JEE Chemistry States Of Matter MCQs Set 15

Answer: (b) the average kinetic energy of the gas increases

Explanation:
1. Average KE is proportional to T, which rises from 293 K to 313 K, so it increases.
2. The speed increases (not decreases) and collisions become more frequent.
3. New pressure = 4 x 313/293 = 4.27 atm, not 8 atm.

Teacher's Note:
1. Average kinetic energy per molecule is (3/2)kT, so it depends only on the absolute temperature.
2. The speeds follow u_rms : u_av : u_mp = 1.225 : 1.128 : 1 and all scale as the square root of T/M.

Answer: (c) 16.7 kPa

Explanation:
1. Total PV = 10 x 25 + 20 x 50 = 1250 kPa m\(^3\).
2. Total volume = 75 m\(^3\), so P = 1250/75 = 16.7 kPa.

Teacher's Note:
1. Dalton's law: each gas in a mixture exerts the pressure it would exert alone in the same volume.
2. Partial pressure equals mole fraction times total pressure.

Answer: (d) 4 kg m\(^{-3}\)

Explanation:
1. \( d = PM/RT = 2\times44/(0.0821\times273) = 3.92 \) g/L, nearly 4 g/L.
2. 4 g/L = 4 kg/m\(^3\).

Teacher's Note:
1. Use PV = nRT and keep the units of P, V, T and R consistent.
2. At fixed amount and temperature P and V are inversely related; at fixed P, V is proportional to T.

Answer: (c) high temperature and low pressure

Explanation:
1. At low pressure the molecules are far apart, so their volume and attraction are negligible.
2. High temperature makes the kinetic energy much larger than the attraction.

Teacher's Note:
1. Real gases follow ideal behaviour at high temperature and low pressure.
2. In the van der Waals equation a/V^2 corrects for attraction and b for the finite molecular volume.

Answer: (c) \( (P + a/V^2) \)

Explanation:
1. The term a/V\(^2\) added to P corrects for the pressure lost because of attraction between molecules.
2. The (V - b) term is for molecular volume.

Teacher's Note:
1. Real gases follow ideal behaviour at high temperature and low pressure.
2. In the van der Waals equation a/V^2 corrects for attraction and b for the finite molecular volume.

Answer: (a) \( \ce{NH3} \) can be liquefied most easily

Explanation:
1. A larger a means stronger intermolecular attraction, which makes liquefaction easier.
2. \( \ce{NH3} \) has the largest a, so it is liquefied most easily.
3. \( \ce{H2} \) has the smallest a, so its forces are the weakest.

Teacher's Note:
1. Real gases follow ideal behaviour at high temperature and low pressure.
2. In the van der Waals equation a/V^2 corrects for attraction and b for the finite molecular volume.

Answer: (c) 0.028

Explanation:
1. Initial moles = 2.24/22.4 = 0.100.
2. At the same V and T, moles are proportional to P: final = 0.100 x 550/760 = 0.0724.
3. Escaped = 0.100 - 0.0724 = 0.028 mol.

Teacher's Note:
1. Use PV = nRT and keep the units of P, V, T and R consistent.
2. At fixed amount and temperature P and V are inversely related; at fixed P, V is proportional to T.

Answer: (d) 0.1 and 0

Explanation:
1. \( P/T = nR/V = 10\times0.0821/8.21 = 0.1 \) (a constant).
2. The plot is a horizontal line, so the intercept is 0.1 and the slope is 0.

Teacher's Note:
1. Use PV = nRT and keep the units of P, V, T and R consistent.
2. At fixed amount and temperature P and V are inversely related; at fixed P, V is proportional to T.

Answer: (b) 4 : 1

Explanation:
1. \( P = dRT/M \).
2. \( P_A/P_B = (d_A/d_B)(M_B/M_A) = 2\times2 = 4 \).
3. The ratio is 4 : 1.

Teacher's Note:
1. Use PV = nRT and keep the units of P, V, T and R consistent.
2. At fixed amount and temperature P and V are inversely related; at fixed P, V is proportional to T.

Answer: (b) 40

Explanation:
1. At constant P and V, \( n \propto 1/T \).
2. Fraction left = 300/500 = 0.6, so 40 per cent escapes.

Teacher's Note:
1. Use PV = nRT and keep the units of P, V, T and R consistent.
2. At fixed amount and temperature P and V are inversely related; at fixed P, V is proportional to T.

Answer: (c) \( d = \dfrac{(m_1 + m_2)}{(n_1 + n_2)}\dfrac{P}{RT} \)

Explanation:
1. \( V = (n_1 + n_2)RT/P \).
2. \( d = \text{mass}/V = (m_1 + m_2)P/[(n_1 + n_2)RT] \).

Teacher's Note:
1. Use PV = nRT and keep the units of P, V, T and R consistent.
2. At fixed amount and temperature P and V are inversely related; at fixed P, V is proportional to T.

Answer: (b) \( \ce{C2H2} \)

Explanation:
1. KOH absorbs \( \ce{CO2} \) = 20 mL, so each molecule gives 2 \( \ce{CO2} \) and x = 2.
2. Pyrogallol absorbs \( \ce{O2} \) = 75 mL, so 25 mL of \( \ce{O2} \) was used for 10 mL, that is 2.5 per molecule.
3. \( x + y/4 = 2.5 \) with x = 2 gives y = 2, so \( \ce{C2H2} \).

Teacher's Note:
1. Write the balanced equation and use volume ratios for gases at the same T and P (Avogadro's law).
2. Liquid water takes up negligible volume, so it is ignored when the volume contraction is found.

