Daily Olympiad: Chemistry - States of Matter [20260701]

Solution

Correct: D

Real gases behave most like ideal gases when the intermolecular forces between molecules are negligible and the volume occupied by the gas molecules themselves is negligible compared to the total volume of the container. These conditions are best met at low pressure (molecules are far apart, reducing interactions) and high temperature (high kinetic energy overcomes attractive forces, and the overall volume of the gas is large, making molecular volume relatively insignificant).

Solution

Correct: B

Critical temperature (Tc) is the temperature above which a gas cannot be liquefied, no matter how much pressure is applied. A higher Tc indicates stronger intermolecular attractive forces, meaning the gas is more easily liquefiable. Since Tc(Y) (304 K) is higher than Tc(X) (190 K), Gas Y has stronger intermolecular forces and is more easily liquefiable than Gas X. Therefore, option B is correct. Stronger intermolecular forces are represented by a larger 'a' value in the Van der Waals equation, so Gas Y would have a higher 'a' constant.

Solution

Correct: B

According to Dalton's Law of Partial Pressures, the partial pressure of a gas in a mixture is equal to its mole fraction multiplied by the total pressure. Total moles of gas = Moles of A + Moles of B = 2 moles + 3 moles = 5 moles. Mole fraction of gas B (XB) = Moles of B / Total moles = 3 moles / 5 moles = 0.6. Partial pressure of gas B (PB) = XB × Total pressure (P_total) = 0.6 × 5 atm = 3 atm.

Solution

Correct: A

The ideal gas law is based on two main assumptions: (1) gas molecules have negligible volume compared to the container volume, and (2) there are no intermolecular forces between gas molecules. Real gases deviate from ideal behavior because, in reality, gas molecules do occupy a finite volume (accounted for by the 'b' parameter in the Van der Waals equation) and there are attractive and repulsive forces between them (accounted for by the 'a' parameter). High kinetic energy (iii) actually promotes ideal behavior by helping molecules overcome attractive forces, and negligible volume (iv) is an ideal assumption, not a cause of deviation. Thus, the primary reasons for deviation are (i) and (ii).

Solution

Correct: A

Boiling point is primarily determined by the strength of intermolecular forces (IMFs). 1. CH4 (Methane): Nonpolar molecule, only exhibits weak London Dispersion Forces (LDF). Its molar mass is 16 g/mol. 2. H2S (Hydrogen Sulfide): Polar molecule, exhibits LDF and Dipole-Dipole interactions. Its molar mass is 34 g/mol. 3. H2O (Water): Highly polar molecule, exhibits LDF, Dipole-Dipole interactions, and very strong Hydrogen Bonding due to the presence of highly electronegative oxygen bonded to hydrogen. Its molar mass is 18 g/mol. Hydrogen bonds are significantly stronger than dipole-dipole interactions, which in turn are stronger than LDF (for comparable molecular sizes, though H2O also has stronger LDF than CH4 due to more electrons). Despite H2S having a higher molecular weight than H2O, the extensive hydrogen bonding in water leads to a much higher boiling point. Thus, the order of increasing boiling points is CH4 < H2S < H2O.

Solution

Correct: C

Let's analyze each statement: A) Viscosity of a liquid decreases with an increase in temperature: Correct. Increased kinetic energy allows molecules to overcome intermolecular forces more easily, leading to less resistance to flow. B) Viscosity is primarily caused by intermolecular forces in liquids: Correct. Stronger attractive forces between molecules impede their movement relative to each other. C) Viscosity of gases decreases with an increase in temperature: Incorrect. Viscosity of gases increases with an increase in temperature. In gases, viscosity is due to momentum transfer between layers of gas moving at different speeds. At higher temperatures, molecules move faster and collide more frequently, leading to greater momentum transfer and thus increased viscosity. D) Stronger intermolecular forces generally lead to higher viscosity in liquids: Correct. More attractive forces mean more resistance to flow.

Solution

Correct: A

A negative slope for the solid-liquid coexistence line in a phase diagram indicates that the melting point of the substance decreases as the external pressure increases. This is a unique characteristic of substances that expand upon freezing (i.e., the solid phase is less dense than the liquid phase, like water or bismuth). When pressure is applied, it generally favors the denser phase. If the liquid is denser than the solid, increasing pressure shifts the equilibrium towards the liquid state, thereby lowering the temperature at which the solid can melt (or liquid can freeze). So, 'the substance expands on freezing' is the correct implication.

