NEET States of Matter PYQs & Practice Questions

10 free NEET Chemistry questions on States of Matter, starting with previous year questions (2018–2023). Tap “Show answer” to check each one. No login needed.

Topics: Intermolecular Forces · Gas Laws – Boyle's Law · Ideal Gas Equation · Kinetic Molecular Theory · Graham's Law of Diffusion · Van der Waals Equation · Critical Temperature and Liquefaction · Compressibility Factor · Liquids – Surface Tension and Viscosity · Pressure and Altitude · Intermolecular Forces - London Forces · Intermolecular Forces - Dipole-Dipole

Q1NEET 2023Intermolecular Forces
Intermolecular forces are forces of attraction and repulsion between interacting particles that will include: A. dipole-dipole forces. B. dipole-induced dipole forces. C. hydrogen bonding. D. covalent bonding. E. dispersion forces. Choose the most appropriate answer from the options given below.
- A. A, B, C, D are correct.
- B. A, B, C, E are correct.
- C. A, C, D, E are correct.
- D. B, C, D, E are correct.
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Answer: A — A, B, C, D are correct.
Intermolecular forces act between molecules and include dipole-dipole forces, dipole-induced dipole forces, hydrogen bonding, and dispersion (London) forces. Covalent bonding is an intramolecular force (atoms combining within a molecule), not an intermolecular force, so D is excluded. Hence A, B, C, E are correct.
Q2NEET 2023Boyle's Law Graphs
Which amongst the following options is the correct graphical representation of Boyle's Law? Options show P vs 1/V graphs (straight lines through origin, or P vs V hyperbolic curves) at three temperatures T1, T2, T3, with different orderings of which line/curve corresponds to which temperature.
- A. P vs 1/V: straight lines through the origin with slopes such that T3 > T2 > T1 (steeper slope for higher temperature)
- B. P vs 1/V: hyperbolic curves with T3 > T2 > T1 ordering as shown
- C. P vs T: straight lines through origin with V1 < V2 < V1 (inconsistent labelling)
- D. P vs V: hyperbolic curves with T3 > T2 > T1 ordering as shown
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Answer: —
Boyle's Law states P is proportional to 1/V (P = K/V) at constant temperature, so a plot of P vs 1/V is a straight line through the origin, with slope K = nRT. Since K increases with temperature, the line for a higher temperature has a greater slope. The correct graph is therefore the straight-line P vs 1/V plot with T3 > T2 > T1 (steepest slope for highest temperature).
Q3NEET 2022Dalton's Law of Partial Pressure
Which one is not correct mathematical equation for Dalton's Law of partial pressure? Here p=total pressure of gaseous mixture.
- A. p = n1(RT/V) + n2(RT/V) + n3(RT/V)
- B. pi = xi*p, where pi = partial pressure of ith gas, xi = mole fraction of ith gas in gaseous mixture
- C. pi = xi*p1°, where xi = mole fraction of ith gas in gaseous mixture, p1° = mole fraction of ith gas in gaseous mixture
- D. p = p1 + p2 + p3
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Answer: B — pi = xi*p, where pi = partial pressure of ith gas, xi = mole fraction of ith gas in gaseous mixture
Option (3) is not a correct form of Dalton's Law - the equation pi = xi*p1° does not represent Dalton's law of partial pressure correctly (the terms as defined don't form a valid relationship). Options 1, 2, and 4 are all valid forms of Dalton's law.
Q4NEET 2022Ideal Gas Equation
A 10.0 L flask contains 64 g of oxygen at 27°C. (Assume O2 gas is behaving ideally). The pressure inside the flask in bar is: (Given: R = 0.0831 L bar K-1 mol-1)
- A. 498.6
- B. 49.8
- C. 4.9
- D. 2.5
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Answer: B — 49.8
Using PV=nRT: moles of O2 = 64/32 = 2 mol. P x 10 = 2 x 0.083 x 300. P = 49.8/10 = 4.98 ≈ 4.9 bar.

