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🌳 Class XI · Biology · Chapter 14

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

Complete NCERT solutions for Class 11 Biology Chapter 14 Breathing and Exchange of Gases — all 14 exercise questions with detailed topper-style answers, every NCERT figure (14.1–14.5) redrawn as an original diagram with PDF page links, respiratory-volume tables, flow charts, exam tips and full chapter-at-a-glance revision notes.

14 questions10 diagrams22 tablesFull revision notes

Chapter at a glance — complete revision notes

Why we breathe. Oxygen (O2) is used by organisms to indirectly break down simple molecules like glucose, amino acids and fatty acids to derive energy for various activities. Carbon dioxide (CO2), which is harmful, is released during these catabolic reactions. O2 must therefore be continuously supplied to the cells and CO2 produced by the cells must be released. The exchange of O2 from the atmosphere with CO2 produced by the cells is called breathing, commonly known as respiration.

14.1 Respiratory organs

Animal groupMode of gas exchange
Sponges, coelenterates, flatworms (lower invertebrates)Simple diffusion over the entire body surface
EarthwormMoist cuticle
InsectsNetwork of tracheal tubes carrying atmospheric air within the body
Aquatic arthropods and molluscsGills (branchial respiration) — vascularised structures
Terrestrial formsLungs (pulmonary respiration) — vascularised bags
FishesGills
Amphibians, reptiles, birds, mammalsLungs; frogs also respire through moist skin (cutaneous respiration)

14.1.1 Human respiratory system

  • Path of air: external nostrils (above the upper lip) → nasal chamber via nasal passage → pharynx (common passage for food and air) → larynx (cartilaginous box, sound box) → trachea → right and left primary bronchi → secondary and tertiary bronchi → bronchioles → very thin terminal bronchioles → alveoli.
  • Epiglottis: a thin elastic cartilaginous flap that covers the glottis during swallowing to prevent food from entering the larynx.
  • Trachea is a straight tube up to the mid-thoracic cavity; it divides at the level of the 5th thoracic vertebra. Trachea, primary, secondary and tertiary bronchi and initial bronchioles are supported by incomplete cartilaginous rings.
  • Alveoli are very thin, irregular-walled, vascularised bag-like structures arising from each terminal bronchiole. The branching network of bronchi, bronchioles and alveoli comprises the lungs.
  • Pleura: two lungs, each covered by a double-layered pleura with pleural fluid between the layers, which reduces friction. The outer pleural membrane touches the thoracic lining and the inner one touches the lung surface.
Diagrammatic view of human respiratory systemHuman respiratory system (sectional view of the left lung also shown)heartEpiglottisLarynxTracheaBronchusLungCut end of rib(rib cage)Pleural membranesAlveoliPleural fluidBronchioleDiaphragm© iBioKaare
Fig. — Diagrammatic view of the human respiratory system with sectional view of the left lung (original iBioKaare schematic). Compare with NCERT textbook — Fig. 14.1, PDF page 4 (book page 184) ↗
PartExtentFunction
Conducting partExternal nostrils up to the terminal bronchiolesTransports air to the alveoli, clears it of foreign particles, humidifies it and brings it to body temperature
Exchange (respiratory) partAlveoli and their ductsActual diffusion of O2 and CO2 between blood and atmospheric air

Thoracic chamber. The lungs lie in the anatomically air-tight thoracic chamber, formed dorsally by the vertebral column, ventrally by the sternum, laterally by the ribs and below by the dome-shaped diaphragm. Any change in the volume of the thoracic cavity is reflected in the pulmonary cavity, which is essential because we cannot directly alter the pulmonary volume.

Steps of respiration:

  • (i) Breathing or pulmonary ventilation — atmospheric air is drawn in and CO2-rich alveolar air is released.
  • (ii) Diffusion of gases (O2 and CO2) across the alveolar membrane.
  • (iii) Transport of gases by the blood.
  • (iv) Diffusion of O2 and CO2 between blood and tissues.
  • (v) Utilisation of O2 by cells for catabolic reactions and the resultant release of CO2 (cellular respiration).

14.2 Mechanism of breathing

Breathing has two stages: inspiration (atmospheric air drawn in) and expiration (alveolar air released). Air moves because of a pressure gradient between lungs and atmosphere. Inspiration occurs when intra-pulmonary pressure is less than atmospheric pressure (negative pressure in the lungs); expiration occurs when it is higher. The diaphragm and the external and internal intercostal muscles generate these gradients.

  • Inspiration: contraction of the diaphragm increases thoracic volume in the antero-posterior axis; contraction of external intercostal muscles lifts the ribs and sternum, increasing volume in the dorso-ventral axis. Pulmonary volume increases, intra-pulmonary pressure falls below atmospheric and air rushes in.
  • Expiration: relaxation of the diaphragm and intercostal muscles returns the diaphragm and sternum to normal positions, reducing thoracic and pulmonary volume. Intra-pulmonary pressure rises slightly above atmospheric pressure and air is expelled.
  • Additional abdominal muscles can strengthen inspiration and expiration. A healthy human breathes 12–16 times per minute. A spirometer estimates the volume of air involved and helps clinical assessment of pulmonary function.
Mechanism of breathing showing inspiration and expirationMechanism of breathing: (a) inspiration, (b) expiration (sagittal view of thorax)Air entering lungsRibs and sternumraisedRib cageDiaphragm contracted (flattened)Volume ofthorax increasedAir expelled from lungsRibs and sternumreturned tooriginal positionDiaphragm relaxed and arched upwardsVolume ofthorax decreased(a) Inspiration(b) Expiration© iBioKaare
Fig. — Mechanism of breathing showing inspiration and expiration (original iBioKaare schematic). Compare with NCERT textbook — Fig. 14.2, PDF page 6 (book page 186) ↗

14.2.1 Respiratory volumes and capacities

VolumeMeaningApproximate value (NCERT)
Tidal volume (TV)Air inspired or expired in a normal respirationAbout 500 mL (6000–8000 mL per minute)
Inspiratory reserve volume (IRV)Extra air that can be inspired by forcible inspiration2500–3000 mL
Expiratory reserve volume (ERV)Extra air that can be expired by forcible expiration1000–1100 mL
Residual volume (RV)Air remaining in the lungs even after forcible expiration1100–1200 mL
CapacityDefinitionComponents
Inspiratory capacity (IC)Total air a person can inspire after a normal expirationTV + IRV
Expiratory capacity (EC)Total air a person can expire after a normal inspirationTV + ERV
Functional residual capacity (FRC)Air remaining in lungs after a normal expirationERV + RV
Vital capacity (VC)Maximum air a person can breathe in after a forced expiration, or breathe out after a forced inspirationERV + TV + IRV
Total lung capacity (TLC)Total air accommodated in the lungs at the end of a forced inspirationRV + ERV + TV + IRV = VC + RV

14.3 Exchange of gases

Alveoli are the primary sites of gas exchange; exchange also occurs between blood and tissues. O2 and CO2 move by simple diffusion mainly based on pressure/concentration gradient. Solubility of the gases and thickness of the diffusion membranes are also important. The pressure contributed by an individual gas in a mixture is its partial pressure (pO2, pCO2).