Answer: (d) A and D only

Explanation:
1. The condition means \( T_A/M_A = T_B/M_B \).
2. \( u_{rms} = \sqrt{3RT/M} \) is therefore the same (D true).
3. For equal masses \( PV = (m/M)RT \) is the same (A true).
4. KE per mole is \( \tfrac32RT \), which differs as the temperatures differ, and pressure is not fixed by the condition.

Teacher's Note:
1. Average kinetic energy per molecule is (3/2)kT, so it depends only on the absolute temperature.
2. The speeds follow u_rms : u_av : u_mp = 1.225 : 1.128 : 1 and all scale as the square root of T/M.

Answer: (a) 3 atm and 5 atm

Explanation:
1. Partial pressure = mole fraction x total pressure: \( P_A = (3/8)\times8 = 3 \) atm, \( P_B = (5/8)\times8 = 5 \) atm.
2. Each gas alone in the same container exerts its partial pressure.

Teacher's Note:
1. Dalton's law: each gas in a mixture exerts the pressure it would exert alone in the same volume.
2. Partial pressure equals mole fraction times total pressure.

Answer: (c) 1

Explanation:
1. Atoms: 0.75 x 4 = 3 mol X and 2 x 2 = 4 mol O, giving 1 mol of \( \ce{X3O4} \).
2. The solid is not counted, so the initial gas is 2 mol \( \ce{O2} \) at 300 K and the final gas is 1 mol at 600 K.
3. \( P_f/P_i = (1\times600)/(2\times300) = 1 \).

Teacher's Note:
1. Use PV = nRT and keep the units of P, V, T and R consistent.
2. At fixed amount and temperature P and V are inversely related; at fixed P, V is proportional to T.

Answer: (b) 648 K

Explanation:
1. The line has slope (4 - 16)/(15 - 6) = -4/3, so \( P = 24 - 4V/3 \).
2. \( T = PV/nR = (24V - 4V^2/3)\times12/2 \), which is maximum when \( 24 - 8V/3 = 0 \), that is V = 9 L, P = 12 atm.
3. \( T_{max} = 12\times9\times12/2 = 648 \) K.

Teacher's Note:
1. Use PV = nRT and keep the units of P, V, T and R consistent.
2. At fixed amount and temperature P and V are inversely related; at fixed P, V is proportional to T.

Answer: (b) 4

Explanation:
1. \( R = 6.023\times1.380 = 8.31174 \).
2. A product keeps the number of significant digits of the least precise factor; 1.380 has 4 digits and 6.023 has 4.
3. So R has 4 significant digits (8.312).

Teacher's Note:
1. Use PV = nRT and keep the units of P, V, T and R consistent.
2. At fixed amount and temperature P and V are inversely related; at fixed P, V is proportional to T.

Answer: (a) increased collision frequency of X with the inert gas as compared to that of Y

Explanation:
1. A smaller d means that X travels less far than Graham's law predicts, so X is slowed more than Y.
2. The light, fast molecules of X collide more often with the inert gas molecules than the heavy Y molecules do.
3. So the increased collision frequency of X with the inert gas explains the smaller d.

Teacher's Note:
1. Graham's law: the rate of effusion or diffusion is inversely proportional to the square root of the molar mass at the same T and P.
2. Compare rates only under identical conditions.

Answer: (a) 9

Explanation:
1. Mass number falls by 32, so 8 alpha particles (helium atoms) are emitted per U atom; charge balance (92 - 82 = 10 = 2 x 8 - 6) needs 6 beta particles, which are not gas.
2. Gas initially: 1 mol air. Finally: 1 mol air + 8 mol He = 9 mol.
3. At fixed V and T the pressure ratio is 9.

Teacher's Note:
1. Use PV = nRT and keep the units of P, V, T and R consistent.
2. At fixed amount and temperature P and V are inversely related; at fixed P, V is proportional to T.

Answer: (a) 4

Explanation:
1. Mean free path is proportional to \( T/P \) and mean speed to \( \sqrt{T} \).
2. \( D \propto T^{3/2}/P \).
3. Factor = \( 4^{3/2}/2 = 8/2 = 4 \).

Teacher's Note:
1. Average kinetic energy per molecule is (3/2)kT, so it depends only on the absolute temperature.
2. The speeds follow u_rms : u_av : u_mp = 1.225 : 1.128 : 1 and all scale as the square root of T/M.

Answer: (a) 2.22

Explanation:
1. Moles are proportional to PV/T: A = 5 x 1/400 = 0.0125, B = 1 x 3/300 = 0.0100 (same units).
2. Both gases are O\(_2\), so energy is conserved: \( T_f = (0.0125\times400 + 0.01\times300)/0.0225 = 355.6 \) K.
3. At equilibrium P and T are equal, so volumes are in the ratio of moles: \( V_A = 4\times0.0125/0.0225 = 2.22 \) m\(^3\).

Teacher's Note:
1. Use PV = nRT and keep the units of P, V, T and R consistent.
2. At fixed amount and temperature P and V are inversely related; at fixed P, V is proportional to T.

Answer: (d) 29.5

Explanation:
1. Each half is 450 mm long. Let the mercury move down x: the upper air expands to 450 + x and the lower air is compressed to 450 - x.
2. Upper: \( P_1 = 760\times450/(450 + x) \). Lower: \( P_2 = 760\times450/(450 - x) = P_1 + 100 \).
3. Solving gives x = 29.5 mm.

Teacher's Note:
1. Use PV = nRT and keep the units of P, V, T and R consistent.
2. At fixed amount and temperature P and V are inversely related; at fixed P, V is proportional to T.