JEE Chemistry States Of Matter MCQs Set 17

Answer: (c) 4 g/mol

Explanation:
1. For an ideal gas \( d=\dfrac{pM}{RT} \), so \( \dfrac{d}{p}=\dfrac{M}{RT} \), independent of p (hence the horizontal line).
2. \( M=\dfrac{d}{p}RT=\dfrac{8}{49.26}\times0.0821\times300=4.0 \) g/mol.

Teacher's Note:
1. 49.26 = 2 x 24.63 = 2RT, so M = 8/2 = 4.
2. The gas is helium.

Answer: (a) 1

Explanation:
1. \( p=\dfrac{nRT}{V}=\dfrac{10\times0.0821}{0.0821}\,T=10T \) (V = 0.0821 L).
2. \( \log p=1+1\cdot\log T \): slope a = 1, y-intercept c = 1.
3. At \( \log p=0 \): \( \log T=-1 \), so b = -1.
4. a + b + c = 1 - 1 + 1 = 1.

Teacher's Note:
1. Isochoric: p is proportional to T, giving slope 1 on a log-log plot.
2. The x-intercept is the value of log T at which log p = 0.

Answer: (b) (C) and (D) only

Explanation:
1. At constant volume \( p\propto T \), so pressure falls with temperature and extrapolates to zero at 0 K (not infinity).
2. At constant pressure \( V\propto T \), so the volume extrapolates to zero at 0 K.
3. Hence (C) and (D) are true; (A) and (B) are false.

Teacher's Note:
1. Real gases liquefy before reaching 0 K.
2. Both lines start from the origin of the absolute scale.

Answer: (b) (B) and (C) only

Explanation:
1. At constant V: \( p=\dfrac{nR}{V}T \), so \( pT=\dfrac{nR}{V}T^2 \): a straight line (not a parabola) against \( T^2 \). (A) is false.
2. At constant p: \( V=\dfrac{nR}{p}T \), so \( VT\propto T^2 \): a parabola against T. (B) is true.
3. \( p/T=nR/V \) is constant, a horizontal straight line against \( T^2 \). (C) is true.

Teacher's Note:
1. A graph y = kx^2 against x is a parabola, but y against x^2 is a straight line.
2. Always substitute the gas law first.

Answer: (a) (A), (B) and (C) are correct, (D) is wrong

Explanation:
1. Boyle's law: pV = constant at fixed T, so pV against V or against p is horizontal.
2. p = constant x (1/V): a straight line through the origin.
3. A line with a positive slope for pV against V contradicts Boyle's law.

Teacher's Note:
1. The product pV = nRT is fixed when n and T are fixed.
2. A hyperbola in p-V becomes a straight line in p versus 1/V.

Answer: (c) (C) a straight line of log p against log(1/V) at 135 degrees

Explanation:
1. Boyle's law: \( p=\dfrac{k}{V} \), so \( \log p=\log k+\log\dfrac1V \): a straight line of slope +1, i.e. an angle of 45 degrees.
2. An angle of 135 degrees means slope -1, which would imply p proportional to V; this does not follow Boyle's law.

Teacher's Note:
1. tan 45 = 1 and tan 135 = -1.
2. pV = k gives a horizontal line against V.

Answer: (d) \( A=v_{mp},\ B=v_{av},\ C=v_{rms} \)

Explanation:
1. \( v_{mp}=\sqrt{2RT/M}:v_{av}=\sqrt{8RT/\pi M}:v_{rms}=\sqrt{3RT/M}=1:1.128:1.224 \).
2. The line through the peak is v_mp, the next is v_av and the largest is v_rms.

Teacher's Note:
1. Order: v_mp < v_av < v_rms.
2. The curve is not symmetric; the tail to the right pulls the average up.

Answer: (b) 19 cm

Explanation:
1. Because the inner level is 5 cm below the outer surface, the gas pressure is \( 70+5=75 \) cm of Hg.
2. Boyle's law: \( 70\times20=75\times L \), so \( L=18.7\approx19 \) cm.

Teacher's Note:
1. If the inner level were above the outer one the gas pressure would be below 70 cm of Hg.
2. Pressure at the same level in a connected liquid is the same.

