Math questions

Question 1

How much O2 is absorbed each breath & per min?

Answer 1

1) Calculate inspiration. Use Pio2 of 149 mmHg –> O2= 149 mmHg / 760 mmHg= 19.61%.
2) 350 mL x 19.61%= 68.62 mL
3) Calculate expiration of O2: PO2 is 104 mmHg –> O2= 104 mmHg / 760 mmHg= 13.68%.
4) 350 mL x 13.68%= 47.89 mL.
5) 68.62 mL – 47.89 mL= 20.73 mL O2/breath.
6) 20.73 mL O2/breath x 12 bpm= 248.76 mL O2/min

Question 2

How many mL of O2 are absorbed per 1 dL?

Answer 2

Around 5 - 6 mL (depending on rounding)

Calculation:
- O2 inhaled per breath= 68.6 mL
- O2 exhaled per breath= 47.6 mL
- 21 mL absorbed per breath
- 21 mL x 12= 252 mL
- Minute ventilation is 4,200 mL
- 4,200 mL dived by 100 (dL)= 42 dL
- 252 mL divided by 42 dL= 6 mL

Question 3

Calculate FRC with the following values: Spirometer= 20L + 15% of that is Helium, final He concentration is 14%?

Answer 3

1) 20 L x 0.15%= 3L
2) X= 3 L / 0.14% = 21.43 L –> FRC= 1.43 L

Question 4

Calculate dead space:
Vt= 500cc
PaCO2= 44
PeCO2= 30

Answer 4

Vd / 500 mL = (44 mmHg - 30 mHg) / 44 mmHg
–> Vd = (14 mmHg / 44 mmHg) x 500 mL
–> Vd = 0.318 mmHg x 500 mL
–> Vd = 159.01 mL

Question 5

Calculate the oxygen tension at sea level:
inspired O2= 0.7
PaCO2= 40
Assume normal V/Q ratio

Answer 5

PAO2 = FiO2 (PB - PH2O) - (PaCO2 / R)
–> PAO2 = 0.70% (760 mmHg - 47 mmHg) - (40 mmHg / 0.8)
–> PAO2 = (0.70% x 711 mmHg) - 50 mmHg
–> PAO2 = 499.1 mmHg - 50 mmHg
–> PAO2 = 449.1 mmHg

Question 6

A patient’s mean arterial blood pressure is 100 mm Hg and his right atrial pressure is 2 mm Hg. His mean pulmonary artery pressure and pulmonary capillary wedge pressure (≈left atrial pressure) determined using a Swan-Ganz catheter, are 15 and 5 mm Hg, respectively.
- If his cardiac output is 5 L/min, calculate his pulmonary vascular resistance and systemic vascular resistance?

Answer 6

• PVR= 2 mm Hg/L/min
• SVR= 19.6 mmHg/L/min

Question 7

A subject starts at her FRC and breathes 100% O2 through a 1-way valve. The expired air is collected in a very large spirometer (called a Tissot spirometer). The test is continued until the expired N2 concentration, as measured by a nitrogen analyzer, is virtually zero. At this time, there are 36 L of gas in the spirometer, of which 5.6% is N2.

What is the subject’s FRC?

Answer 7

2.5 L

• The volume of N2 in the spirometer is 0.056 × 36 L, or 2.0 L. This is the volume of N2 in the subject’s lungs when the test began (at her FRC). Since N2 constituted 80% of her FRC, her FRC is equal to 100 ÷ 80 × 2.0 L, or 1.25 × 2.0 L, which is equal to 2.5 L.

Question 8

A patient on a ventilator has a rate of 10 breaths per minute and a tidal volume (VT) of 500 mL.

a. What is the patient’s VE?
b. If the patient’s anatomic dead space is estimated to be 150 mL, what is his VA?
c. If his rate is increased to 15 breaths per minute with VT remaining at 500 mL, what will his new VE and VA be?
d. If his VT is increased to 750 mL, with his rate remaining at 10 breaths per minute, what will his new VE and VA be?

Answer 8

a) VE= 10bpm x 500mL= 5L/min

b) VA= VE= VA + VD –> VA= VE - VD –> 10bpm x (500 mL - 150 mL)= 3,500 mL/min

c) VE= 15bpm x 500 mL= 7,500 mL/min
VA= 15bpm x (500 mL - 150mL)= 5,250 mL/min

d) VE= 10bpm x 750 mL= 7,500 mL/min
VA= 10bpm x (750 mL - 150 mL)= 6,000 mL/min

Question 9

A postoperative patient whose respiratory muscles have been paralyzed with pancuronium bromide, a curare-like drug, is maintained by a positive-pressure respirator. At end expiration (when alveolar pressure equals 0), intrapleural pressure, as measured by an esophageal balloon, is equal to –3 cm H2O. At the peak of inspiration, alveolar pressure is +20 cm H2O and intrapleural pressure is +10 cm H2O. Tidal volume is 500 mL.

• a. What is the patient’s pulmonary compliance?
• b. What is the patient’s total compliance?
• c. What is the patient’s chest wall compliance?

Answer 9

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