GasAtmospheric airAlveoliBlood (deoxygenated)Blood (oxygenated)Tissues
O2 (mm Hg)159104409540
CO2 (mm Hg)0.340454045
Diagrammatic representation of exchange of gases at the alveolus and body tissuesExchange of gases at the alveolus and at the body tissuesAlveolar airpO2 = 104 mm HgpCO2 = 40 mm HgInspired airExpired airAlveolusPulmonaryarteryPulmonaryveinCO2O2RHLHSystemic veins(carrying deoxygenated blood)pO2 = 40 mm HgpCO2 = 45 mm HgSystemic arteries(carrying oxygenated blood)pO2 = 95 mm HgpCO2 = 40 mm HgBody tissuesIn tissues: O2 passes from blood to cells; CO2 passes from cells to blood© iBioKaare
Fig. — Exchange of gases at the alveolus and body tissues with transport of oxygen and carbon dioxide (original iBioKaare schematic). Compare with NCERT textbook — Fig. 14.3, PDF page 8 (book page 188) ↗
  • There is a pO2 gradient from alveoli to blood and blood to tissues, and a pCO2 gradient in the opposite direction (tissues to blood, blood to alveoli).
  • The solubility of CO2 is 20–25 times that of O2, so far more CO2 diffuses per unit difference in partial pressure.
  • The diffusion membrane has three layers — thin squamous epithelium of alveoli, endothelium of alveolar capillaries, and the basement substance between them — with a total thickness much less than a millimetre.
Diagram of a section of an alveolus with a pulmonary capillarySection of an alveolus with a pulmonary capillaryO2CO2Alveolar cavitySquamous epithelium of alveolar wall(one-celled thick)Basement substanceEndothelium of blood capillaryRed blood cellBlood capillary (lumen)AirDiffusion membrane = three thin layers:1. squamous epithelium of alveolus2. basement substance between them3. endothelium of the capillary(total thickness much less than 1 mm)© iBioKaare
Fig. — Section of an alveolus with a pulmonary capillary (original iBioKaare schematic). Compare with NCERT textbook — Fig. 14.4, PDF page 8 (book page 188) ↗

14.4 Transport of gases

GasTransport formShare
O2Bound to haemoglobin in RBCs (oxyhaemoglobin)About 97%
O2Dissolved in plasmaAbout 3%
CO2As bicarbonate (HCO3−)About 70%
CO2As carbamino-haemoglobin in RBCs20–25%
CO2Dissolved in plasmaAbout 7%

14.4.1 Transport of oxygen

  • Haemoglobin is a red, iron-containing pigment in RBCs. O2 binds reversibly to form oxyhaemoglobin; each haemoglobin molecule can carry a maximum of four O2 molecules.
  • Binding depends primarily on pO2; pCO2, H+ concentration and temperature also interfere.
  • Percentage saturation of haemoglobin plotted against pO2 gives a sigmoid curve — the oxygen dissociation curve.
  • In the alveoli (high pO2, low pCO2, lower H+, lower temperature) conditions favour oxyhaemoglobin formation; in the tissues (low pO2, high pCO2, high H+, higher temperature) they favour dissociation of oxygen.
  • Every 100 mL of oxygenated blood delivers around 5 mL of O2 to the tissues under normal conditions.
Oxygen dissociation curveOxygen dissociation curve (percentage saturation of haemoglobin vs pO2)002020404060608080100100Partial pressure of oxygen, pO2 (mm Hg)Percentage saturation of haemoglobin with oxygenATA = alveoli (pO2 about 104 mm Hg): about 97% saturated T = tissues (pO2 about 40 mm Hg): about 75% saturatedNormal curve (sigmoid)Right shift: high pCO2, H+, temperatureLeft shift: low pCO2, H+, temperature© iBioKaare
Fig. — Oxygen dissociation curve (original iBioKaare schematic). Compare with NCERT textbook — Fig. 14.5, PDF page 9 (book page 189) ↗

14.4.2 Transport of carbon dioxide

  • About 20–25% of CO2 is carried by haemoglobin as carbamino-haemoglobin; binding is related to pCO2 and pO2. When pCO2 is high and pO2 low (tissues) more CO2 binds; when pCO2 is low and pO2 high (alveoli) CO2 dissociates.
  • RBCs contain a very high concentration of carbonic anhydrase (traces in plasma), which catalyses the reaction in both directions.
  • At the tissues CO2 diffuses into blood and forms HCO3− and H+; at the alveoli the reaction runs in reverse, releasing CO2. Every 100 mL of deoxygenated blood delivers about 4 mL of CO2 to the alveoli.

CO2 + H2O ⇌ H2CO3 ⇌ HCO3− + H+  (carbonic anhydrase)

14.5 Regulation of respiration

  • The respiratory rhythm centre in the medulla is primarily responsible for regulation.
  • The pneumotaxic centre in the pons can moderate the rhythm centre; its signals can reduce the duration of inspiration and so alter the respiratory rate.
  • A chemosensitive area adjacent to the rhythm centre is highly sensitive to CO2 and hydrogen ions; an increase activates it and the rhythm centre makes adjustments to eliminate them.
  • Receptors on the aortic arch and carotid artery also detect changes in CO2 and H+ and signal the rhythm centre.
  • The role of oxygen in regulating respiratory rhythm is quite insignificant.

14.6 Disorders of the respiratory system

  • Asthma: difficulty in breathing causing wheezing due to inflammation of bronchi and bronchioles.
  • Emphysema: chronic disorder in which alveolar walls are damaged, reducing the respiratory surface; cigarette smoking is a major cause.
  • Occupational respiratory disorders: in industries involving grinding or stone-breaking, so much dust is produced that the body’s defence cannot cope; long exposure causes inflammation leading to fibrosis (proliferation of fibrous tissue) and serious lung damage. Workers should wear protective masks.
Q1

Define vital capacity. What is its significance?

Answer

Concept. The volume of air moved in and out of the lungs is measured with a spirometer. Different volumes (tidal, inspiratory reserve, expiratory reserve, residual) can be added to obtain pulmonary capacities, which are used in clinical diagnosis. Vital capacity is the most important of these because it measures the “usable” part of the lung volume — the air that can actually be exchanged with the atmosphere.

