CBSE · NCERT · Class 11 Biology · Chapter 14

NCERT Solutions: Class 11 Biology Chapter 14 - Breathing and Exchange of Gases

13 textbook Q&A13 verifiedFree Content

Chapter-wise NCERT intext questions and exercise answers for Breathing and Exchange of Gases, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
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Exercises 13
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1Exercises13 questions
Q.1Define vital capacity. What is its significance?v
Solution

It equals IRV + TV + ERV. Vital capacity indicates the maximum usable capacity of the lungs and is clinically useful for assessing pulmonary function and respiratory health.

Answer:

Vital capacity is the maximum volume of air a person can breathe out after a forced inspiration, or breathe in after a forced expiration.

Q.2State the volume of air remaining in the lungs after a normal breathing.v
Solution

Using NCERT values, ERV is about 1000-1100 mL and RV is about 1100-1200 mL. Therefore, air remaining after normal breathing is about 2100-2300 mL.

Answer:

The volume remaining after normal expiration is functional residual capacity, equal to ERV + RV.

Q.3Diffusion of gases occurs in the alveolar region only and not in the other parts of respiratory system. Why?v
Solution

The alveolar membrane is extremely thin and closely associated with capillaries, giving a large surface area and short diffusion distance. Other respiratory passages have thicker walls and are not specialised for diffusion.

Answer:

Alveoli have the thin, vascular diffusion membrane required for gas exchange; the conducting parts mainly transport, clean, humidify and warm air.

Q.4What are the major transport mechanisms for CO2? Explain.v
Solution

About 70% of CO2 is transported as bicarbonate. In tissues, CO2 enters RBCs and carbonic anhydrase catalyses its conversion to carbonic acid, which dissociates into HCO3- and H+. About 20-25% of CO2 binds haemoglobin as carbamino-haemoglobin. About 7% remains dissolved in plasma. In alveoli, the reactions reverse and CO2 is released for exhalation.

Answer:

CO2 is transported mainly as bicarbonate, also as carbamino-haemoglobin and in dissolved form.

Q.5What will be the pO2 and pCO2 in the atmospheric air compared to those in the alveolar air ? (i) pO2 lesser, pCO2 higher (ii) pO2 higher, pCO2 lesser (iii) pO2 higher, pCO2 higher (iv) pO2 lesser, pCO2 lesserv
  1. i. pO2 lesser, pCO2 higher
  2. ii. pO2 higher, pCO2 lesser
  3. iii. pO2 higher, pCO2 higher
  4. iv. pO2 lesser, pCO2 lesser
Solution

Atmospheric air has pO2 about 159 mm Hg and pCO2 about 0.3 mm Hg. Alveolar air has pO2 about 104 mm Hg and pCO2 about 40 mm Hg. Thus atmospheric pO2 is higher and pCO2 is lower than alveolar air.

Answer:

(ii) pO2 higher, pCO2 lesser.

Q.6Explain the process of inspiration under normal conditions.v
Solution

Diaphragm contraction increases the thoracic chamber in the antero-posterior axis. External intercostal muscles lift ribs and sternum, increasing the dorso-ventral axis. The resulting increase in pulmonary volume lowers intrapulmonary pressure below atmospheric pressure, so air flows into the lungs.

Answer:

Normal inspiration occurs when diaphragm and external intercostal muscles contract, increasing thoracic and pulmonary volume and lowering intrapulmonary pressure.

Q.7How is respiration regulated?v
Solution

The respiratory rhythm centre in the medulla maintains rhythm. The pneumotaxic centre in the pons moderates inspiration and respiratory rate. A chemosensitive area near the rhythm centre detects increased CO2 and H+ and signals adjustments. Receptors in the aortic arch and carotid artery also send signals; oxygen has a comparatively minor role.

Answer:

Respiration is regulated mainly by neural centres in the medulla and pons, with feedback from chemoreceptors sensitive to CO2 and H+.

Q.8What is the effect of pCO2 on oxygen transport?v
Solution

In tissues, high pCO2, high H+ concentration and higher temperature shift conditions toward dissociation of oxyhaemoglobin. In alveoli, low pCO2 and high pO2 favour formation of oxyhaemoglobin.

Answer:

High pCO2 reduces haemoglobin's affinity for oxygen and promotes oxygen release in tissues; low pCO2 favours oxygen loading in alveoli.

Q.9What happens to the respiratory process in a man going up a hill?v
Solution

At high altitude, lower pO2 reduces oxygen diffusion into blood. Chemoreceptor feedback increases ventilation to improve oxygen uptake. Initially the person may feel breathlessness; acclimatisation gradually improves oxygen delivery.

Answer:

Breathing becomes faster and deeper because atmospheric oxygen pressure decreases at higher altitude.

Q.10What is the site of gaseous exchange in an insect?v
Solution

Insects have a network of tracheal tubes that transport atmospheric air directly within the body. Diffusion occurs between the tracheoles and tissue cells.

Answer:

In insects, gaseous exchange occurs through the tracheal system, especially fine tracheoles that reach tissues.

Q.11Define oxygen dissociation curve. Can you suggest any reason for its sigmoidal pattern?v
Solution

When one oxygen molecule binds to haemoglobin, the affinity of remaining binding sites increases, producing a steep middle part of the curve. The curve shows loading of oxygen in lungs and unloading in tissues.

Answer:

The oxygen dissociation curve plots percentage saturation of haemoglobin with oxygen against pO2; it is sigmoid because oxygen binding to haemoglobin is cooperative.

Q.13Distinguish between (a) IRV and ERV (b) Inspiratory capacity and Expiratory capacity. (c) Vital capacity and Total lung capacity.v
Solution

(a) IRV is extra air inspired by forceful inspiration after normal inspiration, about 2500-3000 mL; ERV is extra air expired by forceful expiration after normal expiration, about 1000-1100 mL. (b) Inspiratory capacity = TV + IRV, total air inspired after normal expiration; expiratory capacity = TV + ERV, total air expired after normal inspiration. (c) Vital capacity = IRV + TV + ERV; total lung capacity = vital capacity + residual volume.

Answer:

The pairs differ by whether they describe extra inspired air, extra expired air, combined inspiratory/expiratory capacity, or total lung volume including residual volume.

Q.14What is Tidal volume? Find out the Tidal volume (approximate value) for a healthy human in an hour.v
Solution

Tidal volume = 500 mL. Breathing rate = 12-16 breaths/min. Per minute air = 500 x 12 to 500 x 16 = 6000-8000 mL = 6-8 L. Per hour = 6-8 L/min x 60 min = 360-480 L.

Answer:

Tidal volume is about 500 mL per normal breath; in one hour a healthy human moves about 360-480 L of air at 12-16 breaths per minute.