Low concentration of oxygen in the blood and tissues of people living at high altitude is due to ______. (i) l

NCERT Exemplar · Q6

Q. Low concentration of oxygen in the blood and tissues of people living at high altitude is due to ____________.

(i) low temperature

(ii) low atmospheric pressure

(iii) high atmospheric pressure

(iv) both low temperature and high atmospheric pressure

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Concept understanding — Henrys Law

Henry's Law: The Physics of "Fizz"

Imagine you open a cold bottle of soda. You hear that familiar psshhht sound. Bubbles rush out. Now think: why were those bubbles inside the bottle in the first place? The liquid wasn't boiling. The answer is Henry's Law.

The Intuition: Gas Wants to Dissolve

Gases are just molecules flying around. When a gas touches a liquid, some of those molecules get "trapped" inside the liquid — they dissolve. But here's the key: the more you push on the gas, the more of it gets forced into the liquid.

Think of a crowded bus. If you push more people toward the door (higher pressure), more people get squeezed inside. If you let the pressure off (open the bottle), people rush out. That's exactly what happens with gas and liquid.

In the soda bottle, carbon dioxide gas is pumped in at high pressure. That pressure forces a huge amount of CO₂ to dissolve into the liquid. When you open the bottle, the pressure above the liquid drops to normal air pressure. Suddenly, the liquid can't hold all that CO₂ anymore — so it escapes as bubbles. That's the fizz.

The Precise Statement

Henry's Law says:

C = k_H \cdot P

Where:

$C$ = concentration of the dissolved gas in the liquid (usually mol/L or g/L)
$P$ = partial pressure of that gas above the liquid (usually atm or kPa)
$k_H$ = Henry's law constant — a number that depends on the specific gas, the liquid, and the temperature

In words: At a constant temperature, the amount of gas that dissolves in a liquid is directly proportional to the partial pressure of that gas above the liquid.

What the Constant $k_H$ Tells You

$k_H$ is not universal. It's different for every gas-liquid pair. For example:

CO₂ in water has a certain $k_H$
O₂ in water has a different $k_H$ (smaller — oxygen doesn't dissolve as easily)

Temperature matters too. Higher temperature means lower $k_H$ — gases become less soluble in hot liquids. That's why a warm soda goes flat faster than a cold one.

Watch out

Henry's Law works only for dilute solutions and non-reacting gases. If the gas reacts chemically with the liquid (like HCl gas dissolving in water to form hydrochloric acid), Henry's Law does not apply — the concentration will be much higher than predicted.

Real-Life Examples

Situation	What Henry's Law explains
Soda fizz	High pressure forces CO₂ in; releasing pressure lets it out

Why this formula?

Henry's Law: Why the Formula Holds

Henry's Law describes the solubility of a gas in a liquid at a constant temperature. The key formula is:

P = k_H \cdot x

Where:

$P$ = partial pressure of the gas above the liquid
$x$ = mole fraction of the gas dissolved in the liquid
$k_H$ = Henry's constant (depends on gas, liquid, and temperature)

Why This Linear Relationship Exists

1. Dynamic Equilibrium at the Interface

Imagine a gas above a liquid. At the molecular level:

Gas molecules constantly strike the liquid surface and dissolve
Dissolved molecules constantly escape back into the gas phase

At equilibrium, the rate of dissolution equals the rate of escape. This is a dynamic balance, not a static one.

2. The Driving Force for Dissolution

The rate at which gas molecules enter the liquid depends on:

How many gas molecules hit the surface — this is proportional to the partial pressure $P$ of the gas
How easily they dissolve — this is captured by $k_H$

So:

\text{Rate}_{\text{dissolve}} \propto P

3. The Driving Force for Escape

The rate at which dissolved molecules leave the liquid depends on:

How many dissolved molecules are near the surface — this is proportional to the mole fraction $x$ of the gas in the liquid
How easily they escape — also captured by $k_H$

So:

\text{Rate}_{\text{escape}} \propto x

4. Equating the Two Rates

At equilibrium:

\text{Rate}_{\text{dissolve}} = \text{Rate}_{\text{escape}}

Therefore:

P \propto x

Introducing the proportionality constant $k_H$ :

P = k_H \cdot x

Why It's Linear (Not Exponential or Logarithmic)

The linearity arises because:

No saturation effects at low concentrations — the molecules don't "crowd" each other
Ideal behavior is assumed — gas molecules don't interact strongly with each other or with the solvent
Temperature is constant — $k_H$ doesn't change …

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At high altitude, atmospheric pressure drops, reducing the partial pressure of oxygen and thus its solubility in blood according to Henry's Law. The answer is (ii) low atmospheric pressure.

