States of matter practice · Mixed difficulty

Option B (4) is correct because it is derived from equating the root mean square speed of gas X to the most probable speed of gas Y and solving for the molecular weight of gas Y.

Option A (${d_2} = \frac{{{P_2}{T_1}{d_1}}}{{{P_1}{T_2}}}$) is correct because it correctly relates density, pressure, and temperature for an ideal gas based on the ideal gas law.

Option C (Mixture of gases) is correct because Van der Waals' forces are intermolecular forces that exist between all types of molecules, including those in a mixture of gases.

Option D is correct because the average kinetic energy of a gas depends only on the temperature and is independent of the mass of the gas molecules.

Option D ($\frac{3}{8}$) is correct because the compression factor Z at critical conditions is defined as $\frac{p_cV_c}{RT_c}$ which equals 3/8 for van der Waals gases.

Option A (Unchanged) is correct because the reaction between equal volumes of H2 and Cl2 produces HCl, and the total volume of gases remains the same.

Option D is correct because vapor pressure increases with temperature as more molecules gain enough kinetic energy to escape into the gas phase.

Option A ($\frac{1}{6}$) is correct because the ratio of pressures is inversely proportional to the ratio of densities and directly proportional to the ratio of molecular weights.

Option B (1) is correct because the compressibility factor (Z) for an ideal gas is always equal to 1, as it perfectly obeys the ideal gas law.

Option A (1.0043) is correct because it is calculated using Graham's law of diffusion, considering the molar masses of the isotopes.

Option B (4) is correct because equating the RMS speed of gas X to the most probable speed of gas Y and solving for the molecular weight of Y yields 4.

Wet air is lighter than dry air because water vapor (H2O) has a lower molar mass than the average molar mass of dry air (mostly N2 and O2). The reason is correct that dry air is denser than water, but this doesn't explain why wet air is lighter.

Option D (${p_B} + {\chi _A}\left( {{p_A} - {p_B}} \right)$) is correct because it represents the total pressure of an ideal binary solution according to Raoult's law.

Option A (3.5 atm) is correct because when gases are mixed in the same volume, the total pressure is the sum of the individual pressures (Dalton's Law).

Option B ($2{P_0}$) is correct because pressure is proportional to the number of molecules and the square of their speed, so halving the mass and doubling the speed results in a doubling of the pressure.

Option C (22 atm) is correct because the pressure is directly proportional to the number of moles of gas at constant volume and temperature.

Option B (${V_m} < 22.4\,L$) is correct because a compressibility factor less than unity at STP indicates that the real gas volume is less than the ideal gas volume.

Option B ($Sb{H_3}$ because of higher molecular weight) is correct because, among the given options, $SbH_3$ has the highest molecular weight, leading to stronger van der Waals forces and thus a higher boiling point.

Option C ($1\mathrm{ }g\mathrm{ }{L^{ - 1}}$) is correct because the density is calculated using the ideal gas law in the form density = (PM)/(RT).

Option A (Forces of attraction between the gas molecules) is correct because the van der Waals constant 'a' accounts for the intermolecular forces of attraction between gas molecules.

Option C (03:04) is correct because the ratio of kinetic energies is directly proportional to the product of moles and temperature.

Option C (4.0) is correct because it is calculated using Graham's law of diffusion, relating the rate of diffusion to the square root of the molar mass.

Option C is correct because hot air balloons rise due to the principle of Charles' Law, where volume increases with temperature at constant pressure.

Option C (4) is correct because the pressure ratio is directly proportional to the ratio of densities and inversely proportional to the ratio of molecular weights, resulting in a ratio of 4.

Option A (Boiling point increases with increasing pressure) is correct because increasing the pressure in a pressure cooker raises the boiling point of water, allowing the beans to cook at a higher temperature and thus faster.

Option D ($546\,\,\mathrm{K}$) is correct because it is calculated using Charles's Law.

Option A (Boyle’s temperature) is correct because Boyle's temperature is the temperature at which a real gas exhibits ideal behavior over a considerable range of pressure.

Option A ($\sqrt{{2}}: 1$) is correct because the most probable speed is inversely proportional to the square root of the molar mass.

Option B (600 mL) is correct because the new volume is calculated using Charles's Law, which states that volume is directly proportional to temperature at constant pressure.

Option A ($5 \mathrm{~dm}^3$) is correct because it is the volume of SO3 that will diffuse under the given conditions, calculated using Graham's law of diffusion.