Question

(a)The normal values of tidal volume and dead space volume are 500 and 300 ml respectively. If the lung minute respiratory volume is 6.75 litres, find the frequency of respiration (number of breaths per minute). Calculate the alveolar minute respiratory volume.

(b) What is the percentage change in the lung minute respiratory volume if the frequency of respiration (i) increases by 15% or (ii) decreases by 15% from the normal value in part (a)?

Answer #1

A)Lung minute respiratory volume is just the multiplication of tidal volume and respiratory rate(frequency)

lung minute respiratory volume = tidal volume x respiratory rate

therefore respiratory rate = lung minute volume/tidal volume= 6.75/0.5 L=13.5/min.

The lung minute respiratoryvolume is the sum of alveolar ventilation and dead space ventilation

dead space ventilation= 13.5 x dead space volume= 13.5 x 0.3= 4.05L

there fore alveolar ventilation = 6.75-4.05=2.70 L

B) As the lung mnute volume is diectly propotional to the respiratory rate , 15% increarse in the respiratory rate will increase it by15%.

And 15% decrease will decrease the lung minute volume by 15%.

Patient A has a dead space volume of 150ml, a tidal
volume of 450ml and a respiratory frequency of 10. Patient B has a
dead space volume of 150ml, a tidal volume of 300ml and a
respiratory frequency of 15.
Which patient has the most effective
ventilation?
Why?

Use the following data from a patient's medical chart to answer
the next 3 questions
Blood Pressure.................................... 98/60 (mm
Hg)
Heart Rate........................................... 104 BPM
Central Venous Pressure.................... 3 mm Hg
Respiratory Rate (f)............................. 18
breaths/min
Tidal Volume (VT)……………………... 350 ml/breath
Dead Space (VD)................................. 200
ml/breath
Total Lung Capacity (TLC).................. 6.9 L
(Forced) Vital Capacity (FVC)……….. 4.8 L
PaO2 .........[normal O2 = 100] 70 mm Hg
PaCO2 .......[normal CO2 = 40] 50 mm Hg
pHa......................................................
7.32
[HCO3-]................................................ 30
mEq/L
This...

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