Question

Observe the following Z vs P graph. The missing gas in the above graph can be :

• A. He
• B. Ar
• C. C₅H₁₂
• D. All are correct

Answer 1

Answer: (C)

C₅H₁₂

Answer 2

The compressibility factor ($Z$) versus pressure ($P$) graph illustrates the deviation of real gases from ideal behavior. The deviation below $Z=1$ is primarily due to attractive intermolecular forces. A deeper dip in the $Z$ vs $P$ curve indicates stronger attractive forces. The graph shows the curves for H₂, Ne, NH₃, and a missing gas represented by '?'. The curves are ordered from top to bottom, implying an increase in the strength of attractive forces: H₂ < Ne < NH₃ < ?. We need to identify a gas from the options that has stronger attractive forces than NH₃.

Comparing the options:

• He: Has very weak intermolecular forces, weaker than H₂.
• Ar: Has stronger dispersion forces than Ne due to larger size, but generally weaker than the combined forces in NH₃ (which include dipole-dipole interactions and hydrogen bonding).
• C₅H₁₂ (Pentane): A larger hydrocarbon molecule with significant London dispersion forces, which are stronger than the attractive forces in NH₃.

Therefore, C₅H₁₂ would exhibit a deeper dip below $Z=1$ than NH₃, fitting the position of '?'.