Definition. Vital capacity (VC) is the maximum volume of air a person can breathe in after a forced expiration, or equally the maximum volume of air a person can breathe out after a forced inspiration. It includes the expiratory reserve volume (ERV), tidal volume (TV) and inspiratory reserve volume (IRV):

VC = ERV + TV + IRV

ComponentNCERT value
ERV1000–1100 mL
TVabout 500 mL
IRV2500–3000 mL
VC (sum of the above)about 4000–4600 mL (derived from the NCERT values)

Relationship with other capacities. Vital capacity does not include the residual volume (RV) of 1100–1200 mL that always stays in the lungs. When RV is added, the result is the total lung capacity (TLC = VC + RV).

Significance of vital capacity

  1. Index of lung function. VC tells how much air a person can move in and out of the lungs in one maximal breath. A normal VC indicates healthy, elastic lungs and strong respiratory muscles.
  2. Clinical diagnosis. NCERT states that pulmonary capacities are used in clinical diagnosis. A reduced VC points to conditions such as damage to alveolar walls (emphysema), fibrosis from dust exposure, obstruction of air passages or weakness of respiratory muscles.
  3. Assessing the capacity for gas exchange. A larger VC means that more fresh air can be brought to the alveoli with each deep breath, which is useful for sustained physical activity.
  4. Monitoring progress and fitness. VC can be measured repeatedly with a spirometer to follow the course of a disease or treatment, and it is generally higher in trained individuals such as athletes and lower in people with lung disease (general physiology knowledge).

Extra depth. Because the residual volume cannot be expelled, only the vital capacity is under voluntary control; this is why VC, and not TLC, is the measurement commonly recorded by spirometry. The total of VC and RV gives the total lung capacity of about 5100–5800 mL (derived), showing that nearly 20% of the lung volume is never exchanged.

Exam tip: Write the formula VC = ERV + TV + IRV and the definition with both directions (“after a forced expiration” / “after a forced inspiration”) — both are in the NCERT definition.
Exam tip: For “significance” use the keywords clinical diagnosis of pulmonary function and mention spirometer; add that it excludes residual volume.

Conclusion: Vital capacity (ERV + TV + IRV) is the maximum air that can be breathed in after a forced expiration or out after a forced inspiration; being a measure of the usable lung volume, it is clinically significant in assessing and diagnosing the state of pulmonary function.

Q2

State the volume of air remaining in the lungs after a normal breathing.

Answer

Concept. Even in quiet breathing the lungs are never empty. After each normal expiration some air remains, and even after the most forceful expiration an irreducible amount stays behind. Two terms in NCERT describe these quantities: residual volume (air left after forcible expiration) and functional residual capacity (air left after a normal expiration).

Answer. The volume of air remaining in the lungs after a normal breathing (normal expiration) is the Functional Residual Capacity (FRC). It is the sum of the expiratory reserve volume and the residual volume:

FRC = ERV + RV

ComponentMeaningValue
ERVAir that could still be expelled by a forcible expiration after a normal expiration1000–1100 mL
RVAir that remains even after forcible expiration1100–1200 mL
FRC = ERV + RVAir remaining after normal expirationabout 2100–2300 mL (derived)

Why this air is useful. The FRC acts as a reservoir of air in the lungs. Because it is large compared with the tidal volume (500 mL), the composition of alveolar air changes only slightly with each breath, so the O2 and CO2 levels of blood stay steady. It also keeps the alveoli from collapsing completely and permits continuous gas exchange even between breaths.

Do not confuse. Residual volume (1100–1200 mL) is the air left after a forcible expiration, whereas FRC (about 2100–2300 mL) is the air left after a normal expiration. The question asks about normal breathing, so the answer is FRC.

Exam tip: Answer with the term (functional residual capacity), the formula (ERV + RV) and the number (about 2100–2300 mL) — three separate marks.
Exam tip: Do not write only “residual volume” — that is the air left after forcible expiration; the examiner expects FRC for normal breathing.

Conclusion: After a normal expiration the lungs still hold the functional residual capacity, ERV + RV, which is about 2100–2300 mL; it serves as an air reservoir that keeps alveolar gas composition stable.

Q3

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

Answer

Concept. The respiratory system has two functional parts: the conducting part (external nostrils to terminal bronchioles) and the exchange part (alveoli and their ducts). The conducting part only transports atmospheric air to the alveoli, clears it of foreign particles, humidifies it and brings it to body temperature. The alveoli are the primary sites of gas exchange. Gases diffuse across a membrane only if certain conditions are satisfied — these conditions are met in the alveoli but not elsewhere.

  1. Very thin diffusion membrane. The alveolar diffusion membrane consists of just three layers — the thin squamous epithelium of the alveolus, the endothelium of the alveolar capillary and the basement substance between them — and its total thickness is much less than a millimetre. The walls of the nasal passage, trachea and bronchi are thick and are supported by cartilage rings and muscle, so gases cannot cross them quickly.
  2. Rich blood supply. Each alveolus is surrounded by a network of blood capillaries (Fig. 14.4) so blood and air are separated only by the thin membrane. The conducting passages have no such close association with capillaries for gas exchange.
  3. Large surface area. The huge number of thin, bag-like alveoli provides an enormous surface for diffusion (general knowledge: tens of square metres), unlike the narrow tubes of the conducting part.
  4. Maintained partial-pressure gradient. Diffusion depends on the difference in partial pressures. Alveolar air (pO2 104, pCO2 40 mm Hg) and deoxygenated blood (pO2 40, pCO2 45 mm Hg) have a steep gradient, maintained by constant ventilation and blood flow, so O2 diffuses into blood and CO2 into alveoli.
  5. High solubility of CO2. Since the solubility of CO2 is 20–25 times that of O2, CO2 diffuses readily across the membrane even with a small pressure difference.
  6. Moist, single-cell-thick wall. The alveolar surface is moist, allowing gases to dissolve before they diffuse; the conducting part is lined with mucus-secreting ciliated epithelium suited for trapping dust and not for exchange.
Diagram of a section of an alveolus with a pulmonary capillarySection of an alveolus with a pulmonary capillaryO2CO2Alveolar cavitySquamous epithelium of alveolar wall(one-celled thick)Basement substanceEndothelium of blood capillaryRed blood cellBlood capillary (lumen)AirDiffusion membrane = three thin layers:1. squamous epithelium of alveolus2. basement substance between them3. endothelium of the capillary(total thickness much less than 1 mm)© iBioKaare
Fig. — Section of an alveolus with a pulmonary capillary (original iBioKaare schematic). Compare with NCERT textbook — Fig. 14.4, PDF page 8 (book page 188) ↗
FeatureAlveolar regionOther parts (nose to terminal bronchioles)
Wall thicknessLess than 1 mm; three thin layersThick, with cartilage rings and muscle
Capillary networkDense, closely appliedNot specialised for exchange
FunctionGas exchange (primary site)Conduct, filter, humidify, warm air
Surface areaVery large (millions of alveoli)Small, tubular

Extra depth. The same principle applies at the tissues: gas exchange between blood and tissue cells occurs through the thin walls of the capillaries, not through arteries or veins. Efficient diffusion everywhere in the body therefore needs thin membranes, large surfaces, steep gradients and good perfusion.