Answer: (b) \( \dfrac{h(\ell_1+\ell_2)}{(\ell_2-\ell_1)} \)

Explanation:
1. Open end up: \( P_1=P_0+h \); open end down: \( P_2=P_0-h \).
2. Boyle's law: \( (P_0+h)\ell_1=(P_0-h)\ell_2 \).
3. \( P_0(\ell_2-\ell_1)=h(\ell_1+\ell_2) \), so \( P_0=\dfrac{h(\ell_1+\ell_2)}{\ell_2-\ell_1} \).

Teacher's Note:
1. The gas expands when the tube is turned open end down, so \( \ell_2>\ell_1 \).
2. Check: h = 10, l1 = 15, l2 = 25 gives P0 = 10 x 40/10 = 40 cm.

Answer: (d) The partial pressure of each gas becomes 6 times its initial value

Explanation:
1. At constant T and n, pressure is inversely proportional to volume: V falls from 30 L to 5 L, a factor of 6.
2. Every partial pressure and the total pressure (24 atm) become 6 times larger.
3. Mole fractions remain unchanged because the composition is the same.

Teacher's Note:
1. Partial pressure = mole fraction x total pressure.
2. Total pressure becomes 24 atm.

Answer: (c) 4 atm

Explanation:
1. Moles: N2 3, O2 2. O2 is limiting: 2 mol O2 uses 2 mol N2 and forms 4 mol NO.
2. Left: 1 mol N2 + 4 mol NO = 5 mol, the same as before.
3. The total moles do not change in this reaction, so the pressure stays 4 atm.

Teacher's Note:
1. Equal moles of gas on both sides of the equation.
2. At constant V and T, pressure follows total moles.

Answer: (d) III - (iii) - P

Explanation:
1. I-(i)-R: \( \sqrt{4}=2 \) correct. III-(iv)-S: \( \sqrt{\tfrac94\times\tfrac14}=\tfrac34 \) correct. IV-(iv)-P: \( \sqrt{\tfrac49\times\tfrac14}=\tfrac13 \) correct.
2. III-(iii)-P: \( \tfrac23\times\sqrt{\tfrac94}=\tfrac23\times\tfrac32=1 \), not 1/3: incorrect.

Teacher's Note:
1. The combination whose computed ratio differs from the column-3 value is the answer.
2. 1/3 is the value for P.

Answer: (c) (A), (C) and (D) only

Explanation:
1. In a fixed volume each gas exerts its own pressure, so adding C does not change the partial pressures of A and B (A is true).
2. Before: 4 mol; after: 5 mol, so the pressure ratio before : after = 4/5, not 3/5 (B is false).
3. After = 20 x 5/4 = 25 atm (C true); p(C) = 25 - 20 = 5 atm (D true).

Teacher's Note:
1. Dalton's law of partial pressures.
2. Pressure is proportional to the total number of moles at constant V and T.

Answer: (b) (B) and (D) only

Explanation:
1. Graham's law: \( r_{He}/r_X=\sqrt{M_X/4} \).
2. CO: \( \sqrt{28/4}=\sqrt7=2.65 \) (B true, A false). CO2: \( \sqrt{11}=3.32 \) (C false). SO2: \( \sqrt{64/4}=4 \) (D true).

Teacher's Note:
1. 8.65 would be the ratio of M for He : CO inverted incorrectly.
2. The ratio uses the square root of molar masses.

Answer: (d) (B) and (C) only

Explanation:
1. At the same temperature helium atoms are lighter and so faster (\( v\propto1/\sqrt M \)).
2. Faster atoms hit the wall more often.
3. Pressure is the sum of partial pressures; with equal moles, both gases exert the same partial pressure, so (D) is false.

Teacher's Note:
1. The mean kinetic energy is the same for both gases at one temperature.
2. Equal moles give equal partial pressures.

Answer: (a) \( M_2 \) and \( M_5 \)

Explanation:
1. Initial volume = 8 + 12 = 20 mL of \( M_x \), final volume 8 mL of \( M_y \).
2. Equal atoms: \( 20x=8y \), so \( x:y=2:5 \).
3. The simplest formulae are M2 and M5.

Teacher's Note:
1. Avogadro: volumes are proportional to moles of molecules.
2. Atoms are conserved: 20 x = 8 y.