States of matter practice · Mixed difficulty

Option C (More number of gas molecules strike the surface per second) is correct because decreasing the volume increases the collision frequency with the walls.

Option A is correct. The gas constant R is approximately 1.987 cal K⁻¹ mol⁻¹.

Option C (${\mathrm{L}^2}\,\mathrm{atm\;mo}{\mathrm{l}^{ - 2}}\mathrm{\;and\;mo}{\mathrm{l}^{ - 1}}\mathrm{\;L}$) is correct because it represents the correct units derived from the van der Waals equation.

Option A ($Cl_2$ and $SO_2$) is correct because Dalton's Law is not obeyed when the gases react chemically with each other, and chlorine and sulfur dioxide react.

Option D (If both the Assertion and Reason are incorrect) is correct because ice melts slower at high altitude due to lower atmospheric pressure and temperature.

Option C is correct because the parameter 'a' in the van der Waals equation accounts for the intermolecular attractive forces between gas molecules.

Option C ($23\mathrm{ }\,mL$) is correct because it is calculated using Henry's Law, considering the partial pressure of hydrogen in the gaseous mixture.

Option A (${\left( {\partial V/\partial T} \right)_P} = K\;$) is correct because Charles's Law states that at constant pressure, the volume of a gas is directly proportional to its temperature, meaning the rate of change of volume with respect to temperature at constant pressure is a constant.

Option B (Temperature of the gas) is correct because the total kinetic energy of a gaseous assembly is directly proportional to its temperature.

Option C (606 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 C (4 dm³) is correct because when both pressure and temperature are doubled, the volume remains constant according to the ideal gas law.

Option C (0.50 atm) is correct because equal masses of nitrogen and ethylene result in equal moles, leading to equal partial pressures.

Option B (Same as that of hydrogen) is correct because the average kinetic energy of a gas molecule depends only on the temperature, and both gases are at the same temperature.

Option B is correct because the critical temperature is the temperature above which a gas cannot be liquefied by applying pressure alone.

Option A (3) is correct because the root mean square speed formula is used to solve for the unknown speed X.

Option C ($N\,s\,{m^{ - 2}}$) is correct because it represents the force per unit area per unit velocity gradient, which is the definition of viscosity coefficient.

Option B (Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I) is correct because while both statements are true, surface tension alone doesn't explain why the capillary fills completely.

Option D ($\mathrm{HI}$) is correct because, at the same conditions, the gas with the highest molar mass will have the least volume (since number of moles is the same).

Option A (Hole theory) is correct because the hole theory explains the effect of temperature on viscosity in liquids by considering the availability of "holes" for molecules to move into.

Option B ($Equal\mathrm{ }to\mathrm{ }1\;$) is correct because the compressibility factor (Z) for an ideal gas is always equal to 1.

Option D (Volume and shape of container; no intermolecular attractions) is correct because gases assume the volume and shape of their container and ideally have negligible intermolecular attractions.

Option C ($N{H_3}(g)$) is correct because $NH_3$ exhibits the strongest intermolecular forces (hydrogen bonding), leading to the greatest deviation from ideal behavior.

Option B ($\frac{{4 - R}}{{2R}}$) is correct because it is derived from the ideal gas law, solving for x after substituting the given values.

Option A is correct because, using Charles's Law (V1/T1 = V2/T2), and converting to Kelvin, the final temperature is calculated to be -26.6°C.

Option C (450 K) is correct because it uses the Gay-Lussac's Law ($P_1/T_1 = P_2/T_2$) to find the maximum temperature before the cylinder explodes.

Option B ($Sb{H_3}$ because of higher molecular weight) is correct because as the molecular weight increases down the group, the strength of the London dispersion forces increases, leading to a higher boiling point. $NH_3$ has hydrogen bonding which would normally make it have the highest boiling point, but SbH3 has a significantly higher molecular weight which overcomes the hydrogen bonding in NH3.

Option D (Statement II is correct but Statement I is incorrect) is correct because mean free path decreases as pressure increases, and the formula for mean free path is correctly stated.

Option B is correct because it accurately represents the increasing order of intermolecular attractive forces: dimethyl ether < ethanol < water, based on hydrogen bonding capabilities.

Option D (${V_t} = {V_0}\left( {1 + \frac{t}{{273}}} \right)$) is correct because it represents Charles's Law, which describes the relationship between volume and temperature.

Option A (${\mathrm{N}_2}\mathrm{O}$) is correct because it has the same molar mass as CO2, leading to the same rate of diffusion.