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# The lung volume for a subject is 5000mL, the functional residual capacity is 2100mL, the tidal volume is 582 mL. Calcula...

biomedical instrumentation question:

The lung volume for a subject is 5000mL, the functional residual capacity is 2100mL, the tidal volume is 582 mL. Calculate the inspiratory reserve volume in terms of mL.

Lung volumes or respiratory volumes is the volume of gas present in the lung during a given period of time. There are different types of lung volumes and the addition of these lung volumes can derive the different lung capacities of the lung. The average lung capacity of a human body is about 6 liters and the total lung capacity of the lung is the sum of functional residual capacity, tidal volume and inspiratory reserve volume. The measurement of different lung capacity is important in the physiological diagnosis.

Tidal volume is the amount of air that can be inhaled or exhaled during a respiratory cycle. The tidal volume of the individual is 582 ml.

Inspiratory reserve volume is the amount of air that can be forcefully inhaled after a normal tidal volume. It is used only during the deep breathing and is kept as a reserve volume of the lungs during the normal breathing.

Expiratory reserve volume is the amount of air that can be forcefully exhaled after a normal tidal volume. Residual volume is the amount of air remaining in the lungs after a maximum  exhalation . The functional residual capacity is the amount of air remaining in the lungs after a normal exhalation. The functional residual capacity is the sum of the expiratory reserve volume  and residual volume = 2100 ml

The total lung capacity or the lung volume is the maximum amount of air that a lung can accommodate. It is the sum of all the volume of air in the lungs after a maximum inhalation. That is the sum of Tidal volume + Inspiratory reserve volume+ Expiratory reserve volume +Residual volume= 5000ml

Lung volume = Tidal volume + Inspiratory reserve volume+ Expiratory reserve volume +Residual volume= 5000ml

or Lung volume =Tidal volume + Inspiratory reserve volume+ functional residual capacity = 5000 ml

Therefore,

The Inspiratory reserve volume= Lung volume - (Tidal volume +functional residual capacity )

= 5000 ml - ( 582ml +2100ml )

= 5000ml- 2682 ml

= 2318 ml

The Inspiratory reserve volume= 2318 ml

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