The volume of a persons lungs is 600ml at a pressure of 760mmHg. When the lungs expand to 610 ml by how much will the pressure in the lungs drop (boyle’s law)
Answer (1)
Boyle's Law (P₁V₁ = P₂V₂)where:P₁ = 760 mmHgV₁ = 600 mlV₂= 610 mlFormula: [tex]P_{2}= \frac{P_{1}+V_{1} }{V_{2}}[/tex]P₂ = (760 × 600) / 610P₂ ≈ 746.23 mmHgThe pressure in the lungs drops by:760 mmHg - 746.23 mmHg= 13.77 mmHg
