JEE Chemistry States Of Matter MCQs Set 16

Answer: (c) increase

Explanation:
1. Attractions pull molecules back as they approach the wall, lowering the pressure.
2. Without them, \( P_{ideal} = P_{real} + \frac{an^{2}}{V^{2}} \) would be the observed pressure, which is larger.

Teacher's Note:
1. Attractions reduce the wall pressure.
2. Remove them and pressure rises.

Answer: (a) smaller than that of an ideal gas

Explanation:
1. Taking both corrections into account, \( P = \frac{nRT}{V - nb} - \frac{an^{2}}{V^{2}} \), and for most conditions the attraction term lowers the pressure.

Teacher's Note:
1. Attraction lowers the pressure.
2. Molecular volume raises it.

Answer: (d) 1.335 Angstrom

Explanation:
1. b = 4 N_A (4/3 pi r³), so \( r^{3} = \frac{3b}{16 \pi N_{A}} = \frac{72}{16\pi \times 6.022\times10^{23}} = 2.38\times10^{-24} \) cm³.
2. r = 1.335 × 10^-8 cm = 1.335 Angstrom.

Teacher's Note:
1. b is four times the actual molecular volume per mole.
2. Take the cube root carefully.

Answer: (a) One mole of G occupies 22.4 L at a pressure higher than that at STP (at constant temperature)

Explanation:
1. Z = PV_m/RT > 1 at 1 atm means V_m is larger than 22.4 L at 273 K and 1 atm.
2. To bring V_m down to 22.4 L, the pressure must be raised.

Teacher's Note:
1. Z > 1 means the gas is less compressible.
2. It needs more pressure to reach the ideal molar volume.

Answer: (a) (1) and (2) only

Explanation:
1. Above T_c the liquid and vapour cannot be distinguished, so T_c is the highest temperature for coexistence.
2. The slope is zero at the critical point; at T_c the densities of the two phases are equal, so (3) is false.

Teacher's Note:
1. At the critical point, dP/dV = 0.
2. Densities become equal.

Answer: (a) All four

Explanation:
1. The term an²/V² corrects for attractions, and b is four times the actual volume of the molecules per mole.
2. The units follow from the equation: b in L/mol, a in atm L²/mol².

Teacher's Note:
1. Check units from the equation.
2. Both a and b are characteristic constants.

Answer: (d) (1) and (3) only

Explanation:
1. At the Boyle temperature the effects of a and b cancel at low pressure, so Z is close to 1.
2. At high pressure the molecules are close together and the excluded volume (b) dominates; a gas above critical conditions is not necessarily ideal.

Teacher's Note:
1. T_B = a/Rb.
2. Ideal behaviour needs low P and high T.

Answer: (a) at very high pressure for any temperature, and at low pressure when T > T_b

Explanation:
1. Z > 1 means the gas is harder to compress than an ideal gas.
2. Z > 1 at high pressure (b dominates) and also at low pressure when T > T_b (b effect exceeds a effect).

Teacher's Note:
1. Z > 1 means lower compressibility.
2. Below T_b, Z < 1 at low pressure.

Answer: (a) 4

Explanation:
1. At high pressure Z = 1 + Pb/RT, so the slope b/RT = 0.01.
2. The intercept of PV against P is nRT = 40, so RT = 20 L atm/mol and b = 0.2 L/mol.
3. Excluded volume for 20 mol = nb = 20 × 0.2 = 4 L.

Teacher's Note:
1. Slope gives b/RT and intercept gives nRT.
2. Then multiply b by n.

Answer: (b) 90

Explanation:
1. Z_c = P_cV_c/RT_c = 3/8 for a van der Waals gas.
2. x/24 = 3/8, so x = 9 and 10x = 90.

Teacher's Note:
1. Z_c = (a/27b²)(3b)/(R x 8a/27Rb) = 3/8.
2. Solve x/24 = 3/8.

Answer: (d) 20

Explanation:
1. At low pressure \( Z = 1 - \frac{a}{VRT} = 1 - \frac{20.39}{35 \times 0.0821 \times 350} = 0.98 \).
2. At high pressure \( Z = 1 + \frac{b}{V} = 1 + \frac{0.1383}{35} = 1.004 \).
3. (0.98 + 1.004) × 10 = 19.84, which rounds to 20.

Teacher's Note:
1. Use the limiting forms of Z for the two regions.
2. T = 350 K.

Answer: (b) \( \frac{an^{2}}{V^{2}} \)

Explanation:
1. The equation is \( \left(P + \frac{an^{2}}{V^{2}}\right)(V - nb) = nRT \); the term added to P corrects for attraction.

Teacher's Note:
1. P_ideal = P_real + an²/V².
2. nb corrects the volume.

Answer: (a) the volume occupied by the molecules

Explanation:
1. b is the excluded volume per mole, about four times the actual volume of the molecules.

Teacher's Note:
1. b relates to molecular size.
2. a relates to attraction.

Answer: (d) 1 + pb/RT

Explanation:
1. At high pressure p(V - b) = RT, so pV = RT + pb.
2. Z = pV/RT = 1 + pb/RT.

Teacher's Note:
1. At high pressure b dominates.
2. Z is greater than 1.

Answer: (d) Z = 1 - a/VRT

Explanation:
1. At low pressure V is large, so b can be neglected: (P + a/V²)V = RT, which gives PV = RT - a/V.
2. Dividing by RT gives Z = 1 - a/VRT.

Teacher's Note:
1. Neglect b at low pressure.
2. Z is below 1.

Answer: (d) 1 + pb/RT

Explanation:
1. For large p, a/V² is negligible compared with p, so p(V - b) = RT and Z = 1 + pb/RT.

Teacher's Note:
1. b dominates at high pressure.
2. Z is greater than 1.

Answer: (d) 2P_A = 3P_B

Explanation:
1. P = ZnRT/V.
2. P_A/P_B = (Z_A/Z_B)(V_B/V_A) = 3 × (1/2) = 3/2, so 2P_A = 3P_B.

Teacher's Note:
1. Express P in terms of Z, n, T and V.
2. The ratio of the two gases gives the answer.

Answer: (a) Xe

Explanation:
1. From P(V - b) = RT, Z = 1 + bP/RT, so the slope of Z against P is b/RT.
2. Xe has the largest van der Waals b, so the steepest slope.

Teacher's Note:
1. Slope is proportional to b.
2. Larger atom, larger b.

Answer: (a) Gas C occupies more volume than gas A; gas B is more compressible than gas D

Explanation:
1. C has weaker attractions than A (smaller a) with the same b, so its molecules are less held together and it occupies a larger volume.
2. B and D have the same a but B has a much smaller b (less repulsion), so B is more compressible than D.

Teacher's Note:
1. Large a pulls the gas together.
2. Large b makes the gas resist compression.