Exam tip: Use the phrase “three-layered diffusion membrane less than 1 mm thick” and draw/mention Fig. 14.4; the examiner wants thin membrane + capillary network + conducting-vs-exchange part.
Exam tip: Quote the pressure values (alveolar pO2 104 vs venous 40; pCO2 40 vs 45 mm Hg) to show the gradient — numerical evidence earns extra marks.

Conclusion: Gases diffuse only in the alveoli because only there is a very thin three-layered membrane in intimate contact with a dense capillary network over a large surface with a steep partial-pressure gradient; the rest of the respiratory tract merely conducts, cleans, humidifies and warms the air.

Q4

What are the major transport mechanisms for CO2? Explain.

Answer

Concept. CO2 produced by the cells diffuses into the blood and has to be carried to the lungs. Blood is the transport medium, but CO2 is not carried in just one way. According to NCERT, nearly 70% of CO2 is carried as bicarbonate, 20–25% by red blood cells (as carbamino-haemoglobin) and about 7% in the dissolved state in plasma. Each mechanism has its own chemistry and role.

Diagrammatic representation of exchange of gases at the alveolus and body tissuesExchange of gases at the alveolus and at the body tissuesAlveolar airpO2 = 104 mm HgpCO2 = 40 mm HgInspired airExpired airAlveolusPulmonaryarteryPulmonaryveinCO2O2RHLHSystemic veins(carrying deoxygenated blood)pO2 = 40 mm HgpCO2 = 45 mm HgSystemic arteries(carrying oxygenated blood)pO2 = 95 mm HgpCO2 = 40 mm HgBody tissuesIn tissues: O2 passes from blood to cells; CO2 passes from cells to blood© iBioKaare
Fig. — Exchange of gases at the alveolus and body tissues with transport of oxygen and carbon dioxide (original iBioKaare schematic). Compare with NCERT textbook — Fig. 14.3, PDF page 8 (book page 188) ↗
  1. As bicarbonate (about 70%). RBCs contain a very high concentration of the enzyme carbonic anhydrase (minute quantities also in plasma). At the tissues, where pCO2 is high due to catabolism, CO2 diffuses into blood (RBCs and plasma) and, with water, forms carbonic acid (H2CO3), which dissociates into bicarbonate (HCO3−) and hydrogen ion (H+). The bicarbonate is carried in the blood to the lungs. At the alveolar site, where pCO2 is low, the reaction runs in the opposite direction, releasing CO2 and H2O, and the CO2 diffuses into the alveoli to be exhaled.
  2. As carbamino-haemoglobin (20–25%). CO2 binds to the haemoglobin of RBCs to form carbamino-haemoglobin. Binding is related to the partial pressure of CO2, and pO2 is a major factor affecting it: when pCO2 is high and pO2 low (tissues) more CO2 binds; when pCO2 is low and pO2 high (alveoli) CO2 dissociates from carbamino-haemoglobin and is delivered to the alveoli.
  3. In dissolved form in plasma (about 7%). A small portion of CO2 simply dissolves in plasma and is carried to the lungs; because CO2 is 20–25 times more soluble than O2, this fraction is significant compared with dissolved O2 (3%).

CO2 + H2O ⇌ H2CO3 ⇌ HCO3− + H+  (carbonic anhydrase in RBC)

MechanismShareWhere it happens / key factor
Bicarbonate (HCO3−)About 70%RBCs and plasma; carbonic anhydrase; direction depends on pCO2
Carbamino-haemoglobin20–25%RBCs; high pCO2/low pO2 favours binding
Dissolved CO2About 7%Plasma; high solubility
Tissues: high pCO2→CO2 enters blood / RBC→Forms HCO3− and carbamino-Hb→Venous blood to lungs→Alveoli: low pCO2→CO2 released and exhaled

Extra depth. Every 100 mL of deoxygenated blood delivers about 4 mL of CO2 to the alveoli. The same enzyme works in both directions — the direction is decided by the partial pressure of CO2 — which is why one enzyme can both load CO2 in tissues and unload it in the lungs.

Exam tip: Remember the numbers in order: 70% bicarbonate – 20–25% carbamino-haemoglobin – 7% dissolved; they add up to about 100% and are a common 1-mark question.
Exam tip: Write the reversible reaction with carbonic anhydrase in the RBC and state that at tissues it goes forward, at alveoli it goes backward — this is the key to the full marks.

Conclusion: CO2 is carried by blood mainly as bicarbonate (about 70%, via carbonic anhydrase), as carbamino-haemoglobin (20–25%) and in dissolved form in plasma (about 7%), the direction of each process being governed by the partial pressure of CO2 in tissues and alveoli.

Q5

What 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 lesser

Answer

Concept. The partial pressure of a gas is the pressure contributed by that gas in a mixture of gases; it is written pO2 for oxygen and pCO2 for carbon dioxide. Diffusion of gases in the body is driven by differences in partial pressure, so comparing the values in the atmosphere and in the alveoli explains the direction of movement of the gases. Table 14.1 of NCERT gives the exact values.

GasAtmospheric airAlveolar airComparison (atmosphere vs alveoli)
O2159 mm Hg104 mm HgAtmospheric pO2 is higher
CO20.3 mm Hg40 mm HgAtmospheric pCO2 is lesser

Correct answer: (ii) pO2 higher, pCO2 lesser. Atmospheric air has pO2 = 159 mm Hg (higher than the alveolar 104 mm Hg) and pCO2 = 0.3 mm Hg (lesser than the alveolar 40 mm Hg).

Why each wrong option is wrong

  • (i) pO2 lesser, pCO2 higher — this describes the alveolar air compared with atmospheric air (alveolar pO2 104 is lower, pCO2 40 is higher). It is the reverse of what the question asks, which is the atmosphere compared with the alveoli.
  • (iii) pO2 higher, pCO2 higher — atmospheric pCO2 is only 0.3 mm Hg, far lower than the alveolar 40 mm Hg, so “pCO2 higher” is wrong.
  • (iv) pO2 lesser, pCO2 lesser — atmospheric pO2 (159) is greater than the alveolar value (104), so “pO2 lesser” is wrong.