Why altitude affects oxygen availability

Henry's Law: the solubility of a gas in a liquid is proportional to its partial pressure. When you climb a mountain, total atmospheric pressure falls - and with it, the partial pressure of oxygen in the air you breathe. Even though oxygen still makes up about 21% of the atmosphere at any altitude, the absolute partial pressure of oxygen is lower, so less oxygen dissolves into blood plasma and binds to hemoglobin - hypoxia.

C = k_H \cdot p_{\text{gas}}

Step-by-step reasoning

Atmospheric pressure decreases with altitude (roughly 1 atm at sea level, ~0.33 atm at Everest's summit).
Partial pressure of oxygen falls proportionally: $p_{O_2}=0.21\times p_{atm}$ .
Henry's Law governs oxygen dissolution in blood - lower $p_{O_2}$ means lower dissolved oxygen. …

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Here is the breakdown of the common mistakes students make on this question, along with the correct reasoning.

The Correct Answer

The correct answer is (ii) low atmospheric pressure.

The Core Concept: Partial Pressure of Oxygen

The key to this question is understanding Henry's Law and Partial Pressure.

Henry's Law: The amount of gas dissolved in a liquid (like blood) is directly proportional to the partial pressure of that gas in the atmosphere above the liquid.
Partial Pressure: The pressure exerted by a single gas (like oxygen) in a mixture of gases (like air).

At high altitude, the total atmospheric pressure drops. Since oxygen makes up a constant ~21% of the air, the partial pressure of oxygen also drops significantly. This lower partial pressure means less oxygen is forced into the blood and tissues.

Common Mistakes & How to Avoid Them

Mistake 1: Choosing "Low Temperature" (Option A)

Why students choose this:

Students often think, "It's cold on a mountain, so cold air must have less oxygen." They confuse the density of air with the concentration of oxygen.

The Reality:

Temperature affects the density of air, but the percentage of oxygen in the air remains ~21%.
More importantly, cold temperature actually increases the solubility of gases in liquids. If temperature were the only factor, cold blood would hold more oxygen, not less.
The primary reason for low oxygen in the blood is the low partial pressure, not the cold.

How to Avoid:

Focus on the "why": Ask yourself: "What is the driving force that pushes oxygen into the blood?" The answer is pressure (specifically, the pressure gradient). Temperature is a secondary factor that affects solubility, not the primary driving force.
Remember the context: The question asks about high altitude. The defining characteristic of high altitude is low atmospheric pressure, not just low temperature.

Mistake 2: Choosing "Both Low Temperature and High Atmospheric Pressure" (Option D)

Why students choose this:

This is a trap for students who are guessing or who haven't read the options carefully. They might think "high altitude = cold" and then incorrectly assume pressure must also be high.

The Reality:

Atmospheric pressure decreases with altitude. It is never "high" at high altitude. This option contains a direct factual error.
The combination of "low temperature" and "high pressure" is actually the condition for maximum oxygen solubility, which is the opposite of what happens at high altitude.

How to Avoid:

Know the basic relationship: Altitude $\uparrow$ → Atmospheric Pressure $\downarrow$ .
Eliminate contradictions: As soon as you see "high atmospheric pressure" in an option about high altitude, you can immediately discard it. It violates a fundamental fact.

Mistake 3: Confusing "Atmospheric Pressure" with "Partial Pressure" …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
$K_H$
›Reveal solutionSolution
Applying Henry's law to find methane's mole fraction, then converting to moles dissolved in 1 L of water, gives about $1.38\times10^{-1}$ mol — option (D).

Concept and Intuition

Henry's law states that the partial pressure of a gas above a solution is proportional to its mole fraction in the solution: $p = K_H x$ . A large $K_H$ (here 400 bar) means the gas is not very soluble, so only a small mole fraction dissolves under a given pressure. Once we know the mole fraction, and knowing that water vastly outnumbers the dissolved gas molecules, we can approximate moles of water as the total moles in solution to extract the actual moles of dissolved methane.

Step-by-Step Solution

Henry's law: $p = K_H x \Rightarrow x = \dfrac{p}{K_H} = \dfrac{1\ \text{bar}}{0.4\ \text{kbar}} = \dfrac{1}{400} = 2.5\times10^{-3}$ .

Moles of water in 1 L (mass ≈ 1000 g, taking density ≈ 1 g/mL): $n_{water} = \dfrac{1000}{18} = 55.56$ mol. …
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AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
$\mathrm{N_2}$
›Reveal solutionSolution
Henry's law constant increases as solubility decreases; helium, being the least soluble of the four gases in water, has the highest $K_H$ — option (C).

