Which of the following graph correctly represents the plots of \(K_H\) at 1 bar gases in water versus temperature?
Show Hint
The Correct Option is D
Approach Solution - 1
Approach Solution -2
The solution requires an understanding of Henry's Law and the factors that influence the Henry's Law constant (\( K_H \)).
- Henry's Law: This law relates the partial pressure (\( p \)) of a gas above a liquid to its mole fraction (\( x \)) in the liquid phase: \[ p = K_H \cdot x \] From this equation, the solubility of a gas (represented by \( x \)) is inversely proportional to \( K_H \) for a given partial pressure. A higher \( K_H \) value means lower solubility. \[ \text{Solubility} \propto \frac{1}{K_H} \]
- Effect of Temperature on \( K_H \): The dissolution of gases in liquids is typically an exothermic process. A simplified model suggests that increasing the temperature decreases solubility, which would mean \( K_H \) monotonically increases with temperature. However, the actual behavior for many nonpolar gases in water is more complex. The value of \( K_H \) first increases with temperature, reaches a maximum (corresponding to a minimum in solubility), and then begins to decrease at higher temperatures. This non-monotonic behavior is a more accurate representation.
- Dependence of \( K_H \) on the Nature of the Gas: The value of \( K_H \) is specific to the gas-solvent pair. For nonpolar gases in a polar solvent like water, the solubility is governed by London dispersion forces. Gases that are smaller and less polarizable interact more weakly with water, leading to lower solubility and a higher \( K_H \) value.
Step-by-Step Solution:
Step 1: Analyze the Temperature Dependence shown in the graphs.
The graphs show different trends for \( K_H \) versus temperature (\( t/^\circ\text{C} \)).
- Graphs (2) and (3) show a simple, monotonic increase of \( K_H \) with temperature. This represents the introductory model where gas solubility always decreases with increasing temperature.
- Graphs (1) and (4) show a more complex, non-monotonic relationship: \( K_H \) first increases, reaches a maximum value, and then decreases. This is a more accurate depiction of the behavior of many real gases (like He, N\(_2\), O\(_2\), CH\(_4\)) in water.
Given the options, we should consider the more accurate, non-monotonic trend as the correct representation of the physical phenomenon.
Step 2: Analyze the relative order of \( K_H \) values for the gases in option (4).
Option (4) plots \( K_H \) for Helium (He), Nitrogen (N\(_2\)), and Methane (CH\(_4\)). We need to determine the correct order of their \( K_H \) values.
- Helium (He): A very small noble gas with extremely low polarizability. It has very weak intermolecular interactions with water.
- Nitrogen (N\(_2\)): A nonpolar diatomic molecule, larger and more polarizable than He.
- Methane (CH\(_4\)): A nonpolar molecule, larger and more polarizable than N\(_2\).
Greater polarizability leads to stronger London dispersion forces with water molecules, resulting in higher solubility.
Therefore, the order of solubility is: \( \text{Solubility(CH}_4\text{)} > \text{Solubility(N}_2\text{)} > \text{Solubility(He)} \).
Since \( K_H \) is inversely proportional to solubility, the order of the Henry's Law constants must be the reverse:
\[ K_H(\text{He}) > K_H(\text{N}_2) > K_H(\text{CH}_4) \]
Step 3: Evaluate Graph (4) based on the analysis.
Let's check if Graph (4) is consistent with our findings.
- Shape of the Curves: The graph correctly shows the non-monotonic behavior where \( K_H \) first increases and then decreases with temperature.
- Relative Positions: At any given temperature on the graph, the curve for He is the highest, followed by N\(_2\), and then CH\(_4\) is the lowest. This corresponds to the order \( K_H(\text{He}) > K_H(\text{N}_2) > K_H(\text{CH}_4) \), which we determined to be correct.
Conclusion:
Graph (4) correctly represents both the sophisticated temperature dependence and the relative magnitudes of the Henry's Law constants for He, N\(_2\), and CH\(_4\) in water. The other graphs are incorrect because they either show a simplified temperature dependence (3) or an incorrect relative ordering of the \( K_H \) values (1, 2).
Therefore, the correct representation is given by option (4).
Top Questions on The Kinetic Theory of Gases
For an ideal gas, a cyclic process ABCA as shown in the P–T diagram. When represented in P–V plot, it would be
- WBJEE - 2026
- Physics
- The Kinetic Theory of Gases
- Let \( \bar{v}, v_{rms}, v_p \) denote the mean speed, root mean square speed and most probable speed of molecules of mass \( m \) in an ideal monoatomic gas at temperature \( T \). Which statements are correct?
- WBJEE - 2026
- Physics
- The Kinetic Theory of Gases
- Consider two boxes containing ideal gases $A$ and $B$ such that their temperatures, pressures and number densities are same. The molecular size of $A$ is half of that of $B$ and mass of molecule $A$ is four times that of $B$. If the collision frequency in gas $B$ is $32\times10^8\ \text{s}^{-1}$, then collision frequency in gas $A$ is ___________\,$\text{s}^{-1}$.
- JEE Main - 2026
- Physics
- The Kinetic Theory of Gases
- Consider the transition metal ions \( \text{Mn}^{3+}, \text{Cr}^{3+}, \text{Fe}^{3+} \) and \( \text{Co}^{3+} \) and all form low spin octahedral complexes. The correct decreasing order of unpaired electrons in their respective \(d\)-orbitals of the complexes is
- JEE Main - 2026
- Chemistry
- The Kinetic Theory of Gases
- The charge stored by the capacitor C in the given circuit in the steady state is _________ $\mu$C.
- JEE Main - 2026
- Physics
- The Kinetic Theory of Gases
Questions Asked in JEE Main exam
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.- JEE Main - 2026
- Solutions
- The wavelength of photon 'A' is 400 nm. The frequency of photon 'B' is \(10^{16}\,\text{s}^{-1}\). The wave number of photon 'C' is \(10^{5}\,\text{cm}^{-1}\). The correct order of energy of these photons is:
- JEE Main - 2026
- General Chemistry
- Given below are two statements: Statement I: The increasing order of boiling point of hydrogen halides is HCl<HBr<HI<HF. Statement II: The increasing order of melting point of hydrogen halides is HCl<HBr<HF<HI. In the light of the above statements, choose the correct answer from the options given below:
- JEE Main - 2026
- General Chemistry
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]- JEE Main - 2026
- Ray optics and optical instruments
A small block of mass \(m\) slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration \(a_0\). The angle between the inclined plane and ground is \(\theta\) and its base length is \(L\). Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is _______.
- JEE Main - 2026
- General Physics