States of matter practice · Mixed difficulty

Option C ($533\,\,\mathrm{K}$) is correct because the kinetic energy of a gas is directly proportional to the number of moles and temperature, allowing us to equate the kinetic energies of He and Ar.

Option D ($1/\sqrt d$) is correct because the root mean square velocity is inversely proportional to the square root of the density at constant pressure.

Option B ($0\,^\circ C,2{\mkern 1mu} {\mkern 1mu} \mathrm{atm}$) is correct because density is highest at low temperature and high pressure.

The correct answer is 10.09 mL. This is calculated by finding the volume occupied by the molecules themselves, using the radius and Avogadro's number.

Option B (${\left( {\partial P/\partial V} \right)_T} = - K/{V^2}$) is correct because it represents the derivative of pressure with respect to volume at constant temperature, according to Boyle's Law.

Option D (The product of pressure and volume always remains constant) is correct because Boyle's law states that for a given mass of an ideal gas at constant temperature, the product of pressure and volume is constant.

Option C (1) is correct because the ratio of RMS speeds is calculated using the formula and the given temperatures and molar masses, resulting in a ratio of 1.

Option B is correct because both statements are true, but Statement II doesn't explain Statement I.

Option C (unchanged on changing temperature) is correct because the mean free path depends on the number density of the molecules and their size. In a rigid container, the number density remains constant, and the size of the molecules doesn't change significantly with temperature.

Option D (Vapour pressure) is correct because vapour pressure increases with increasing temperature due to higher kinetic energy of molecules.

Option A (If both A and R are correct and R is the correct explanation of A) is correct because hot air balloons rise due to the hot air being less dense than the surrounding cooler air, which is exactly what the reason states.

Option B (0.082 litre-atmosphere $\mathrm{K}^{-1} \mathrm{~mol}^{-1}$) is correct because it represents the correct value and units for the ideal gas constant R when using liters, atmospheres, Kelvin, and moles.

Option A (${A}^{2+}, {B}^{3+}$) is correct because, in a spinel structure, the oxide ions have a -2 charge, and to maintain charge neutrality, A must have a +2 charge and B must have a +3 charge.

Option C ($\;127\,^\circ C$ and $1\,\,\mathrm{atm}$) is correct because molar volume is maximum at high temperature and low pressure according to the ideal gas law.

Option C (${\mathrm{H}_2}{\mathrm{F}_2}$) is correct because the molar mass calculated from the given volume at NTP is twice the empirical formula mass.

Option C (${H_2}$) is correct because the mean free path is inversely proportional to the pressure and directly proportional to the temperature and the size of the molecule. Since P and T are constant, the smallest molecule, hydrogen, will have the longest mean free path.

Option A (Extremely lower pressure) is correct because at extremely low pressures, the volume occupied by the gas molecules becomes negligible compared to the total volume, and intermolecular forces become insignificant, thus approaching ideal behavior.

Option B (Avogadro’s hypothesis) is correct because Avogadro's hypothesis states that equal volumes of all gases, at the same temperature and pressure, have the same number of molecules, leading to the molar volume concept.

Option A (Both statements are correct and statement-II is correct explanation of statement-I) is correct because the positive slope of the compressibility factor for hydrogen at 0°C is due to the dominance of repulsive forces, even at low pressures.

Option D (Kinetic energy of molecules remains the same) is correct because at constant temperature, the average kinetic energy of gas molecules remains constant.

Option B is correct because the SI unit of viscosity coefficient is Newton-second per square meter, which is $Ns{m^{ - 2}}$.

Option D is correct because both the assertion and the reason are incorrect. Greater critical temperature implies easier liquefaction, and stronger intermolecular forces lead to higher critical temperatures.

Option D (${\mathrm{N}_2}$) is correct because nitrogen constitutes approximately 78% of the Earth's atmosphere, making it the gas with the highest partial pressure.

Option B is correct because the most probable velocity is proportional to $\sqrt{T}$, average velocity is proportional to $\sqrt{T}$, and RMS velocity is proportional to $\sqrt{T}$, with different constants, leading to the temperature relationship.

Option D ($Z = \frac{{p{V_m}}}{{RT}} = 1 - \frac{a}{{RT}}$) is correct because at low pressure, the volume term in the van der Waals equation becomes negligible, simplifying the equation to this form.

Option A (Zero) is correct because the Joule-Thomson coefficient for an ideal gas is zero, indicating no temperature change during throttling.

Option D is correct because it represents the reduced equation of state derived from the van der Waals equation using reduced variables.

The correct answer is Manometer. A manometer is a device used for measuring gaseous pressure based on the principle of balancing the pressure of the gas against the pressure exerted by a column of liquid.

Option D (Hydrogen bonding) is correct because the two strands of DNA are held together by hydrogen bonds between complementary base pairs (adenine with thymine, and guanine with cytosine).

Option B is not true because the square of the mean speed is not equal to the mean squared speed, they are related by a constant factor.