Concept and Intuition

Henry's law states $p = K_H \cdot x$ , where $p$ is the partial pressure of the gas and $x$ is its mole fraction dissolved in the liquid. A larger $K_H$ means a smaller mole fraction dissolves for the same partial pressure — i.e., higher $K_H$ = lower solubility. Noble gases like helium have very weak intermolecular (van der Waals) interactions with water and are notoriously poorly soluble, which is why divers use helium-based gas mixtures (to reduce the amount of gas — like nitrogen — that dissolves in blood and causes decompression sickness).

Step-by-Step Solution

Recall the qualitative solubility order in water for these gases: $\mathrm{O_2}$ is somewhat more soluble than $\mathrm{N_2}$ , and both are more soluble than the very inert, small, monoatomic He.

Since $K_H \propto \dfrac{1}{\text{solubility}}$ , the least soluble gas has the largest $K_H$ . …
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AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
$\mathrm{N_2}$
›Reveal solutionSolution
Henry's law with equal partial pressures gives a mole-fraction ratio inversely proportional to the Henry's law constants, $\approx0.45$ .

Concept and Intuition

Henry's law states that the partial pressure of a gas above a liquid is directly proportional to its mole fraction dissolved in the liquid: $p=K_H\,x$ . A larger $K_H$ means the gas is less soluble (needs a higher partial pressure to dissolve the same mole fraction), so for two gases at the same partial pressure, the one with the larger $K_H$ ends up with the smaller mole fraction dissolved.

Step-by-Step Solution

Write Henry's law for each gas: $p_{N_2}=K_{H,N_2}\,x_{N_2}$ and $p_{O_2}=K_{H,O_2}\,x_{O_2}$ .

Given $p_{N_2}=p_{O_2}$ (equal partial pressures), divide the two equations: $\dfrac{x_{N_2}}{x_{O_2}}=\dfrac{K_{H,O_2}}{K_{H,N_2}}$ . …
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AP EAPCET 2022Set ap-2022-07-11-FN1 markMCQ
$4.27\times10^5$
›Reveal solutionSolution
This is a direct application of Henry's law relating partial pressure of a gas above a liquid to its mole fraction dissolved in it.

Concept and Intuition

Henry's law states $p = K_H\,x$ , where $p$ is the partial pressure of the gas, $x$ is its mole fraction in solution, and $K_H$ is the Henry's law constant (here expressed in mm Hg, so the pressure must also be converted to mm Hg for consistent units).

Step-by-Step Solution

Convert pressure to mm Hg: $2\text{ atm} = 2\times760 = 1520$ mm Hg.

Apply Henry's law: $x = \dfrac{p}{K_H} = \dfrac{1520}{4.27\times10^5}$ . …
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AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
$\mathrm{CO_2}$
›Reveal solutionSolution
Using Henry's law $p = K_H x$ to get the mole fraction of dissolved $\mathrm{CO_2}$ , then converting mole fraction to molarity via the moles of water in 200 mL, gives $\approx 1.11\times10^{-1}\ \mathrm{mol\,L^{-1}}$ .

Concept and Intuition

Henry's law states that the partial pressure of a gas above a solution is proportional to its mole fraction in the solution: $p = K_H\,x_{gas}$ . This lets us find the mole fraction of dissolved gas directly from the applied pressure and the Henry's law constant. Since the mole fraction of a dilute solute is tiny, we can approximate the total moles of solution as just the moles of solvent (water), and convert mole fraction to concentration using the known amount of solvent present.

Step-by-Step Solution

Apply Henry's law: $x_{CO_2} = \dfrac{p}{K_H} = \dfrac{3.4}{1.7\times10^3} = 2\times10^{-3}$ .

Find moles of water in 200 mL of soda water (density $\approx 1\ \mathrm{g/mL}$ , so mass $= 200\ \mathrm{g}$ ): $n_{H_2O} = \dfrac{200}{18} = 11.11\ \mathrm{mol}$ . …
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AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
$K_H$
›Reveal solutionSolution
This is a direct Henry's law + unit-conversion problem: convert the given partial pressure into a mole fraction, then into ppm (mg solute per kg solvent). The answer is 35.50 ppm.

Concept and Intuition

Henry's law states that for a gas dissolved in a liquid at a given temperature, the partial pressure of the gas above the solution is proportional to the mole fraction of the gas dissolved in the liquid: $p = K_H\,x$ . A large $K_H$ (like 50 kbar here) means the gas is poorly soluble — it takes a huge partial pressure to force even a tiny mole fraction into solution, which is exactly the situation for $O_2$ in water. Once we know the mole fraction, we can convert it to ppm because for very dilute solutions the mole fraction is essentially the ratio of moles of solute to (moles of solute + moles of solvent) ≈ moles of solute / moles of solvent.