Why the alveolar values differ. The inspired air mixes with the large volume of air already in the lungs (the functional residual capacity of about 2100–2300 mL), so fresh air is diluted. O2 is continuously removed from alveolar air into blood, which lowers pO2 to 104 mm Hg, and CO2 is continuously added from blood, raising pCO2 to 40 mm Hg. (Humidification of inspired air also lowers its pO2 slightly — general physiology.)

Atmosphere pO2 159→Alveoli 104→Deoxygenated blood 40 / oxygenated blood 95→Tissues 40

Extra depth. The values form a gradient: O2 moves from atmosphere (159) to alveoli (104) to blood (40 → 95) to tissues (40). CO2 moves the other way: tissues (45) → blood (45 → 40) → alveoli (40) → atmosphere (0.3). The gradient for CO2 is small but it diffuses efficiently because CO2 is 20–25 times more soluble than O2.

Exam tip: Memorise Table 14.1: O2 159 → 104 → 40 (deoxygenated) → 95 (oxygenated) → 40 (tissues); CO2 0.3 → 40 → 45 → 40 → 45. Numerical questions are common.
Exam tip: In an MCQ, first decide the direction of comparison (atmosphere relative to alveoli) — options (i) and (ii) are mirror images and many students pick (i) by mistake.

Conclusion: Atmospheric air has a higher pO2 (159 mm Hg) and a lesser pCO2 (0.3 mm Hg) than alveolar air (104 and 40 mm Hg), so option (ii) is correct.

Q6

Explain the process of inspiration under normal conditions.

Answer

Concept. Breathing has two stages: inspiration, during which atmospheric air is drawn in, and expiration, by which alveolar air is released. Air moves in and out of the lungs only when a pressure gradient is created between the lungs and the atmosphere. Inspiration can occur only if the pressure within the lungs (intra-pulmonary pressure) is less than the atmospheric pressure, i.e. there is a negative pressure in the lungs with respect to the atmosphere. Because the lungs are located in an air-tight thoracic chamber and we cannot directly alter pulmonary volume, the change is brought about indirectly by altering the thoracic volume with the help of the diaphragm and the external intercostal muscles.

Steps of normal inspiration

  1. Contraction of the diaphragm. The dome-shaped diaphragm contracts and flattens, increasing the volume of the thoracic chamber in the antero-posterior axis.
  2. Contraction of the external intercostal muscles. These muscles between the ribs contract and lift up the ribs and the sternum, increasing the volume of the thoracic chamber in the dorso-ventral axis.
  3. Increase in thoracic and pulmonary volume. The overall increase in thoracic volume causes a similar increase in the pulmonary (lung) volume because the lungs are attached to the thoracic wall through the pleural membranes.
  4. Fall in intra-pulmonary pressure. An increase in pulmonary volume decreases the intra-pulmonary pressure to less than the atmospheric pressure.
  5. Air enters the lungs. The pressure gradient forces air from outside to flow into the lungs through the air passages until the pressures are equal — this is inspiration (Fig. 14.2a).
Mechanism of breathing showing inspiration and expirationMechanism of breathing: (a) inspiration, (b) expiration (sagittal view of thorax)Air entering lungsRibs and sternumraisedRib cageDiaphragm contracted (flattened)Volume ofthorax increasedAir expelled from lungsRibs and sternumreturned tooriginal positionDiaphragm relaxed and arched upwardsVolume ofthorax decreased(a) Inspiration(b) Expiration© iBioKaare
Fig. — Mechanism of breathing showing inspiration and expiration (original iBioKaare schematic). Compare with NCERT textbook — Fig. 14.2, PDF page 6 (book page 186) ↗
Diaphragm and external intercostals contract→Thoracic volume increases→Pulmonary volume increases→Intra-pulmonary pressure falls→Air flows in
FeatureInspirationExpiration
DiaphragmContracts, flattensRelaxes, arches upwards
External intercostalsContract; ribs and sternum liftedRelax; ribs and sternum return
Thoracic volumeIncreasesDecreases
Intra-pulmonary pressureBelow atmosphericSlightly above atmospheric
Air movementInto lungsOut of lungs
NatureActive (muscular contraction)Normally passive

Extra depth. Under normal (quiet) conditions inspiration is an active process but expiration is mostly passive, resulting from relaxation of the same muscles. The strength of inspiration and expiration can be increased with the help of additional muscles in the abdomen. A healthy human breathes 12–16 times per minute, with a tidal volume of about 500 mL per breath.

Exam tip: Draw a labelled sketch of Fig. 14.2(a) with diaphragm contracted, ribs and sternum raised, volume of thorax increased, air entering lungs, and write the five steps in bullets — both diagram and flow chart are rewarded.
Exam tip: Mention both axes: diaphragm → antero-posterior; external intercostals → dorso-ventral; and the key cause: intra-pulmonary pressure < atmospheric pressure.

Conclusion: In normal inspiration the diaphragm and external intercostal muscles contract, enlarging the thoracic and hence pulmonary volume in both antero-posterior and dorso-ventral axes; the intra-pulmonary pressure falls below atmospheric pressure and air flows into the lungs.

Q7

How is respiration regulated?

Answer

Concept. Human beings can maintain and moderate the respiratory rhythm to suit the demands of body tissues — breathing is faster during exercise and slower at rest. This is achieved by the neural system, which receives information on blood chemistry and adjusts the respiratory muscles accordingly.

  1. Respiratory rhythm centre (medulla). A specialised centre in the medulla oblongata region of the brain is primarily responsible for regulating respiration. It generates the basic rhythm of breathing movements.
  2. Pneumotaxic centre (pons). A centre in the pons region of the brain can moderate the functions of the respiratory rhythm centre. Neural signals from this centre can reduce the duration of inspiration and thereby alter the respiratory rate.
  3. Chemosensitive area. A chemosensitive area situated adjacent to the rhythm centre is highly sensitive to CO2 and hydrogen ions. An increase in these substances activates this centre, which signals the rhythm centre to make the necessary adjustments (faster, deeper breathing) so that these substances are eliminated.
  4. Peripheral receptors. Receptors associated with the aortic arch and carotid artery also recognise changes in CO2 and H+ concentration and send signals to the rhythm centre for remedial action.
  5. Role of oxygen. The role of oxygen in the regulation of the respiratory rhythm is quite insignificant under normal conditions.
Centre / receptorLocationFunction
Respiratory rhythm centreMedullaPrimary regulation of rhythm
Pneumotaxic centrePonsModerates rhythm centre; shortens inspiration
Chemosensitive areaAdjacent to rhythm centre in medullaSensitive to CO2 and H+
Aortic and carotid receptorsAortic arch, carotid arteryDetect CO2 and H+ changes
Oxygen—Insignificant role
Rise in CO2 / H+ in blood→Chemosensitive area and aortic / carotid receptors stimulated→Signal to rhythm centre (medulla)→Rate and depth of breathing increase→CO2 and H+ eliminated

Extra depth. This is a negative-feedback loop: the product (CO2) of the process regulates the process itself. During exercise more CO2 is produced, the centres are stimulated, and ventilation rises to meet the increased demand.

