The problem presented explores the thermodynamics of gas adsorption on a solid surface at a given temperature. Several key concepts in thermodynamics come into play: enthalpy change \((\Delta H)\) and entropy change \((\Delta S)\).
Understanding Adsorption: Adsorption is the process where gas molecules adhere to the surface of solids, creating a layer of adsorbate on the adsorbent. This process is usually exothermic (releases heat), impacting enthalpy and entropy.
Enthalpy Change \((\Delta H)\):
During adsorption, gas molecules lose freedom as they bind to the surface.
This process typically releases energy; hence, \(\Delta H\) is negative, indicating an exothermic process.
Entropy Change \((\Delta S)\):
Entropy, a measure of disorder, usually decreases in adsorption. Initially, gas molecules are more free-moving in the gaseous phase.
When these molecules are adsorbed onto the solid surface, their freedom is significantly reduced, thus reducing entropy. Hence, \(\Delta S\) is negative.
The Correct Option Analysis:
The correct thermodynamic conditions for adsorption where both \(\Delta H\) and \(\Delta S\) are negative match the option: \(\Delta H<0, \Delta S<0\).
The other options do not fit this logical pattern for adsorption:
\(\Delta H>0, \Delta S>0\): Would imply an endothermic process and increasing disorder, which is not typical for adsorption.
\(\Delta H>0, \Delta S<0\): An endothermic process with reducing entropy is less feasible.
\(\Delta H<0, \Delta S>0\): An exothermic process with increasing disorder is not characteristic of adsorption.
Therefore, the correct choice for the condition of a gas adsorbing on a solid surface at a given temperature is: \(\Delta H<0, \Delta S<0\).