Solution

Correct: A

In the Van der Waals equation: - The 'a' parameter accounts for intermolecular attractive forces. A very large 'a' indicates very strong attractive forces between gas molecules. - The 'b' parameter accounts for the finite volume occupied by the gas molecules. A very small 'b' implies that the molecular volume is negligible. When 'a' is very large, the attractive forces are very strong. These strong attractive forces pull molecules closer together, reducing the effective pressure exerted by the gas. This makes the gas more compressible than an ideal gas at moderate pressures. The presence of strong attractive forces also means significant deviation from ideal behavior (which assumes no attractive forces). While 'b' being small would lead to more ideal behavior concerning volume, the large 'a' (strong attractive forces) is the dominant factor for deviation. Therefore, the gas deviates significantly from ideal behavior due to strong attractive forces, and this facilitates compression.

Solution

Correct: D

Real gases behave most like ideal gases when the intermolecular forces between molecules are minimized and the volume occupied by the gas molecules themselves is negligible compared to the total volume of the container. These conditions are best met at low pressure (molecules are far apart, reducing interactions) and high temperature (high kinetic energy overcomes attractive forces, and the overall volume of the gas is large, making molecular volume relatively insignificant).

Solution

Correct: B

Let's analyze the postulates: A) Gas particles are in continuous, random motion: This is a fundamental postulate of KMT, applicable to both ideal and real gases. B) The volume occupied by the gas molecules themselves is negligible compared to the total volume of the container: This is a key postulate specific to *ideal* gases. Real gas molecules do occupy a finite volume, which becomes significant at high pressures, causing deviation from ideal behavior. C) The average kinetic energy of gas molecules is directly proportional to the absolute temperature: This is a fundamental postulate of KMT and holds true for both ideal and real gases. D) Collisions between gas molecules are perfectly elastic, and there is no net loss of kinetic energy: This is also a fundamental postulate of KMT. The question asks for a statement that is NOT a postulate *or* is true only for ideal gases. Option B falls into the latter category, as it is an ideal gas assumption that is not strictly true for real gases, especially under non-ideal conditions.

Solution

Correct: C

Boiling point is directly related to the strength of intermolecular forces (IMFs). Both H2O and H2S are polar molecules and exhibit London Dispersion Forces and Dipole-Dipole interactions. However, water has a unique ability to form extensive hydrogen bonds due to the high electronegativity of oxygen and the small size of hydrogen. Hydrogen bonds are significantly stronger than the dipole-dipole interactions present in H2S (sulfur is less electronegative than oxygen, resulting in weaker bond polarity and thus weaker hydrogen bond donor/acceptor ability, if any significant H-bonding occurred at all for H2S, which it doesn't to a large extent). Therefore, the much stronger hydrogen bonding in water requires significantly more energy to overcome during boiling, leading to its exceptionally high boiling point compared to H2S, even though H2S has a higher molar mass.

Solution

Correct: B

The compressibility factor Z indicates deviation from ideal behavior. For most gases, at moderate pressures, attractive forces dominate, leading to Z < 1. However, for very light gases like H2 and He, their intermolecular attractive forces (Van der Waals 'a' parameter) are extremely weak. At room temperature, the kinetic energy of these molecules is high enough to largely overcome these weak attractive forces. In such cases, the finite volume occupied by the gas molecules themselves (Van der Waals 'b' parameter) becomes the dominant factor causing deviation. The excluded volume effect leads to a higher effective pressure or volume, making the actual volume (V) larger than the ideal volume, thus Z = PV/nRT > 1. This means significant repulsive forces arise when molecules are pushed close together due to their finite size. Therefore, option B is the primary reason for Z > 1 for H2 at these conditions.

Solution

Correct: B

Let's recall the formulas: Average Kinetic Energy per molecule (KE_avg) = (3/2)kT, where k is Boltzmann's constant and T is the absolute temperature. KE_avg depends *only* on the absolute temperature and is independent of the mass or type of gas molecule. Root Mean Square (RMS) speed (u_rms) = √(3RT/M), where R is the gas constant, T is the absolute temperature, and M is the molar mass of the gas. u_rms depends on *both* the absolute temperature and the molar mass of the gas. Therefore, statement B is the correct description of their relationship.

Solution

Correct: B

Boiling is the process where a liquid changes into a vapor throughout the bulk of the liquid, not just at the surface. This occurs when the vapor pressure of the liquid becomes equal to the external pressure acting on the liquid's surface. If the external pressure (e.g., atmospheric pressure) is decreased, the liquid needs to achieve a lower vapor pressure to start boiling. Liquids produce lower vapor pressures at lower temperatures. Therefore, decreasing the external pressure lowers the boiling point of the liquid. A common example is water boiling at a lower temperature at high altitudes where atmospheric pressure is lower.

Solution

Correct: B

The critical constants (critical temperature Tc, critical pressure Pc, and critical volume Vc) are specific properties of a real gas at its critical point. These can be derived from the Van der Waals equation by setting the first and second derivatives of pressure with respect to volume to zero at the critical point. The derived expression for the critical temperature is Tc = 8a / 27Rb. This is a standard formula that NEET aspirants are expected to know for real gases.