Step-by-Step Solution

Apply Henry's law: $x_{O_2} = \dfrac{p}{K_H} = \dfrac{1\ \text{bar}}{50000\ \text{bar}} = 2\times10^{-5}$ .

Since the solution is dilute, $x_{O_2} \approx \dfrac{n(O_2)}{n(H_2O)}$ .

In 1 L of water, mass $\approx 1000$ g, so $n(H_2O) = \dfrac{1000}{18} = 55.56$ mol.

$n(O_2) = x_{O_2}\times n(H_2O) = 2\times10^{-5}\times 55.56 = 1.111\times10^{-3}$ mol. …
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AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
$K_H$
›Reveal solutionSolution
Since solubility varies inversely with Henry's law constant, ranking the given KH values from largest to smallest and reversing gives the increasing-solubility order Ar<CO2<CH4<HCHO. Answer (D).

Concept and Intuition

Henry's law states $p=K_H x$ , where $p$ is the gas's partial pressure above the solution and $x$ is its dissolved mole fraction. For a FIXED partial pressure, a larger $K_H$ forces a SMALLER equilibrium mole fraction $x=p/K_H$ — meaning the gas is less soluble. So solubility and $K_H$ are inversely related.

Step-by-Step Solution

List the given $K_H$ values: $Ar=40.39$ , $CO_2=1.67$ , $HCHO=1.83\times10^{-5}$ , $CH_4=0.413$ .

Sort from largest $K_H$ (least soluble) to smallest $K_H$ (most soluble): $Ar(40.39)>CO_2(1.67)>CH_4(0.413)>HCHO(1.83\times10^{-5})$ . …
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AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
$Q.A gas X is dissolved in water at 2 bar pressure. Its mole fraction in the solution is 0.02. Find the mole fraction of water in the solution when the pressure of the gas is doubled at the same temperature. (A) 0.04 (B) 0.98 (C) 0.96 (D) 0.02$
›Reveal solutionSolution
This tests Henry's law ( $p = K_H x$ ): doubling the gas pressure doubles the gas's mole fraction in solution, so the water's mole fraction drops from $0.98$ to $0.96$ .

Concept and Intuition

Henry's law says the partial pressure of a gas above a dilute solution is directly proportional to its mole fraction in the solution: $p = K_H \cdot x_{\text{gas}}$ , with $K_H$ constant at a given temperature. So if pressure doubles, the mole fraction of dissolved gas also doubles (as long as $K_H$ doesn't change, i.e., temperature is constant, as stated). The mole fraction of water then adjusts to keep the two mole fractions summing to 1 (a binary solution of gas + water).

Step-by-Step Solution

Given: at $p_1 = 2$ bar, mole fraction of gas $x_1 = 0.02$ .

Apply Henry's law to find $K_H$ : $K_H = \dfrac{p_1}{x_1} = \dfrac{2}{0.02} = 100$ bar.

At the same temperature, $K_H$ is unchanged. New pressure $p_2 = 2 \times 2 = 4$ bar.

New mole fraction of gas: $x_2 = \dfrac{p_2}{K_H} = \dfrac{4}{100} = 0.04$ .

Since gas + water make up the whole solution, mole fraction of water $= 1 - x_2 = 1 - 0.04 = 0.96$ . …
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AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
$CO_2$
›Reveal solutionSolution
Apply Henry's law to get the mole fraction of dissolved CO₂, then convert to mass using the moles of water present in 500 mL. Answer: 3.7 g.

Concept and Intuition

Henry's law states that the partial pressure of a dissolved gas is proportional to its mole fraction in solution: $p = K_H \cdot x$ . Once the (small) mole fraction of dissolved gas is known, it can be converted to moles using the (much larger) known amount of solvent, since $x \approx n_{gas}/n_{solvent}$ when the gas is dilute.

Step-by-Step Solution

Convert pressure: $P = 5\ \text{atm} = 5\times101325\ \text{Pa} = 506{,}625$ Pa.

Henry's law: $x_{CO_2} = \dfrac{P}{K_H} = \dfrac{506{,}625}{1.67\times10^8} = 3.034\times10^{-3}$ .

Moles of water in 500 mL (density $\approx 1$ g/mL): $n_{water} = \dfrac{500}{18} = 27.78$ mol. …
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Q.Low concentration of oxygen in the blood and tissues of people living at high altitude is due to ____________.

Why altitude affects oxygen availability

Step-by-step reasoning

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