States of Matter – JEE Advanced Previous Year Questions with Solutions

States of Matter JEE Advanced questions and solutions are available on this page to help students get a clear idea of the question types and the pattern. Detailed solutions solved by experts are provided to every question listed here. States of Matter JEE Advanced previous year questions and solutions will further help students to revise the concepts and related topics before the main exams.

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JEE Advanced Previous Year Questions on the States of Matter

Question 1. A closed tank has two compartments A and B, both filled with oxygen (assumed to be ideal gas). The partition separating the two compartments is fixed and is a perfect heat insulator (Figure 1). If the old partition is replaced by a new partition that can slide and conduct heat but does NOT allow the gas to leak figure 2), the volume(in m³) of compartment A after the system attains equilibrium is _________.

Solution: (2.22)

We know that the ideal gas equation is,

PV = nRT

⇒ n = PV / RT

Given:

If we look at Figure 1, in compartment A the pressure is 5 bar, the volume is 1 m³, and temperature 400 K. So we can write it as:

P_A = 5

V_A = 1

T_A = 400

⇒ n_A = 5 × 1 / 400 × R = 1 / 80 R

Likewise in compartment B, the pressure is 1 bar, the volume is 3 m³, and temperature 400 K. We can write it as:

P_B = 1

V_B = 3

T_B = 300

⇒ n_B = 1 × 3 / 300 × R = 1 / 100 R

After the system attains equilibrium;

P_A / T_A = P_B / T_B

⇒ n_A × R / V_A = n_B × R / V_B

⇒ (1 / 80R) / V_A × R = (1 / 100R) / V_B × R

⇒ V_B = ⅘ V_A ………(i)

Now, the total volume of the containers.

(V) = V_A + V_B = 1 + 3 = 4

⇒ V_A + V_B = 4 ………(ii)

Now, from equations (i) and (ii), we get,

V_A + ⅘ V_A = 4

⇒ 9 / 5 V_A = 4

⇒ V_A = 2.22 m³

Question 2. The surface of copper gets tarnished by the formation of copper oxide. N₂ gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the N₂ gas contains 1 mole % of water vapour as an impurity. The water vapour oxidises copper as per the reaction given below:

2Cu(s) + H₂O(g) → Cu₂O(s) + H₂(g)

ρH₂ is the minimum partial pressure of H₂ (in a bar) needed to prevent the oxidation at 1250 K. The value of ln(ρH₂) is ___.

(Given : total pressure = 1 bar, R (universal gas constant) = 8 JK^-1mol^-1, ln(10) = 2.3. Cu(s) and Cu2O(s) are mutually immiscible.

At 1250 K: 2Cu(s) + ½ O₂(g) → Cu₂O(s); ΔGƟ = − 78,000 J mol^-1

H2(g) + ½ O₂(g) → H₂O(g); ΔGƟ = − 1,78,000 J mol^-1; G is the Gibbs energy)

Solution: (-14.6)

2Cu(s) + ¼ O₂(g) → Cu₂O(s) ΔG^o = −78 kJ

Hence, 2Cu(s) + H₂O(g) → Cu₂O + H₂(g) ΔG^o = +100kJ

ΔG = ΔG^o + RTln Q

0 = +100 + 8 / 1000 × 1250 lnρ × ρH2 / H₂O

− 100 × 1000 / 8 = 1250ln ρH₂ / (1001 × 1)

∴ ln ρH₂ = −14.6

Question 3. The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The absolute temperature of an ideal gas is increased 4 times and its pressure is increased 2 times. As a result, the diffusion coefficient of this gas increases x times. The value of x is.

Solution: (4)

Let D_i = initial value and D_f = final value

D_i / D_f = λ_i √v_i / λ_f √v_f

D_i / D_f = T_i / P_i √T_i / M ÷ T_f / P_f √T_f / M

D_i / D_f = P_f .T_i √T_i ÷ P_i .T_f √T_f

D_i / D_f = 2P_i .T_i √T_i ÷ P_i .4T_i √4T_f

D_i / D_f = 2 / 8

D_i / D_f = ¼

D_f = 4D_i

Therefore, x = 4

Question 4. The figure below is the plot of potential energy versus internuclear distance (𝑑) of H₂ molecule in the electronic ground state. What is the value of the net potential energy 𝐸_o (as indicated in the figure) in kJ mol^-1, for 𝑑 = 𝑑_o at which the electron-electron repulsion and the nucleus-nucleus repulsion energies are absent? As a reference, the potential energy of the H atom is taken as zero when its electron and the nucleus are infinitely far apart. Use Avogadro constant as 6.023 × 10²³ mol^-1.

Solution: (–5242.42)

P. E of 2 H-atoms

Total eng = P.E./2

Potential Energy = 2 Total Energy

E = -13.6 × z²/n² ev/atom

= –2 × 13.6 × z²/n² ev/atom + (–2 × 13.6 × z²/n²) ev/ atom

= –2 × 2 × 13.6 × 1 x ev/atom

= –4 × 13.6 × 1.6 × 10^-19 J/atom × 6.023 × 10²³ atom/mole

= –4 × 13.6 × 1.6 × 6.023 ×10⁴ J/mole

= –5242.42 KJ/mol

Question 5. One mole of a monatomic gas satisfies the equation P(V – b) = RT Where b is a constant. The relationship of interatomic potential V(r ) and interatomic distance r for the gas is given by;

Solution: (C)

Given equation = P(V − b) = RT.