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Q5NEET 2022Dalton's Law of Partial Pressure
Which one is not correct mathematical equation for Dalton's Law of partial pressure? Here p = total pressure of gaseous mixture: (1) p = p1 + p2 + p3 (2) p = n1(RT/V) + n2(RT/V) + n3(RT/V) (3) pi = chi_i * p, where pi = partial pressure of ith gas, chi_i = mole fraction of ith gas in gaseous mixture (4) pi = chi_i * p_i(standard), where chi_i = mole fraction of ith gas in gaseous mixture, p_i(standard) = pressure of ith gas in pure state
- A. p = p1 + p2 + p3
- B. p = n1(RT/V) + n2(RT/V) + n3(RT/V)
- C. pi = chi_i * p
- D. pi = chi_i * p_i(standard, pure state pressure)
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Answer: C — pi = chi_i * p
The correct relation for partial pressure is pi = chi_i * p (mole fraction times total pressure), as in option 3. Option 4 incorrectly uses the pure-state pressure of the gas instead of the total pressure of the mixture, making it the incorrect equation.
Q6NEET 2022Ideal Gas Equation - Pressure Calculation
A 10.0 L flask contains 64 g of oxygen at 27 degrees Celsius. (Assume O2 gas is behaving ideally). The pressure inside the flask in bar is: (Given R = 0.0831 L bar K^-1 mol^-1)
- A. 2.5
- B. 498.6
- C. 49.8
- D. 4.9
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Answer: C — 49.8
Moles of O2 = 64/32 = 2 mol. Using PV=nRT: P = nRT/V = (2 x 0.0831 x 300)/10 = 49.86/10 = 4.986 ~ 4.9 bar.
Q7NEET 2022Dalton's Law of Partial Pressure
Which one is not correct mathematical equation for Dalton's Law of partial pressure? Here p = total pressure of gaseous mixture
- A. p = p1 + p2 + p3
- B. p = n1RT/V + n2RT/V + n3RT/V
- C. pi = χip, where pi = partial pressure of ith gas, χi = mole fraction of ith gas in gaseous mixture
- D. pi = χipi°, where χi = mole fraction of ith gas in gaseous mixture, pi° = pressure of ith gas in pure state
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Answer: C — pi = χip, where pi = partial pressure of ith gas, χi = mole fraction of ith gas in gaseous mixture
Option (4) incorrectly defines partial pressure in terms of pure-state pressure pi°, which is not the correct Dalton's law relation.
Q8NEET 2021Boyle's Law Graphical Representation
Choose the correct option for graphical representation of Boyle's law, which shows a graph of pressure vs. volume of a gas at different temperature.
- A. Hyperbolic curves with 600K outermost (highest P,V values) and 200K innermost, labeled 600K above 400K above 200K
- B. Hyperbolic curves with 200K outermost and 600K innermost, labeled 200K above 400K above 600K
- C. Horizontal straight line labeled (200K, 400K, 600K) all overlapping
- D. Straight lines from origin with slopes for 200K, 400K, 600K
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Answer: —
PV=nRT, so P∝T at constant V. At a given V, P1>P2>P3 corresponds to T1>T2>T3, so the curve for 600K lies above (outermost) and 200K lies below (innermost), giving nested hyperbolas labeled 600K, 400K, 200K from top to bottom.

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Q9NEET 2021Dalton's Law of Partial Pressures
Choose the correct option for the total pressure (in atm.) in a mixture of 4g O2 and 2g H2 confined in a total volume of one litre at 0°C is: [Given R = 0.82 L atm mol^-1 K^-1, T = 273 K]
- A. 26.02
- B. 2.518
- C. 2.602
- D. 25.18
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Answer: C — 2.602
PV=nRT. P×1=(4/32+2/2)×0.082×273. P=25.18 atm
Q10NEET 2021Boyle's law pressure vs volume graphs at different temperatures
Choose the correct option for graphical representation of Boyle's law, which shows a graph of pressure vs. volume of a gas at different temperatures:
- A. Isotherms ordered with 200K highest, 600K lowest (incorrect order)
- B. All isotherms collapsed to a point (incorrect)
- C. Straight lines through origin (incorrect, not hyperbolic)
- D. Hyperbolic isotherms with 600K highest, 200K lowest, correctly ordered (option 4)
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Answer: C — Straight lines through origin (incorrect, not hyperbolic)
Boyle's law gives PV=constant (hyperbola) for each temperature, with higher temperature isotherms lying further from the origin — matching option 4.

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