Exam tip: Name the three brain areas with their regions: rhythm centre – medulla; pneumotaxic centre – pons; chemosensitive area – near rhythm centre, and state that CO2 and H+ (not O2) are the main stimuli.
Exam tip: A flow chart of the feedback (high CO2 → receptors → rhythm centre → faster breathing) is an easy way to show the complete mechanism.

Conclusion: Respiration is regulated by the neural system: the rhythm centre in the medulla, moderated by the pneumotaxic centre in the pons, adjusts breathing in response to CO2 and H+ detected by the chemosensitive area and aortic/carotid receptors, whereas oxygen plays an insignificant role.

Q8

What is the effect of pCO2 on oxygen transport?

Answer

Concept. About 97% of oxygen is carried by haemoglobin in the RBCs as oxyhaemoglobin. The binding of O2 to haemoglobin depends primarily on the partial pressure of O2, but NCERT states that partial pressure of CO2, hydrogen ion concentration and temperature are other factors that can interfere with this binding. The effect of pCO2 is best understood from the oxygen dissociation curve, the sigmoid plot of percentage saturation of haemoglobin against pO2.

Effect of pCO2. A high pCO2 reduces the affinity of haemoglobin for oxygen and favours the dissociation of oxygen from oxyhaemoglobin; a low pCO2 increases the affinity and favours the formation of oxyhaemoglobin. On the graph, a high pCO2 shifts the curve to the right (the same pO2 gives a lower percentage saturation), while a low pCO2 shifts it to the left. (This is known as the Bohr effect.)

Oxygen dissociation curveOxygen dissociation curve (percentage saturation of haemoglobin vs pO2)002020404060608080100100Partial pressure of oxygen, pO2 (mm Hg)Percentage saturation of haemoglobin with oxygenATA = alveoli (pO2 about 104 mm Hg): about 97% saturated T = tissues (pO2 about 40 mm Hg): about 75% saturatedNormal curve (sigmoid)Right shift: high pCO2, H+, temperatureLeft shift: low pCO2, H+, temperature© iBioKaare
Fig. — Oxygen dissociation curve (original iBioKaare schematic). Compare with NCERT textbook — Fig. 14.5, PDF page 9 (book page 189) ↗
SitepCO2Other factorsEffect on haemoglobin and O2
AlveoliLow (40 mm Hg)High pO2, lower H+, lower temperatureFavour formation of oxyhaemoglobin: O2 loading
TissuesHigh (45 mm Hg)Low pO2, higher H+, higher temperatureFavour dissociation of oxyhaemoglobin: O2 unloading

Physiological significance. Active tissues produce more CO2 (and acid and heat), so haemoglobin automatically releases more oxygen exactly where it is needed most. In the lungs, as CO2 is removed, haemoglobin picks up O2 more readily. Under normal conditions every 100 mL of oxygenated blood delivers about 5 mL of O2 to the tissues.

Extra depth — the reverse effect. The binding of CO2 to haemoglobin (carbamino-haemoglobin) is in turn influenced by pO2: low pO2 in the tissues favours CO2 binding and high pO2 in the alveoli favours its release. Thus O2 and CO2 transport are linked and help each other.

Exam tip: State clearly: high pCO2 → lower affinity → O2 released (curve shifts right); low pCO2 → higher affinity → O2 bound (curve shifts left), and give the alveoli vs tissues contrast.
Exam tip: Mention that the other interfering factors are H+ concentration and temperature — the NCERT line lists all three together.

Conclusion: A high pCO2 (as in tissues) lowers the affinity of haemoglobin for oxygen and promotes release of O2, whereas a low pCO2 (as in alveoli) promotes O2 binding, so oxygen is loaded in the lungs and delivered in the tissues.

Q9

What happens to the respiratory process in a man going up a hill?

Answer

Concept. Atmospheric air contains about 21% oxygen at every altitude, but the atmospheric pressure falls as altitude increases. Since partial pressure = percentage × total pressure, the pO2 of the air at high altitude is much lower than the 159 mm Hg at sea level (Table 14.1). The respiratory system of a man climbing a hill therefore faces a shortage of oxygen — a condition called hypoxia. (This question goes beyond the printed text of the chapter; the answer applies the NCERT principles of partial pressure, diffusion gradients and regulation of respiration.)

Immediate effects

  1. Lower alveolar and blood pO2. Lower atmospheric pO2 reduces the pO2 in alveolar air, so the diffusion gradient for O2 from alveoli to blood becomes less steep and less O2 diffuses into the blood.
  2. Lower saturation of haemoglobin. According to the oxygen dissociation curve, a lower pO2 means a lower percentage saturation of haemoglobin, so less oxygen is delivered to the tissues.
  3. Symptoms of hypoxia (mountain sickness). Breathlessness, headache, nausea, weakness, fatigue and in severe cases confusion may occur because brain and muscles get too little O2 for aerobic respiration.

Compensatory responses

  1. Increased rate and depth of breathing. The aortic and carotid receptors sense the changes and signal the respiratory rhythm centre, so ventilation increases to bring in more air. (Normally O2 has an insignificant role in regulation, but at low pO2 peripheral receptors do respond.)
  2. Increased heart rate to pump more blood and carry more oxygen to the tissues.
  3. Increased RBC production and haemoglobin (acclimatisation over days to weeks), improving the oxygen-carrying capacity of blood; capillary networks in tissues may also increase (general physiology knowledge).
  4. Adjustment of the dissociation curve: the factors that favour O2 release in tissues (higher H+, pCO2, temperature) help unload oxygen to the cells.
Change at altitudeResultCompensation
Atmospheric pressure fallspO2 of air fallsFaster, deeper breathing
Alveolar pO2 fallsLess O2 diffuses into bloodGreater heart rate
Hb saturation fallsHypoxia in tissuesMore RBCs and haemoglobin over time
Climbing: air pressure and pO2 fall→Alveolar and blood pO2 fall→Hypoxia→Faster breathing and heart rate→More RBCs (acclimatisation)
Exam tip: Write the chain: low atmospheric pressure → low pO2 → less haemoglobin saturation → hypoxia → faster breathing, higher heart rate, more RBCs. Also say that the percentage of O2 stays the same but the partial pressure falls.
Exam tip: Use the term hypoxia (linked to Q12) and mention the receptors on the aortic arch and carotid artery for extra marks.

