Karmakar
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