4. Subject 4 has the following values: Minute ventilation = 60 liters/minute. Re

Question

4. Subject 4 has the following values: Minute ventilation = 60 liters/minute. Respiratory frequency = 60 Briminute, PaCO2-30 mm Hg, PECO,-24 mm Hg, VCO2 -1668 ml/minute. Identify whether each of the following is correct or incorrect. a. the tidal volume = 1 liter b. the VD/VT is 0.45 C. the alveolar ventilation = 28 liters/minute d. the subject's metabolic rate is higher than normal e. this subject is probably exercising 5. Subject 5 is a 21 year old BSU student who has the following values at baseline: PaO2 54 mm Hg, PaCO2- 60 mm Hg while breathing FIO2- 0.21 Ps 690 mm Hg. ARespiratory Therapist administers additional oxygen with an FIO20.4. Identify whether each of the following are correct or incorrect. a. At baseline, the PaCO2 is within normal limits b. At baseline, the PaO2 is within normal limits c. At baseline, the value for PaO2 explaied by hypoventilation alone d. The alA ratio is consistent with low VIQ or shunting e. Administration of oxygen will result in a decrease in PaCO2 proportional to the increase in PaO2 Administration of 40% oxygen would be expected to result in a PAO2-257 mm Hg f. 6. Subject 6 has the following values: FIO2 0.7, PaO2 250 mm Hg, Hb 6 g/dl, SV02-50%, CO-6 lpm, PaCO2-35 mm Hg. Estimate the SaO2 and PVO2 using the normal oxyhemoglogin dissociation curve, and identify whether each of the following is correct or incorrect. a. The subject has low VIQ b. The Sao,-100% c. The PVO2-26 mm Hg d. The oxygen consumption is about 280 ml/minute e· The subject must be exercising f. Dissolved CO2 = 0.105 meQL

Explanation / Answer

Question 4

Answer

a. The tidal volume = 1 L is correct

The tidal volume is calculated by following formula

V=VT*f

here V=minute ventilation/minute volume

VT= tidal volume

f= respiratory frequency

VT= V/f =60/60 =1 liter.

up on calculation the tidal volume is 1 liter.

B. VD/VT is 0.45 it is incorrect

because VD/VT is the dead space over tidal voume

VD/VT calculated by (PaCO2 - PECO2) / PaCO2

30 - 24 /30 =0.2

C. The alveolar ventilation = 28L/minute is incorrect

it is calculated by (Tidal volume - Dead space) * respiration rate

1 - 0.2 * 60 =0.8 * 60 = 48 L/minute

D. Metabolic rate is higher than normal. Incorrect

here the tidal volume is less that indicate the metabolic rate is not higher than normal. because the tidal volume increases when the metabolic rate is more.

and also justified by VE / VCO2 value

VE is the minute volume

from that 60 / 1668 = 0.03.

normally the value is 0.8. but during the exercise it is increased to 1.10. So there is no increased metabolic rate.

E. is also Incorrect

from the above data the peron is not doing any exercise.

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