JEE Chemistry States Of Matter MCQs Set 05

Answer: (c) 1 : 81

Explanation:
1. Graham's law: \(\frac{r_X}{r_Y}=\sqrt{\frac{d_Y}{d_X}}=9\)
2. So \(\frac{d_Y}{d_X}=81\)
3. Hence \(d_X:d_Y=1:81\)

Teacher's Note:
1. The faster gas is the lighter one, so its density is smaller. Square the rate ratio and invert it for the density ratio.

Answer: (c) 1 : 2

Explanation:
1. Molecular speed is proportional to \(\sqrt T\)
2. \(T_1=300\) K and \(T_2=1200\) K
3. \(\frac{v_1}{v_2}=\sqrt{\frac{300}{1200}}=\frac12\)

Teacher's Note:
1. Always convert temperatures to kelvin before using the speed formulas. A fourfold rise in temperature only doubles the speed.

Answer: (c) 1 atm

Explanation:
1. Let each gas weigh \(w\) g. Moles of the lighter gas \(=w/4\), of the heavier gas \(=w/40\)
2. Mole fraction of the lighter gas \(=\frac{1/4}{1/4+1/40}=\frac{10}{11}\)
3. \(p=\frac{10}{11}\times1.1=1\) atm

Teacher's Note:
1. Equal masses do not mean equal moles. Partial pressure equals mole fraction times total pressure.

Answer: (c) 0.5 L

Explanation:
1. For a fixed amount of gas \(V\propto T/P\)
2. Pressure doubles and temperature halves, so the volume becomes \(\frac{1}{2}\times\frac12=\frac14\) of the original
3. \(V=\frac{2}{4}=0.5\) L

Teacher's Note:
1. Use the combined gas law when both pressure and temperature change. Volume rises with temperature and falls with pressure.

Answer: (c) 1 : 1

Explanation:
1. \(v_{rms}\propto\sqrt{T/M}\)
2. \(\sqrt{50/2}=5\) for hydrogen and \(\sqrt{800/32}=5\) for oxygen
3. The ratio is 1 : 1

Teacher's Note:
1. Check the value of T/M for each gas before taking the ratio. Different gases at different temperatures can still have the same speed.

Answer: (a) 1340 m/s

Explanation:
1. \(v_{rms}=\sqrt{\frac{3RT}{M}}\) with \(M=4\times10^{-3}\) kg/mol and \(T=300\) K
2. \(v_{rms}=\sqrt{\frac{3\times8\times300}{4\times10^{-3}}}=\sqrt{1.8\times10^{6}}\)
3. \(\approx1340\) m/s

Teacher's Note:
1. Molar mass must be in kg/mol when R is in J/(K mol). Forgetting this changes the answer by a factor of about 30.

Answer: (c) 1.9 atm

Explanation:
1. At constant volume \(P_1/T_1=P_2/T_2\)
2. \(T_1=273\) K, \(T_2=523\) K, \(P_1=1\) atm
3. \(P_2=\frac{523}{273}=1.9\) atm

Teacher's Note:
1. Gay-Lussac's law: pressure is proportional to absolute temperature at constant volume. The volume 2.5 L is not needed.

Answer: (c) He, \(\ce{CH4}\) and \(\ce{NH3}\)

Explanation:
1. X rises linearly above the ideal line from the start, so repulsion (molecular size) dominates throughout: this is He, which has almost no attraction
2. Y dips slightly below the ideal line at low pressure and then rises, so it has weak attraction: \(\ce{CH4}\)
3. Z dips deepest, so it has the strongest attraction: \(\ce{NH3}\)
4. Order X, Y, Z = He, \(\ce{CH4}\), \(\ce{NH3}\)

Teacher's Note:
1. A dip below Z = 1 means attractive forces dominate; a rise above 1 means molecular volume dominates. The stronger the attraction, the deeper the dip.

Answer: (d) \(Z\neq1\) : for 10 moles of an ideal gas

Explanation:
1. For an ideal gas \(Z=PV/nRT=1\) at all pressures, whatever the number of moles
2. \(Z<1\) means attractions dominate and the gas is more compressible; \(T_c=\frac{8a}{27Rb}\) and \(P_c=\frac{a}{27b^2}\) are correct
3. Only the last match is wrong

Teacher's Note:
1. For an ideal gas Z is exactly 1 for any amount. Remember the critical constants: \(V_c=3b\), \(P_c=a/27b^2\), \(T_c=8a/27Rb\).

Answer: (d) All of these

Explanation:
1. Start with \(\left(P+\frac{an^2}{V^2}\right)(V-nb)=nRT\)
2. Divide by \(V-nb\): \(P+\frac{an^2}{V^2}=\frac{nRT}{V-nb}\)
3. Rearranging gives the second and third forms, so all three are the same equation

Teacher's Note:
1. The pressure term corrects for attraction and the volume term corrects for molecular size. Algebraically equivalent forms are all correct.