States Of Matter

From PV and VT plots, higher temperature corresponds to higher volume at given pressure and vice versa. Analyze graph slopes accordingly.

Positive deviation from ideal behavior means compressibility factor at all pressures. From the graph, curves for b and c lie completely above the line , indicating that they show only positive deviation throughout.

For an ideal gas: Real gases deviate from ideal behavior due to:
- Intermolecular forces
- Finite volume of molecules
Hence:
- Real gas pressure is higher than ideal gas at the same volume, especially at high pressure or low temperature.
Thus, real gas curve lies above the ideal gas curve on a vs. graph.

The standard molar enthalpy of vaporization ( ) is the enthalpy change required to vaporize one mole of a liquid at its standard state (usually 1 bar pressure, specified temperature - often the normal boiling point). A higher enthalpy of vaporization indicates that more energy is required to overcome the intermolecular attractive forces in the liquid state and convert the liquid into a gas. Therefore, a higher generally implies stronger intermolecular forces. The question asks for the order of dipole-dipole attractive forces. While reflects the *total* strength of all types of intermolecular forces (London dispersion forces, dipole-dipole forces, hydrogen bonding), if we assume that dipole-dipole forces are a significant contributor and other factors are comparable or scale similarly, then a higher would suggest stronger dipole-dipole forces. This is a common simplifying assumption in such comparative questions unless other information (like molecular structure or polarity) is given. Given values: Liquid A: Liquid B: Liquid C: Arranging these values in decreasing order: (B)> (C)> (A). Since higher implies stronger intermolecular forces, the order of strength of these forces (and by inference, dipole-dipole forces if they are dominant or representative of the total) is: B>C>A. This matches option (a).

XeF has 2 bond pairs and 3 lone pairs on the central atom (Xe). Total 5 electron pairs form a trigonal bipyramidal arrangement. Since lone pairs are more than bond pairs on Xe, this satisfies the condition.

According to Graham's Law of Diffusion, the rate of diffusion ( ) of a gas is inversely proportional to the square root of its molar mass ( ) or its vapor density ( ), assuming constant temperature and pressure. or (since Molar Mass ). Rate of diffusion ( ) can be expressed as Volume diffused ( ) per unit time ( ): . Given: For gas A: Volume mL, time minute. Rate of diffusion of A: . For gas B: Volume mL, time minute. Rate of diffusion of B: . Let be the vapor density of gas A and be the vapor density of gas B. From Graham's Law: . We are given . We need to find . Substitute the rates of diffusion: . . So, . Square both sides: . . Now, solve for : . This matches option (c).

According to Graham’s law of diffusion: - Molar mass of H = 2
- Molar mass of D = 4
- Molar mass of T = 6
So: Rounded to:

Given: - Weight % of H = 20% - Weight % of He = 16% - So, O = 100 - 20 - 16 = 64% Let’s convert to moles: Total moles =
Partial pressure mole fraction:
As the question asks for , the correct order is:

Use the ideal gas equation:
Given: K, L, g, g/mol
mol
atm

Step 1: Assume 100 g of mixture.
Given that H is 33.33% by weight, then: Step 2: Convert masses to moles.
Step 3: Use mole ratio to find partial pressure ratio.
Total moles =
Step 4: Calculate the pressure ratio.

Step 1: Recall the Formulas for RMS Velocity and Most Probable Velocity
The root mean square (rms) velocity ( ) is given by: The most probable velocity ( ) is given by: Where:
= Ideal gas constant
= Absolute temperature in Kelvin
= Molar mass of the gas in kg/mol
Step 2: Convert Temperatures to Kelvin and Calculate Molar Masses
For SO : Temperature Molar mass of SO ( ): S = 32 g/mol, O = 16 g/mol. For O : Temperature Molar mass of O ( ): O = 16 g/mol. Step 3: Express x and y using the Velocity Formulas
Given is the rms velocity of SO : Given is the most probable velocity of O : Step 4: Calculate the Ratio x : y
Combine the square roots and cancel common terms ( and ): Simplify the terms inside the square root: Notice that is half of : Simplify the fraction: Therefore, the ratio is . Step 5: Analyze the Options
\begin{itemize} \item Option (1): 3 : 4. This matches our calculated ratio. \item Option (2): 4 : 3. Incorrect. \item Option (3): 2 : 3. Incorrect. \item Option (4): 3 : 2. Incorrect. \end{itemize}

\textbf{Step 1: Recall the Ideal Gas Law} The ideal gas law is given by: $ P V n R T n_1 T_1 n_2 T_2
\text{H}_2 m_A = 5.6 \, \text{g} M_A \text{H}_2 \text{H}_2 m_{\text{H}_2} = 1 \, \text{g} \text{H}_2 M_{\text{H}_2} = 2 \, \text{g mol}^{-1} \text{H}_2 \frac{n_1}{T_1} = \frac{n_2}{T_2} M_A 56 28 44 60 $
Incorrect — does not match the calculated value.

- Statement I is incorrect: Real gases do **not** behave ideally at high pressure and low temperature — they deviate more significantly. - Statement II is incorrect: At high pressures, the compressibility factor can be greater than 1 (due to repulsive forces), not just ≤1.