If we compare it with van der Waals’ equation, i.e., [P + a / V²] [V−b] = RT, we get a = 0.

Therefore, only repulsive forces are present that contribute only at a very close distance.

As a result, the potential energy will increase abruptly indicating that graph (C) is correct. The dominance of repulsive force can be confirmed using the compressibility factor.

P(V − b) = RT

PV = Pb + RT

PV / RT = Pb / RT +1

Z = Pb / RT +1, i.e., Z>1 (Repulsive forces).

So, repulsive tendencies will be dominant when interatomic distances are small. It also means that the interatomic potential is never negative but becomes positive at small interatomic distances.

Question Paragraph: X and Y are two volatile liquids with molar weights of 10 gmol^-1 and 40 gmol^-1 respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L = 24 cm, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of 300K. Vapours of X and Y react to form a product which is first observed at a distance of d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.

Question 6. The value of d in cm (shown in the figure) as estimated from Graham’s law is

A. 8

B. 12

C. 16

D. 20

Solution: (C)

Rate of diffusion α 1 / Molecular − weight

Let distance covered by X be ′d′ and distance covered by Y is ′24 − d′

Now,

r_x / r_y= d / 24 − d

⟹ d / 24 − d = √40 / 10

⟹ d = 16cm

Since the time taken must be the same for both the gases, the distance travelled will be in the ratio 2:1. We divide the tube into 3 parts, each part has a length of 8 cm. Hence, the distance d = 16 cm (because of the ratio of distances/velocities being 2:1).

Hence, option C is the correct answer.

Question 7. The experimental value of d is found to be smaller than the estimate obtained using Graham’s law. This is due to;

A. Larger mean free path for X as compared to that of Y

B. Larger mean free path for Y as compared to that of X

C. Increased collision frequency of Y with the inert gas as compared to that of X with the inert gas

D. Increased collision frequency of X with the inert gas as compared to that of Y with the inert gas

Solution: (D)

The experimental value of d is found to be smaller than the estimate obtained using Graham’s law. This is due to the increased collision frequency of X with the inert gas as compared to that of Y with inert gas.

With the increase in the collision frequency, the molecular speed decreases than expected so the distance covered will be less.

Question 8. If the distribution of molecular speeds of gas is as per the figure shown below, then the ratio of the most probable, the average, and the root mean square speeds, respectively, is.

A. 1 : 1 : 1

B. 1 : 1 : 1.224

C. 1 : 1.128 : 1.224

D. 1 : 1.128 : 1

Solution: (C)

The graph represents a symmetrical distribution of speed and hence, the most probable and the average speed should be the same. But the root mean square speed must be greater than the average speed.

According to distribution of molecular speed curve,

U_mps = √2RT / mol.wt. U_avg = √8RT / 𝜋mol.wt. U_rms = √3RT / mol.wt.

U_mps : U_avg : U_rms = √2 : √8 / 𝜋 : √3

= 1 : 1.128 : 1.224

Question 9. The qualitative sketches I, II and III given below show the variation of surface tension with molar concentration of three different aqueous solutions of KCl, CH₃OH and CH₃(CH₂)₁₁OSO₃^–Na⁺ at room temperature. The correct assignment of the sketches is:

A. I : KCl; II : CH₃OH; II I : CH₃(CH₂)₁₁OSO₃^–Na⁺

B. I : CH₃(CH₂)₁₁OSO₃^–Na⁺ II : CH3OH ; III : KCl

C. I : KCl ; II : CH₃(CH₂)₁₁OSO₃^–Na⁺; III : CH3OH

D. I : C₃OH; II : KCl ; III : CH₃(CH₂)₁₁OSO₃^–Na⁺

Solution: (D)

C₃OH; II : KCl ; III : CH₃(CH₂)₁₁OSO₃^–Na⁺

The surface tension of a liquid is affected to some extent by impurities. Some of them which tend to concentrate on the surface of liquids, compared to its bulk lower the surface tension. Compounds such as detergents and soaps CH₃(CH₂)₁₁OSO₃^–Na⁺ decrease the surface tension considerably.

On the other hand, alcohol (eg.CH₃OH,C₂H₅OH) tends to lower the surface tension only slightly. It is mainly due to CH₃OH having a smaller dielectric constant. The dielectric constant is directly proportional to surface tension. If we add CH₃OH in water, the overall dielectric constant decreases and thereby surface tension also decreases.

Inorganic impurities present in the bulk of a liquid also have a tendency to increase the surface tension of water. Strong electrolytes such as KCl increase the surface tension slightly.

Question 10. Which of the following statement(s) is (are) correct regarding the root mean square speed (U_rms) and average translational kinetic energy (∈_av) of a molecule in a gas at equilibrium?

A. U_rms is doubled when its temperature is increased four times

B. ∈_av is doubled when its temperature is increased four times

C. ∈_av at a given temperature does not depend on its molecular mass

D. U_rms is inversely proportional to the square root of its molecular mass

Solution: (A, C and D)

The root mean square speed (Urms) and average translational kinetic energy (ϵ_av) has following relation with temperature and molecular mass

ϵ_av = √3 / 2 RT, U_rms = √3RT / M and U_rms ∝ √1 / M

∴ ϵ_av doesn’t depend on its molecular mass.

If we consider the above-given relation, ϵ_av gets doubled when the temperature is increased by 2 times. So we can say that option B is incorrect.

Thus, the correct options are A, C, and D.

Gaseous State