States of matter practice · Medium

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 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 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 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.

Option A ($1:1.128:1.224$) is correct because the ratio of most probable speed, average speed, and RMS speed is derived from their respective formulas.

The correct answer is C (921.58 m / s) because the average velocity is calculated from the root mean square velocity using the appropriate formula.

Option D (All of these) is correct because the ratio a/b in van der Waals' equation can be expressed in multiple equivalent units, including atm litre mol$^{-1}$, atm dm$^3$ mol$^{-1}$, and dyne cm mol$^{-1}$.

Option D (Increases both the average kinetic energy and rms velocity, but not significantly) is correct because the temperature change is small relative to the absolute temperature, leading to only a slight increase in both average kinetic energy and rms velocity.

Option B (1.73) is correct because the root mean square speed is proportional to the square root of the temperature, so tripling the temperature increases the speed by a factor of $\sqrt{3}$ ≈ 1.73.

Option C ($u_{ {r.m.s. }}=\sqrt{\frac{2 E}{M}}$) is correct because the root mean square velocity is related to the average kinetic energy by the formula E = (1/2) * M * u_rms².

Option B (5.26%) is correct because it is the percentage increase in pressure required to decrease the volume by 5% according to Boyle's Law.

Option D (All of these) is correct because Boyle's law, Charles's law, and Avogadro's law can all be derived from the kinetic theory of gases.

Option A (960) is correct because the total kinetic energy is calculated by summing the kinetic energies of each gas and dividing by the total number of moles.

Option A (Benzo) is correct because the benzo group ($\mathrm{C_6H_5C<}$) has three valencies available for bonding, making it trivalent.

Option A (4.32 litres) is correct because it is the original volume calculated using Charles's Law, which relates volume and temperature at constant pressure.

Option B (${CCl}_4$ and ${CH}_4$) is correct because instantaneous dipole-induced dipole interactions (London dispersion forces) exist between non-polar molecules.

Option D ($81.8 \%$) is correct because it is calculated by finding the fraction of nitrogen remaining and subtracting it from 1 to find the fraction escaped.

Option C is correct. A gas can be easily liquefied by applying pressure when it is cooled to below its critical temperature. Above the critical temperature, no amount of pressure can liquefy the gas.

Option D (As temperature increases, pressure increases at constant volume) is correct because Gay-Lussac's law states that pressure is directly proportional to temperature at constant volume.

Option B (40) is correct because the percentage of air escaped is calculated using Charles's Law, and the result is 40%.

Option A ($\mathrm{N}_1$) is correct because the number of gas molecules remains the same when the kinetic energy doubles and the temperature doubles.