JEE Main · 2019 · 10 Apr 2019 · Shift 1 · States of Matter

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Consider the following table:

\begin{tabular}{lll}
Gas & & \\
A & 642.32 & 0.05196 \\
B & 155.21 & 0.04136 \\
C & 431.91 & 0.05196 \\
D & 155.21 & 0.4382
\end{tabular}
and a van der Waals constants. The correct statement about the gases is:
  1. (A)Gas C will occupy more volume than gas A ; gas B will be lesser compressible than gas D
  2. (B)Gas C will occupy lesser volume than gas A ; gas B will be more compressible than gas D.
  3. (C)Gas C will occupy lesser volume than gas A ; gas B will be lesser compressible than gas D
  4. (D)Gas C will occupy more volume than gas A; gas B will be more compressible than gas D.✓ Correct

Answer: (D)

Solution

The van der Waals constant measures intermolecular attraction and the excluded volume. Gases A and C share the same but A has the larger ( against ), so A's molecules are held more tightly and C ends up occupying the greater volume. Gases B and D share the same but D has the far larger ( against ), so D has less free space and resists compression most, leaving B the more compressible of the two. So C occupies more volume than A, and B is more compressible than D.

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option

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