Liquefaction of Gases — Joule-Thomson effect
States of Matter
4
JEE Qs
8%
Hard
75
min
Master the concept of inversion temperature and the conditions (T < T_i) under which real gases show cooling in the Joule-Thomson effect, as this is a frequent point of testing.
🧮 Key Formulas
✅ Key Points for JEE
- 1The Joule-Thomson effect describes the temperature change of a real gas when it expands isenthalpically (ΔH=0) from a high pressure to a low pressure region through a porous plug.
- 2For a gas to cool upon expansion (μ_JT > 0), its initial temperature must be below its inversion temperature (T < T_i). Above T_i, the gas heats up upon expansion.
- 3The inversion temperature (T_i = 2a / (Rb) for a van der Waals gas) is a specific temperature where μ_JT = 0; it depends on the van der Waals constants 'a' (intermolecular attraction) and 'b' (finite molecular volume).
- 4Ideal gases do not exhibit the Joule-Thomson effect (μ_JT = 0) because they have no intermolecular forces, hence no potential energy change upon expansion.
- 5Liquefaction processes like Linde's and Claude's cycles utilize the Joule-Thomson effect for cooling, often involving compression, pre-cooling, and repetitive expansion to reach liquefaction temperatures.
⚠️ Common Mistakes
- ✕Confusing the Joule-Thomson effect (isenthalpic expansion) with adiabatic or isothermal expansion, leading to incorrect predictions of temperature change.
- ✕Assuming all gases cool upon expansion, without considering the critical role of the inversion temperature; gases expand and heat up if their initial temperature is above T_i.
- ✕Not understanding that the attractive intermolecular forces (represented by 'a' in van der Waals equation) are crucial for cooling in the J-T effect, as work is done against these forces, reducing internal energy and thus temperature.
- ✕Incorrectly equating critical temperature (T_c) with inversion temperature (T_i); T_i is generally much higher than T_c for most gases.
📝 Practice Questions
See allQ49.A container of fixed volume contains a gas at 27∘C. To double the pressure of the gas, the temperature of gas should be raised to ______ ∘C.
Q42.The ratio of vapour densities of two gases at the same temperature is 4 , then the ratio of r.m.s. velocities will 25 be: (1) 25 (2) 2 4 5 (3) 5 (4) 4 2 25
Q74.Some CO2 gas was kept in a sealed container at a pressure of 1 atm and at 273 K . This entire amount of CO2 gas was later passed through an aqueous solution of Ca(OH)2 . The excess unreacted Ca(OH)2 was later neutralized with 0.1 M of 40 mL HCl . If the volume of the sealed container of CO2 was x, then x is ________ cm3 (nearest integer). [Given : The entire amount of CO2( g) reacted with exactly half the initial amount of Ca(OH)2 present in the aqueous solution.] 2025 (22 Jan Shift 1) JEE Main Previous Year Paper
Q36.At 600 K, the root mean square (rms) speed of gas X (molar mass = 40) is equal to the most probable speed of gas Y at 90 K. The molar mass of the gas Y is _____ g mol–1 . (Nearest integer)
Q53.At constant temperature, a gas is at a pressure of 940 . 3 mm Hg. The pressure at which its volume decreases by 40% is ______ mm Hg. (Nearest integer) +2 . 2 V +0 . 70 V -0 . 45 V 2 - ⟶ Fe3 + ⟶ Fe2 + ⟶ Fe0
Q36. Three bulbs are filled with CH4, CO2 and Ne as shown in the picture. The bulbs are connected through pipes of zero volume. When the stopcocks are opened and the temperature is kept constant throughout, the pressure of the system is found to be______ atm. (Nearest integer).
NCERT Chapters
- Class 11 Chemistry Ch 5: States of Matter
- Class 11 Chemistry Ch 6: Thermodynamics