1. Henry’s law relation

   For a gas dissolved in water, $p = K_H x$ where:

   • $p$ = partial pressure of gas
   • $x$ = mole fraction of gas in solution
   • $K_H$ = Henry’s law constant

   Thus, if we plot $p$ vs $x$, the graph is a straight line through origin with slope equal to $K_H$.

2. Given Henry’s law constants

   The gases have: $K_H(A)=30.40 \text{ kbar}$ $K_H(B)=2.34 \text{ kbar}$ $K_H(C)=1.56\times 10^{-5} \text{ kbar}$ $K_H(D)=0.513 \text{ kbar}$

3. Compare slopes

   Since slope $= K_H$, larger $K_H$ means steeper line.

   Ordering of $K_H$ values: $30.40 > 2.34 > 0.513 > 1.56\times 10^{-5}$

   So slope order is: $A > B > D > C$

4. Identify lines in the graph

   In the usual $p$ vs $x$ graph:

   • the steepest line corresponds to the largest $K_H$ $o A$
   • the least steep line corresponds to the smallest $K_H$ $o C$

   Therefore, if line p is the steepest and line s is the flattest, then: $p = A, \quad s = C$

5. Match with options

   This corresponds to Option A.

Comparison with stored answer

Stored correct answer = A

My derived answer = A

So, they agree.