Conclusion: On going up a hill the falling atmospheric pressure lowers the pO2 of inspired and alveolar air, causing hypoxia; the body compensates by faster breathing, increased heart rate and, on prolonged stay, an increased number of red blood cells.

Q10

What is the site of gaseous exchange in an insect?

Answer

Concept. Mechanisms of breathing vary among animal groups depending on habitat and level of organisation. NCERT notes that earthworms use their moist cuticle, aquatic arthropods and molluscs use gills, but insects have a network of tubes (tracheal tubes) to transport atmospheric air within the body.

Answer. In an insect the gaseous exchange takes place in the tracheal system, i.e. at the fine ends of the tracheal tubes (tracheoles) which reach the tissues and cells directly.

  1. Spiracles. The body surface has small openings called spiracles on the sides of the thorax and abdomen through which atmospheric air enters (and leaves).
  2. Tracheae. Each spiracle leads to a tube, the trachea, whose wall is kept open by chitinous thickenings, and the tracheae branch repeatedly through the body.
  3. Tracheoles. The finest branches, the tracheoles, are fluid-filled and extend right up to individual cells and tissues, so oxygen diffuses directly into cells and CO2 diffuses out.
  4. Role of blood. Because air is delivered right to the tissues, the insect’s blood (haemolymph) does not carry oxygen; the exchange is directly between tracheoles and cells (general zoology knowledge).
Atmospheric air→Spiracles→Tracheae→Tracheoles→Body cells (O2 in, CO2 out)
AnimalSite of gaseous exchange
Sponges, coelenterates, flatwormsEntire body surface (simple diffusion)
EarthwormMoist cuticle
InsectsTracheal tubes (tracheoles)
Aquatic arthropods, molluscs, fishesGills
FrogLungs and moist skin

Extra depth. The tracheal system is an extremely efficient arrangement for small active animals because diffusion through air is fast; it is one reason insects can sustain very rapid flight metabolism. Larger body movements of the abdomen help ventilate the tubes.

Exam tip: Answer in one phrase — “tracheal tubes (tracheal system) with tracheoles, opening by spiracles” — and add that air reaches the cells directly, without blood.
Exam tip: Remember the comparison list: sponges/flatworms – body surface; earthworm – moist cuticle; insects – tracheal tubes; aquatic arthropods/molluscs/fishes – gills; amphibians – lungs + skin.

Conclusion: Gaseous exchange in an insect occurs through the network of tracheal tubes, whose fine tracheoles carry atmospheric air from the spiracles directly to the tissues.

Q11

Define oxygen dissociation curve. Can you suggest any reason for its sigmoidal pattern?

Answer

Concept. Haemoglobin is a red, iron-containing pigment present in RBCs. O2 binds with haemoglobin in a reversible manner to form oxyhaemoglobin, and each haemoglobin molecule can carry a maximum of four molecules of O2. The extent of binding mainly depends on the partial pressure of O2.

Definition. The oxygen dissociation curve is the sigmoid curve obtained when the percentage saturation of haemoglobin with O2 is plotted against the partial pressure of oxygen (pO2). It is highly useful in studying the effect of factors such as pCO2, H+ concentration and temperature on the binding of O2 with haemoglobin.

Oxygen dissociation curveOxygen dissociation curve (percentage saturation of haemoglobin vs pO2)002020404060608080100100Partial pressure of oxygen, pO2 (mm Hg)Percentage saturation of haemoglobin with oxygenATA = alveoli (pO2 about 104 mm Hg): about 97% saturated T = tissues (pO2 about 40 mm Hg): about 75% saturatedNormal curve (sigmoid)Right shift: high pCO2, H+, temperatureLeft shift: low pCO2, H+, temperature© iBioKaare
Fig. — Oxygen dissociation curve (original iBioKaare schematic). Compare with NCERT textbook — Fig. 14.5, PDF page 9 (book page 189) ↗

Reason for the sigmoidal pattern. The S-shape arises because the binding of oxygen to haemoglobin is cooperative:

  1. A haemoglobin molecule has four oxygen-binding sites (each of the four polypeptide chains carries a haem group with iron), and so it can carry a maximum of four O2 molecules (NCERT).
  2. At low pO2 the first O2 molecule binds with difficulty (the molecule is in a low-affinity, tense state), so saturation rises slowly — the flat initial part of the curve.
  3. Once one O2 binds, it changes the shape of the haemoglobin molecule and increases the affinity of the remaining sites; the second, third and fourth O2 molecules bind progressively more easily, so saturation rises steeply over a small increase of pO2 — the steep middle part.
  4. As haemoglobin approaches full saturation, only a few vacant sites remain, so the curve flattens into a plateau at nearly 100% saturation.
Part of curvepO2 rangeExplanationPhysiological meaning
Initial slow riseVery lowFirst O2 binds with difficulty—
Steep middle partAbout 20–40 mm HgCooperative binding: affinity rises after each O2Small fall in pO2 in tissues releases much O2
PlateauAbout 90–100 mm HgAlmost all sites occupiedAlveolar blood nearly fully saturated

Extra depth. This shape is physiologically ideal: in the alveoli (pO2 about 104 mm Hg) haemoglobin is on the plateau, so it loads oxygen fully even if pO2 varies a little; in the tissues (pO2 about 40 mm Hg) it is on the steep part, so a small drop gives up a large amount of oxygen. If binding were not cooperative the curve would be a simple hyperbola with poor unloading in the tissues.

Exam tip: Define the curve with both axes (% saturation of haemoglobin vs pO2), draw the S-curve, and give the reason in one phrase: cooperative binding of four O2 molecules per haemoglobin.
Exam tip: Link the three parts of the curve to alveoli (plateau) and tissues (steep part) — this shows understanding beyond the definition.

Conclusion: The oxygen dissociation curve is the sigmoid plot of percentage saturation of haemoglobin against pO2; its S-shape results from the cooperative binding of up to four O2 molecules to one haemoglobin molecule, each bound O2 making the next easier to bind.

Q12

Have you heard about hypoxia? Try to gather information about it, and discuss with your friends.

Answer

Concept. Hypoxia is a condition in which the body or a region of the body is deprived of an adequate supply of oxygen at the tissue level. It is not mentioned in the printed chapter text, but it follows directly from the principles of gas exchange and transport described in Sections 14.3–14.4. (The information below is general physiology knowledge, compiled as the question asks.)

