States of matter practice · Mixed difficulty

Option A (Boiling point of water involved in cooking is increased) is correct because a pressure cooker increases the pressure inside, which raises the boiling point of water, allowing food to cook at a higher temperature and thus faster.

Option A (150 K) is correct because the RMS velocity is inversely proportional to the square root of the molar mass and directly proportional to the square root of the temperature. Setting the RMS velocities equal allows solving for the unknown temperature.

Option A (0.39) is correct because critical volume is the inverse of critical density multiplied by the molar mass.

Option C is correct because oxygen has a higher critical temperature and a higher inversion temperature than hydrogen, making it easier to liquefy.

Option A ($2\mathrm{ }:\mathrm{ }1$) is correct because the pressure ratio is directly proportional to the mole ratio, which is derived from the mass and molar mass of each gas.

Option A ($0.25$) is correct because it is calculated using Raoult's Law, where the partial pressure of a component is equal to its mole fraction multiplied by its vapor pressure in the pure state.

Option B (Near the hydrogen chloride bottle) is correct because HCl has a higher molar mass and diffuses slower than ammonia.

Option B ($\mathrm{H}_2$) is correct because, according to Graham's law of diffusion, the rate of diffusion is inversely proportional to the square root of the molar mass. Hydrogen has the lowest molar mass among the given options.

Option C ($28$) is correct. The molecular weight is calculated using the ideal gas law, relating pressure, volume, temperature, and the number of moles.

Option A (${\mathrm{H}_2} > {\mathrm{N}_2} > {\mathrm{O}_2} > HBr$) is correct because the root mean square velocity is inversely proportional to the square root of the molar mass. Therefore, the gas with the lowest molar mass will have the highest RMS velocity.

Option B is correct because compressibility less than unity implies that the molar volume is less than the ideal molar volume (22.4 L at STP).

Option C ($25\mathrm{ }\,\,mL$) is correct because it is calculated using Charles's Law, which relates volume and temperature at constant pressure.

Option B is correct because vapors of a liquid can only exist below the critical temperature. Above the critical temperature, the substance exists as a gas.

Option A (Increase in average molecular speed) is correct because as temperature increases, the average kinetic energy of the gas molecules increases, leading to higher speeds and more frequent/forceful collisions with the container walls, thus increasing pressure.

Option A (298 K, 1 bar) is correct because the standard state is defined as 298 K (25°C) and 1 bar.

Option A (Contain average $KE$ proportional to absolute temperature) is correct because this is a fundamental postulate of the kinetic theory of gases.

Option A (Attract one another) is correct because at high pressure, the intermolecular distances are small, and the attractive forces between the gas molecules become significant, causing deviation from ideal behavior.

Option A (82.82 atm) is correct because it is calculated using the van der Waals equation with the given parameters.

Option A (${p_i} = {\chi _i}p_i^ \circ$) is correct because it represents Raoult's Law, not Dalton's Law. Dalton's Law relates partial pressure to the total pressure and mole fraction.

Option B is correct because, using the ideal gas law and the given information, the final temperature is calculated to be 500 K.

Option D (One litre of chlorine) is correct because chlorine has the highest molar mass among the given gases, and at NTP, equal volumes contain equal numbers of moles.

Option C is correct because it is the rearranged form of the van der Waals equation, explicitly solving for pressure (P).

Option D ($\frac{2}{{\sqrt \pi }}$) is correct because it represents the ratio of most probable velocity to average velocity.

Option A is correct. The correction factor 'a' in the ideal gas equation accounts for the forces of attraction between gas molecules.

Option A ($300\,\,\mathrm{m}{\mathrm{s}^{ - 1}}$) is correct because the RMS velocity can be calculated directly from the pressure and density using the formula $v_{rms} = \sqrt{\frac{{3P}}{\rho}}$.

Option A ($1.24 \times {10^2}\mathrm{J}$) is correct because it uses the formula for kinetic energy based on temperature and the number of moles of oxygen.

Option A (Inversion temperature) is correct because the Joule-Thomson coefficient is zero at the inversion temperature of a gas.

Option C (0.35 times that of He) is correct because the rate of diffusion is inversely proportional to the square root of the molar mass.

Option C ($8 \times {10^4}\,N/\,{m^2}$) is correct because it represents the partial pressure of oxygen calculated from its mole fraction in the mixture.

Option C (1/4) is correct because the compressibility factor is the ratio of the actual volume to the ideal volume.