Critical Constants — Tc, Pc, Vc
States of Matter
4
JEE Qs
8%
Hard
75
min
Master the derivation of critical constants from the Van der Waals equation using calculus, and understand the physical significance of each constant, especially Tc, as these are frequently tested.
🧮 Key Formulas
✅ Key Points for JEE
- 1Critical temperature (Tc) is the maximum temperature above which a gas cannot be liquefied, regardless of the applied pressure.
- 2The critical point (Tc, Pc, Vc) corresponds to the point of inflection on the P-V isotherm for a real gas, mathematically defined by (∂P/∂V)T = 0 and (∂²P/∂V²)T = 0.
- 3Gases with higher 'a' values (stronger intermolecular attractive forces) have higher critical temperatures and are consequently easier to liquefy.
- 4The compressibility factor at the critical point, Zc = PcVc/(RTc), is approximately 3/8 for ideal Van der Waals gases and provides insight into the deviation from ideal behavior at critical conditions.
- 5The Van der Waals constants 'a' and 'b' can be directly calculated from critical constants, establishing a link between microscopic molecular properties and macroscopic critical phenomena.
⚠️ Common Mistakes
- ✕Errors in algebraic manipulation or differentiation when deriving critical constants (Tc, Pc, Vc) from the Van der Waals equation.
- ✕Incorrectly using or interchanging the formulas for 'a' and 'b' in terms of critical constants, leading to calculation errors.
- ✕Confusing the physical significance of 'a' (intermolecular forces) and 'b' (effective molecular volume) and their impact on liquefaction with their numerical values.
- ✕Forgetting to use consistent units for the gas constant 'R' (e.g., L atm mol⁻¹ K⁻¹ vs. J mol⁻¹ K⁻¹) when performing calculations involving critical constants.
📝 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