JEE Chemistry States of Matter MCQs Set 13

Answer: (c) both small

Explanation:
1. Ideal behaviour needs negligible attraction (small a) and negligible molecular volume (small b).
2. With both small, the equation reduces to PV = nRT.

Teacher's Note:
1. The a term corrects pressure for attraction and the nb term corrects volume for molecular size.
2. Helium has very small a and b, so it is the nearest to ideal.

Answer: (a) intermolecular forces of attraction

Explanation:
1. Attraction slows molecules just before they strike the wall, so each impact is weaker.
2. This lowers the measured pressure: \( P_{ideal}=P+\frac{an^2}{V^2} \).

Teacher's Note:
1. Molecular size works in the opposite direction, it reduces the free volume and raises pressure.
2. Attraction dominates at moderate pressure, size at high pressure.

Answer: (d) However great the pressure applied, a gas cannot be liquefied below its critical temperature

Explanation:
1. Below the critical temperature a gas can be liquefied if enough pressure is applied, so the statement as written is false.
2. The other three statements correctly describe critical behaviour.

Teacher's Note:
1. T below Tc is the necessary condition for liquefaction by compression.
2. Above Tc no pressure can produce a distinct liquid phase.

Answer: (b) \( \ce{Z=1-\frac{a}{V_mRT}} \)

Explanation:
1. At low pressure \( V_m \) is large, so b is negligible but \( a/V_m^2 \) is not.
2. \( (P+a/V_m^2)V_m=RT \) gives \( PV_m=RT-a/V_m \).
3. Dividing by RT: \( Z=1-\frac{a}{V_mRT} \), which is less than 1.

Teacher's Note:
1. Z below 1 means the gas is more compressible than an ideal gas.
2. This is the region where attractions dominate.

Answer: (b) \( \ce{Z=1+\frac{Pb}{RT}} \)

Explanation:
1. At high pressure \( V_m \) is small and \( a/V_m^2 \) is negligible compared with P.
2. \( P(V_m-b)=RT \) gives \( PV_m=RT+Pb \).
3. So \( Z=1+\frac{Pb}{RT}>1 \).

Teacher's Note:
1. Z above 1 means the gas is harder to compress than an ideal gas.
2. Molecular size (b) is the cause.

Answer: (d) \( \ce{Z=1+\frac{Pb}{RT}} \)

Explanation:
1. For H\(_2\) and He the constant a is extremely small (weak attractions).
2. The equation becomes \( P(V_m-b)=RT \), so \( Z=1+\frac{Pb}{RT} \).
3. Hence Z is greater than 1 at all pressures.

Teacher's Note:
1. These gases are non polar and very small, so attraction is almost absent.
2. Their isotherms never dip below Z = 1.

Answer: (a) 2.24 mL

Explanation:
1. \( V=\frac{ZnRT}{P}=\frac{0.9\times0.001\times0.0821\times273}{9} \) L.
2. \( V=2.24\times10^{-3} \) L = 2.24 mL.

Teacher's Note:
1. Use \( PV=ZnRT \) for a real gas.
2. Convert millimole to mole before substituting.

Answer: (c) \( \ce{\frac{R}{V-b}} \)

Explanation:
1. \( P=\frac{RT}{V-b} \).
2. At constant V, \( \frac{dP}{dT}=\frac{R}{V-b} \), a positive constant.

Teacher's Note:
1. The isochore is a straight line, as for an ideal gas, but with V replaced by V - b.
2. The slope is positive.

Answer: (d) \( \ce{\frac{n^2a}{V^2}} \)

Explanation:
1. Attraction lowers the pressure by an amount proportional to the square of the number density, \( n^2/V^2 \).
2. The correction is \( +\frac{n^2a}{V^2} \) added to the observed pressure.

Teacher's Note:
1. nb is the volume correction.
2. Both a and b are constants characteristic of each gas.

Answer: (b) decreases

Explanation:
1. b is four times the actual volume of the molecules in one mole: \( b=4N_A v \).
2. S\(_2\) molecules are much smaller than S\(_8\) molecules, so the excluded volume per mole of molecules falls.

Teacher's Note:
1. b tracks molecular size.
2. The number of molecules rises but each is far smaller.

Answer: (a) 0.5

Explanation:
1. With b = 0: \( \left(P+\frac{an^2}{V^2}\right)V=nRT \), i.e. \( 0.1V+\frac{0.4}{V}=0.4 \).
2. \( 0.1V^2-0.4V+0.4=0 \) gives \( (V-2)^2=0 \), so V = 2 L.
3. \( Z=\frac{PV}{nRT}=\frac{0.1\times2}{0.02\times20}=0.5 \).

Teacher's Note:
1. Solve the quadratic for V first; here it is a perfect square.
2. Z below 1 shows attraction dominates.

Answer: (c) Z

Explanation:
1. \( T_c=\frac{8a}{27Rb}\propto\frac{a}{b} \).
2. a/b: W 148, X 267, Y 188, Z 444.