Types and causes of hypoxia

TypeCauseExample
Hypoxic hypoxiaLow pO2 in arterial bloodHigh altitude, lung diseases like emphysema or asthma, airway obstruction
Anaemic hypoxiaReduced oxygen-carrying capacity of bloodAnaemia, loss of blood, carbon monoxide poisoning (CO binds to Hb)
Stagnant (circulatory) hypoxiaInadequate blood flowHeart failure, shock, cold exposure
Histotoxic hypoxiaCells cannot use O2Cyanide poisoning (blocks cellular respiration)

Effects. Shortness of breath, rapid breathing and heart rate, headache, dizziness, bluish discolouration (cyanosis), fatigue, confusion and, if severe or prolonged, loss of consciousness and damage to the brain and heart, since these organs need a continuous supply of O2 for aerobic respiration.

Link with the chapter. Hypoxia reduces the pO2 gradient for diffusion (Section 14.3), reduces haemoglobin saturation (oxygen dissociation curve, Section 14.4), and stimulates the aortic and carotid receptors to increase breathing (Section 14.5). Chronic lung damage, such as emphysema from smoking or fibrosis from dust, causes hypoxia (Section 14.6).

Prevention and treatment. Treat the underlying cause; give supplemental oxygen; descend from high altitude and acclimatise gradually; stop smoking; wear protective masks in dusty industries; treat anaemia.

Cause (low pO2, anaemia, poor circulation, poisoning)→Less O2 reaches cells→Hypoxia→Compensation: faster breathing and heart rate→Treatment: oxygen and treating cause
Exam tip: Cover four points: definition, types/causes, symptoms, remedy; link one cause to the chapter (high altitude, emphysema, CO poisoning).
Exam tip: Explain that hypoxia is about tissue oxygen supply, not just breathing — four types (hypoxic, anaemic, stagnant, histotoxic) is a good scoring list.

Conclusion: Hypoxia is an inadequate supply of oxygen to the tissues, arising from low environmental pO2, lung disease, reduced haemoglobin function, poor circulation or cellular poisoning, and it is relieved by correcting the cause and providing oxygen.

Q13

Distinguish between (a) IRV and ERV (b) Inspiratory capacity and Expiratory capacity (c) Vital capacity and Total lung capacity

Answer

Concept. Respiratory volumes are the individual amounts of air handled in breathing movements; by adding a few of them we obtain pulmonary capacities. The terms are measured with a spirometer and used in clinical diagnosis. The values are NCERT’s; sums are derived from them.

(a) IRV and ERV

PointInspiratory reserve volume (IRV)Expiratory reserve volume (ERV)
MeaningAdditional air that a person can inspire by a forcible inspiration after a normal inspirationAdditional air that a person can expire by a forcible expiration after a normal expiration
Average value2500–3000 mL1000–1100 mL
Phase of breathingInspirationExpiration
Part of capacityPart of IC and VCPart of EC, FRC and VC

(b) Inspiratory capacity and Expiratory capacity

PointInspiratory capacity (IC)Expiratory capacity (EC)
DefinitionTotal volume of air a person can inspire after a normal expirationTotal volume of air a person can expire after a normal inspiration
ComponentsTV + IRVTV + ERV
Approximate value (derived)500 + (2500–3000) = 3000–3500 mL500 + (1000–1100) = 1500–1600 mL
Starting pointEnd of normal expirationEnd of normal inspiration

(c) Vital capacity and Total lung capacity

PointVital capacity (VC)Total lung capacity (TLC)
DefinitionMaximum air a person can breathe in after forced expiration (or out after forced inspiration)Total volume of air accommodated in the lungs at the end of a forced inspiration
ComponentsERV + TV + IRVRV + ERV + TV + IRV
Includes residual volume?NoYes (TLC = VC + RV)
Approximate value (derived)4000–4600 mL5100–5800 mL
Can be measured by spirometer directly?YesNo (RV cannot be expelled)

Extra depth. Remember the pattern: capacities are always sums of volumes. FRC = ERV + RV; IC = TV + IRV; EC = TV + ERV; VC = ERV + TV + IRV; TLC = VC + RV. Tidal volume (500 mL) is the central volume common to IC, EC and VC.

Exam tip: Write each definition and the formula in the table (IRV vs ERV; TV+IRV vs TV+ERV; ERV+TV+IRV vs VC+RV) — a three-part “distinguish” question needs all three tables to score fully.
Exam tip: The one-line giveaway: TLC = VC + residual volume; IC starts after normal expiration, EC after normal inspiration.

Conclusion: IRV (2500–3000 mL) is the extra air inspired and ERV (1000–1100 mL) the extra air expired by forcible effort; IC = TV + IRV while EC = TV + ERV; and VC = ERV + TV + IRV while TLC = VC + residual volume.

Q14

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

Answer

Concept. Tidal volume (TV) is the volume of air inspired or expired during a normal respiration. It is approximately 500 mL. It is one of the four respiratory volumes measured with a spirometer, and it is the volume that forms the common part of the inspiratory, expiratory and vital capacities.

Calculation of tidal volume of air breathed in one hour. NCERT states that a healthy man breathes 12–16 times per minute and that he can inspire or expire approximately 6000 to 8000 mL of air per minute.

  1. Tidal volume per breath = 500 mL.
  2. Breaths per minute = 12 to 16.
  3. Air per minute = 500 × 12 = 6000 mL (lower limit) to 500 × 16 = 8000 mL (upper limit).
  4. Time = 60 minutes.
  5. Air per hour = 6000 × 60 = 3,60,000 mL = 360 litres (lower limit) to 8000 × 60 = 4,80,000 mL = 480 litres (upper limit).

Volume per hour = TV × breaths per minute × 60 = 500 mL × (12–16) × 60 = 360–480 litres

Breaths/minAir per minute (mL)Air per hour (L)
126000360
14 (average)7000420
168000480

Extra depth. The air actually reaching the alveoli is less than this because part of each breath stays in the conducting passages (the dead space, about 150 mL: general physiology). Also, during exercise both the rate and the depth of breathing rise, so the volume per hour is much greater. Moreover, the tidal volume is only a part of the total lung capacity (5100–5800 mL, derived), which shows how much reserve the lungs have.

Exam tip: Show the working: 500 mL × 12–16 breaths/min × 60 min = 3,60,000 to 4,80,000 mL (360–480 L). Writing units and the range is essential for full marks.
Exam tip: Give the definition (“air inspired or expired in a normal respiration”) before the calculation, and quote NCERT’s “6000–8000 mL per minute” as a cross-check.

Conclusion: Tidal volume is the air inspired or expired in a normal breath, about 500 mL; at 12–16 breaths per minute a healthy human therefore breathes about 6000–8000 mL per minute, i.e. approximately 360–480 litres in an hour.

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