JEE Chemistry States of Matter MCQs Set 04

Answer: (b) \( \frac{r_{O_2}}{r_{CO_2}}=1.17 \)

Explanation:
1. \( \frac{r_{O_2}}{r_{CO_2}}=\sqrt{\frac{44}{32}}=\sqrt{1.375}=1.17 \).

Teacher's Note:
1. The lighter gas O₂ must be faster, so the ratio is above 1.
2. Compute the root of 44/32.

Answer: (d) At high temperature and low pressure

Explanation:
1. At low pressure the molecules are far apart, so their volume is negligible.
2. At high temperature kinetic energy overcomes attractions.
3. Both conditions together make the ideal gas model valid.

Teacher's Note:
1. Remember the pair: high T, low P.
2. Ideal gas assumptions need weak attraction and small molecular volume.

Answer: (c) volume occupied by the molecules

Explanation:
1. The term \( (V-nb) \) corrects the volume for the space taken up by the molecules.
2. So \( b \) (about four times the actual molecular volume per mole) measures the excluded volume.

Teacher's Note:
1. a is for attraction; b is for volume.
2. b has units of L per mole.

Answer: (c) \( \left(P+\frac{a}{V^2}\right) \)

Explanation:
1. Attractive forces reduce the measured pressure, so \( \frac{a}{V^2} \) is added to P.
2. The bracket \( \left(P+\frac{a}{V^2}\right) \) is the pressure correction.

Teacher's Note:
1. Pressure correction is the one with a.
2. Volume correction is the one with b.

Answer: (d) intermolecular forces

Explanation:
1. Molecules near the wall are pulled inward by their neighbours.
2. This weakens the impact on the wall, so the measured pressure is lower than that of an ideal gas.

Teacher's Note:
1. Attractive forces reduce wall impact.
2. This is the origin of the correction \( an^2/V^2 \).

Answer: (c) SO₂

Explanation:
1. A larger value of 'a' means stronger intermolecular attraction, so easier liquefaction.
2. SO₂ has the largest 'a' (6.71).

Teacher's Note:
1. Compare the 'a' values directly.
2. Higher a means easier liquefaction.

Answer: (c) PV = RT + Pb

Explanation:
1. At high pressure V is small but \( b \) is not negligible, while \( a/V^2 \) can be neglected against P.
2. So \( P(V-b)=RT \), i.e. \( PV=RT+Pb \).

Teacher's Note:
1. Decide which correction dominates.
2. At high pressure the volume correction wins.

Answer: (c) 0.5 atm and 500 K

Explanation:
1. Ideal behaviour needs low pressure and high temperature.
2. 0.5 atm at 500 K has the lowest pressure and highest temperature among the options.

Teacher's Note:
1. Pick the lowest P and highest T.
2. Check both factors, not one.

Answer: (b) 1

Explanation:
1. \( Z=\frac{PV}{nRT} \) and an ideal gas obeys \( PV=nRT \), so \( Z=1 \).

Teacher's Note:
1. Z measures deviation from ideal behaviour.
2. Z equal to 1 means no deviation.

Answer: (b) \( V_m<22.4 \) L

Explanation:
1. \( Z=\frac{PV_m}{RT}<1 \) at STP, so the molar volume is less than the ideal value of 22.4 L.

Teacher's Note:
1. Compare with the ideal molar volume at STP.
2. Z below 1 means the gas is more compressible than ideal.

Answer: (c) NH₃

Explanation:
1. Stronger attraction (larger 'a') means easier liquefaction.
2. NH₃ has the largest 'a'.

Teacher's Note:
1. Pick the largest a value.
2. Ammonia also has hydrogen bonding.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Assertion is true: attractions reduce the impact of molecules on the wall.
2. Reason is true: an ideal gas has no attraction, a real gas has some, and this lowers the pressure. So R explains A.

Teacher's Note:
1. Real gas pressure is corrected by adding \( an^2/V^2 \).
2. Ideal gases have zero attraction by definition.

Answer: (c) A is true, but R is false.

Explanation:
1. Assertion is true: CO₂ has a larger 'a' (3.59 against 0.244), so it liquefies much more easily than H₂.
2. Reason is false: the cause is the larger 'a', not a higher Z; at ordinary pressures Z of CO₂ is below 1, while for H₂ it is above 1.

Teacher's Note:
1. Link liquefaction to the value of 'a'.
2. Z is a result of the forces, not their cause.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. At low pressure the volume V is large, so \( b \) can be dropped from \( (V-b) \).
2. Then \( PV+\frac{a}{V}=RT \) and \( Z=1-\frac{a}{VRT} \). So R explains A.

Teacher's Note:
1. Low pressure: neglect b.
2. High pressure: neglect a.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( T_c=\frac{8a}{27Rb} \) increases with \( a \).
2. A larger 'a' means stronger attraction, which makes liquefaction easier; so R explains A.

Teacher's Note:
1. Higher Tc means the gas can be liquefied at higher temperatures.
2. Tc is directly linked to 'a'.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. \( Z_c=\frac{P_cV_c}{RT_c}=\frac{\frac{a}{27b^2}\times3b}{R\times\frac{8a}{27Rb}}=\frac38 \), so A is true.
2. R is also true at the critical point (Z is below 1), but it does not explain the value 3/8, which follows only from the critical constants.

Teacher's Note:
1. Substitute the three critical constants.
2. 3/8 is a number, not a statement about forces.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. A gas cannot be liquefied by pressure alone above its critical temperature, so A is true.
2. R is a true general statement, but it is not why the gas stays gaseous above Tc.

Teacher's Note:
1. Above Tc no pressure can liquefy a gas.
2. Do not link a true statement to the wrong cause.

Answer: (d) A is false, but R is true.

Explanation:
1. PV of a real gas first decreases (for gases showing attraction), reaches a minimum and then increases; it is not linear throughout, so A is false.
2. R is false: PV does depend on pressure for a real gas (it is constant only for an ideal gas, Boyle's law).

Teacher's Note:
1. Boyle's law holds only for ideal gases.
2. Look at the shape of the PV against P curve.