Teacher's Note:
1. Compare the ratio a/b because R is common.
2. Largest a/b means easiest to liquefy and highest Tc.

Answer: (c) \( \ce{b-\frac{a}{RT}} \)

Explanation:
1. From \( Z=\frac{1}{1-b/V}-\frac{a}{VRT} \) expand: \( Z\approx1+\frac1V\left(b-\frac{a}{RT}\right) \).
2. Comparing with the virial form gives \( B=b-\frac{a}{RT} \).

Teacher's Note:
1. B is the second virial coefficient and depends on temperature.
2. B = 0 at \( T_B=\frac{a}{Rb} \), the Boyle temperature.

Answer: (a) \( \ce{4vN_0} \)

Explanation:
1. Excluded volume per molecule is \( \frac12\cdot\frac43\pi(2r)^3=4\cdot\frac43\pi r^3=4v \).
2. For one mole, \( b=4vN_0 \).

Teacher's Note:
1. b is four times the actual volume of the molecules in a mole.
2. It is not just the volume, because of the excluded region around each pair.

Answer: (d) NH\(_3\)

Explanation:
1. A larger a means stronger intermolecular attraction.
2. Stronger attraction means a higher critical temperature, so easier liquefaction.
3. NH\(_3\) has the largest a (4.17).

Teacher's Note:
1. NH\(_3\) also hydrogen bonds.
2. Ease of liquefaction rises with a.

Answer: (b) 0.5 atm and 500 K

Explanation:
1. Ideal behaviour is approached at low pressure and high temperature.
2. Among the choices, 0.5 atm and 500 K combines the lowest pressure with the highest temperature.

Teacher's Note:
1. Low pressure makes molecules far apart; high temperature reduces the effect of attraction.
2. Deviations grow with pressure and fall with temperature.

Answer: (b) 0.65, CO\(_2\) is more compressible than an ideal gas

Explanation:
1. \( Z=\frac{PV}{RT}=\frac{40\times0.4}{0.0821\times300}=0.65 \).
2. Z < 1 means attraction dominates, so the gas is more compressible than an ideal gas.

Teacher's Note:
1. Z below 1 means the real volume is less than the ideal volume.
2. Always check which of P, V, T are given.

Answer: (a) (ii) and (iii)

Explanation:
1. B is temperature dependent, for example \( B=b-\frac{a}{RT} \), so (i) is false.
2. B = 0 at \( T_B=\frac{a}{Rb} \), so (ii) is true.
3. B/V is dimensionless so B has the units of V_m, so (iii) is true.

Teacher's Note:
1. B is negative at low temperature and positive at high temperature.
2. It changes sign at the Boyle temperature.

Answer: (c) the effect of molecular attraction compensates the effect of molecular volume

Explanation:
1. At \( T_B=\frac{a}{Rb} \), \( b-\frac{a}{RT}=0 \).
2. Reduction of pressure by attraction is exactly cancelled by the increase due to size.

Teacher's Note:
1. a and b are not zero at the Boyle temperature.
2. Their effects on Z are equal and opposite.

Answer: (d) \( 3.4\times10^{-9} \) cm

Explanation:
1. b = 0.4 cm\(^3\) mol\(^{-1}\) and \( b=4N_A\cdot\frac43\pi r^3 \).
2. \( r^3=\frac{3b}{16\pi N_A}=\frac{3\times0.4}{16\pi\times6\times10^{23}}=3.98\times10^{-26} \) cm\(^3\).
3. \( r=3.4\times10^{-9} \) cm.

Teacher's Note:
1. Convert litres to cm\(^3\) (1 L = 1000 cm\(^3\)).
2. Take the cube root at the end.

Answer: (b) (2), (3) and (4) only

Explanation:
1. (1) is false, gases can be compressed (and liquefied) below \( T_c \).
2. (2) and (3) are correct.
3. \( T_B=\frac{a}{Rb} \) and \( T_c=\frac{8a}{27Rb} \), so \( T_B/T_c=\frac{27}{8} \), (4) is correct.

Teacher's Note:
1. The ratio 27/8 holds for a van der Waals gas.
2. Boyle temperature is about 3.4 times the critical temperature.

Answer: (a) (2) and (4) only

Explanation:
1. \( P_c=\frac{a}{3V_c^2} \) from the critical conditions, so \( a=3P_cV_c^2 \).
2. \( T_c=\frac{8a}{27Rb} \) and \( P_c=\frac{a}{27b^2} \) give \( \frac{T_c^2}{P_c}=\frac{64a}{27R^2} \), so \( a=\frac{27R^2T_c^2}{64P_c} \).
3. \( \frac{RT_c}{8P_c} \) equals b, not a, and \( V_c/3 \) is also b.

Teacher's Note:
1. Expressions (1) and (3) are both b.
2. Check units to avoid confusing a and b.