JEE Chemistry States Of Matter MCQs Set 14

Answer: (c) A is at temperature \( T_1 \) and B at \( T_2 \), with \( T_1 > T_2 \)

Explanation:
1. At T: \( \sqrt{8RT/\pi M_A} = \sqrt{3RT/M_B} \), so \( M_A/M_B = 8/3\pi = 0.85 \), A is the lighter gas.
2. Mean speeds are equal when \( T_A/M_A = T_B/M_B \), that is \( T_A/T_B = M_A/M_B < 1 \), so A must be at the lower temperature.
3. If A is hotter than B (\( T_1 > T_2 \)) the mean speed of A is even larger than that of B, and equality is impossible. The other three cases can be arranged to give equal speeds.

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) the gas does work against the atmosphere at the cost of the kinetic energy of its molecules, so it cools

Explanation:
1. On release the gas expands rapidly into the low-pressure surroundings and does work pushing the atmosphere back.
2. This work is done at the expense of the kinetic energy of the molecules, so the temperature falls (Joule-Thomson type cooling).
3. The volume increases (not decreases), so the compression statement is wrong, and the cooling lets solid \( \ce{CO2} \) form.

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) 3.24 L

Explanation:
1. 15.0 g of KOH is 15.0/56.1 = 0.267 mol.
2. From the equation 2 mol KOH absorb 1 mol \( \ce{CO2} \), so 0.134 mol of \( \ce{CO2} \).
3. \( V = nRT/P = 0.134\times0.0821\times295/1 = 3.24 \) L.

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) the collisions between the molecules increase

Explanation:
1. At constant temperature the average speed (and kinetic energy) stays the same.
2. Compression puts the same molecules in a smaller volume, so they strike each other and the walls more often.
3. The number of collisions increases.

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: (d) The pressure and volume of a gas are inversely related

Explanation:
1. The inverse relation of P and V is Boyle's law, an experimental gas law that the theory explains.
2. The other three statements are postulates of the theory.

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: (b) \( \ce{H2} > \ce{N2} > \ce{O2} > \ce{HBr} \)

Explanation:
1. \( \overline{u^2} = 3RT/M \), so at the same temperature it is inversely proportional to M.
2. Molar masses: \( \ce{H2} \) 2, \( \ce{N2} \) 28, \( \ce{O2} \) 32, HBr 81.
3. The order is \( \ce{H2} > \ce{N2} > \ce{O2} > \ce{HBr} \).

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: (b) 10/3 g/mol

Explanation:
1. \( n(\ce{H2}) = 3\times16.42/(0.0821\times300) = 2 \) mol and \( n(\ce{D2}) = 6\times16.42/(0.0821\times300) = 4 \) mol.
2. Average molar mass = (2 x 2 + 4 x 4)/6 = 20/6.
3. This is 10/3 g/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: (c) 96.7

Explanation:
1. Total moles = \( PV/RT = 1\times2/(0.0821\times299) = 0.0814 \) mol.
2. Per cent \( \ce{H2} \) = 0.0788/0.0814 x 100 = 96.7.

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) 40.1 mL

Explanation:
1. The mercury is higher inside, so the gas pressure = 750 - 100 = 650 torr.
2. \( V_{STP} = 50\times(650/760)\times(273/291) \).
3. This is 40.1 mL.

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) in a straight line path

Explanation:
1. Ideal gas molecules exert no forces on each other except during collision.
2. With no force acting between collisions, a molecule moves with constant velocity in a straight line.

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: (b) \( \ce{C2H4O3} \)

Explanation:
1. Mass ratio C : H = 6 : 1 gives atom ratio 6/12 : 1/1 = 1 : 2, so \( Y = 2X \).
2. Burning \( \ce{C_XH_{2X}} \) needs \( X + X/2 = 1.5X \) molecules of \( \ce{O2} \), that is 3X atoms of oxygen.
3. The compound has half of this, \( Z = 1.5X \). For X = 2: Y = 4 and Z = 3, giving \( \ce{C2H4O3} \).

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: (b) its absolute temperature

Explanation:
1. Kinetic energy of one mole is \( \tfrac32 RT \).
2. It depends only on the absolute temperature.

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) five times that of \( \ce{O2} \)

Explanation:
1. Equal volumes at the same T and P hold equal moles.
2. Mass ratio = 154/32 = 4.8, which is nearly 5.

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) 58

Explanation:
1. \( M = dRT/P = 2.380\times0.0821\times288/(736/760) \).
2. P = 0.968 atm, so M = 56.3/0.968 = 58.1 g/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: (c) 327 C

Explanation:
1. Equal \( u_{rms} \) needs \( T/M \) equal: \( T/64 = 300/32 \).
2. T = 600 K = 327 C.

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) inversion temperature

Explanation:
1. Below the inversion temperature intermolecular attraction dominates, so the gas cools on expansion.
2. Above it the gas warms on expansion.

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) 25%

Explanation:
1. At constant pressure and volume \( n \propto 1/T \).
2. Fraction left = 300/400 = 0.75, so 25 per cent is expelled.

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) Plot IV

Explanation:
1. Boyle's law: PV is constant at fixed T and n.
2. So P vs 1/V is a straight line through the origin (I), PV vs P is horizontal (II) and P vs V is a rectangular hyperbola (III).
3. PV does not rise with P for an ideal gas, so IV is wrong.

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) the quantity of the gas

Explanation:
1. PV = nRT, so at a fixed T the constant \( C = nRT \).
2. It depends on the amount of gas, not on P, V or the molar mass.

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) the density of the gas

Explanation:
1. The mass and volume are fixed, so the density is unchanged.
2. Pressure, average energy and average speed all fall with temperature.

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) V against 1/T

Explanation:
1. V is proportional to T and to 1/P, and T is proportional to P, giving straight lines.
2. V is proportional to T, so V against 1/T is a hyperbola, not a line.

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.