Compressibility Factor — Z, deviation from ideal
States of Matter
4
JEE Qs
8%
Hard
60
min
Master the Z vs P graphs for various gases (especially H₂, He, N₂, CO₂) at different temperatures and relate their shape directly to the dominance of attractive vs. repulsive forces and the impact of Boyle temperature.
🧮 Key Formulas
✅ Key Points for JEE
- 1Compressibility factor (Z) quantifies the deviation of a real gas from ideal behavior. Z=1 for an ideal gas under all conditions.
- 2If Z < 1, attractive forces dominate between gas molecules, making the gas more compressible than an ideal gas (P_real < P_ideal for same V, or V_real < V_ideal for same P). This occurs typically at low pressure and moderate temperature.
- 3If Z > 1, repulsive forces (finite volume of molecules) dominate, making the gas less compressible than an ideal gas (P_real > P_ideal for same V, or V_real > V_ideal for same P). This occurs typically at high pressure.
- 4Graphs of Z vs P show characteristic minima for most gases (except H₂ and He) at lower temperatures, indicating initial dominance of attractive forces, followed by an increase above 1 at higher pressures due to repulsive forces.
- 5At Boyle Temperature (T_B = a/Rb), a real gas behaves ideally (Z ≈ 1) over a wider range of pressures, as attractive and repulsive force deviations effectively cancel out.
⚠️ Common Mistakes
- ✕Confusing the conditions under which Z < 1 (attractive forces dominant) vs. Z > 1 (repulsive forces dominant) based on pressure and temperature.
- ✕Incorrectly interpreting Z vs P graphs, especially the significance of the minimum and the behavior of H₂/He.
- ✕Not applying the correct approximated Van der Waals equation for Z based on whether low or high pressure conditions are given in a problem.
📝 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