Question: 76: Four gas cylinders containing \mathrm{He}, \mathrm{N}_{2}, \mathrm{CO}_{2} and \mathrm{NH}_{3} gases separately are gradually cooled from a temperature of 500 K . Which gas will liquify first?
(Given \mathrm{T}_{\mathrm{C}} in \mathrm{K} – \mathrm{He}: 5.3, \mathrm{N}_{2}: 126, \mathrm{CO}_{2}: 304.1 and \mathrm{NH}_{3}: 405.5 )
(1) He
(2) \mathrm{N}_{2}
(3) \mathrm{CO}_{2}
(4) \mathrm{NH}_{3}
Answer: Option (4)
Explanation:
Liquefaction of a gas becomes possible when the temperature is brought below its critical temperature \mathrm{T_C}.
The higher the critical temperature, the easier it is to liquefy the gas on cooling.
All gases are initially at 500 K and are cooled gradually. The gas having the highest critical temperature will reach its liquefaction condition first.
The given critical temperatures are: \mathrm{He} = 5.3\,K,
\mathrm{N}_2 = 126\,K, \mathrm{CO}_2 = 304.1\,K,
and \mathrm{NH}_3 = 405.5\,K.
Among these gases, \mathrm{NH}_3 has the highest critical temperature.
Therefore, on cooling from 500 K, ammonia will liquefy first.
Hence, the correct answer is option (4).