## CHAPTER 2

• $Compliance = ΔVΔP$

• Remember to use transmural pressure differences in calculations of compliance:

• $Transmural pressure = pressure inside−pressure outside$

• Compliance of lungs is

• $CL = ΔV(PA − Pip)end inspiration − (PA − Pip)pre inspiration$

• $where PA = alveolar pressurePip =intrapleural pressure$

• $CL = 0.500 L[0 − (− 10) cm H2O] − [0 − (−5) cm H2O]CL = 0.10 L/cm H2O$

1. $CL = 0.500 L(20 − 10 cm H2O) − [0 − (−3) cm H2O]CL = 0.07 L/cm H2O$

2. $CT = ΔV(PA − PB)end inspiration−(PA − PB)pre inspirationCT = 0.500 L(20 − 0) cm H2O − (0 −0) cm H2OCT = 0.500 L20 cm H2OCT = 0.025 L/cm H2O$

where PB = pressure outside the chest wall. It is considered to be 0 cm H2O in this calculation.

3. $1CT=1CL + 1CW1CW=1CT − 1CLCW=10.025 L/cm H2O − 10.07 L/cm H2O1CW=(40 − 14.3) / L/cm H2O1CW=25.7/L/cm H2OCW=0.039/L/cm H2O$

2–3. The correct answer is e.

• All of the conditions lead to decreased compliance. Surgical removal of one lobe would decrease pulmonary compliance because the lobes of the lung are in parallel and compliances in parallel add directly.

2–4. The correct answer is e.

2–5. The correct answer is e.

2–6. The correct answer is e.

2–7. The correct answer is d.

• Alveolar elastic recoil is greater at high lung volumes, which helps to oppose dynamic compression and decrease airways resistance by traction on small airways. During a forced expiration, as soon as ...

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