Gases possess characteristic critical temperature which depends upon the magnitude of intermolecular forces between the particles. Following are the critical temperatures of some gases.
Gases
Critical temperature in Kelvin
H2
33.2
He
5.3
O2
154.3
N2
126
From the above data what would be the order of liquefaction of these gases? Start writing the order from the gas liquefying first
(i) H2, He, O2, N2
(ii) He, O2, H2, N2
(iii) N2, O2, He, H2
(iv) O2, N2, H2, He
Gases possess characteristic critical temperature which depends upon the magnitude of intermolecular forces between the particles. Following are the critical temperatures of some gases.
|
Gases |
Critical temperature in Kelvin |
|
H2 |
33.2 |
|
He |
5.3 |
|
O2 |
154.3 |
|
N2 |
126 |
From the above data what would be the order of liquefaction of these gases? Start writing the order from the gas liquefying first
(i) H2, He, O2, N2
(ii) He, O2, H2, N2
(iii) N2, O2, He, H2
(iv) O2, N2, H2, He
This is a multiple choice answer as classified in NCERT Exemplar
Option (iv) O2, N2, H2, He
Higher the critical temperature of gases more easily, the gas will be liquefied. So, the order of liquefying of gases will be O2 > N2 > H2 >He.
Similar Questions for you
CH3COOH + NaOH → CH3COONa + H2O
ΔH = –50.6 kJ/mol
NaOH + SA [HCl] → NaCl + H2O
ΔH = –55.9 kJ/mol
the value of ΔH for ionisation of CH3COOH
⇒ ΔH = +55.9 – 50.6
5.3 kJ/mol
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Chemistry NCERT Exemplar Solutions Class 11th Chapter Five 2025
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