States Of Matter – JEE Main ChemistryPractice Questions & PYQs

The behavior of an ideal gas is governed by the equation $PM = dRT$, which defines the relationship between density, pressure, molar mass, and temperature. By rearranging this expression to isolate density, we arrive at $d = \left[ \frac{P M}{R} \right] \frac{1}{T}$, demonstrating that density is inversely proportional to temperature rather than linearly related. Because this reciprocal relationship dictates a hyperbolic curve, any graphical representation depicting density and temperature as having a straight-line dependency is physically incorrect.

The behavior of real gases is characterized by the compressibility factor, which is defined as $Z = \frac{PV}{nRT}$. Since the gases are at the same temperature and contain an equal number of moles, the ratio of their compressibility factors can be expressed as $\frac{Z_A}{Z_B} = \frac{P_A V_A}{P_B V_B}$.

By substituting the known relations $Z_A = 3Z_B$ and $V_A = 2V_B$ into this ratio, we get $3 = \frac{P_A \cdot 2V_B}{P_B \cdot V_B}$. Canceling the volume terms and simplifying leads to the relationship between the pressures of the two gases:
$2P_A = 3P_B$

The most probable velocity of gas molecules is defined by the relationship $V_{mp} = \sqrt{\frac{2RT}{M}}$, which indicates that velocity is determined by the square root of the ratio of temperature to molar mass. Calculating this factor for each gas gives $\sqrt{\frac{300}{28}} = \sqrt{10.7}$ for nitrogen at 300 K, $\sqrt{\frac{400}{32}} = \sqrt{12.5}$ for oxygen at 400 K, and $\sqrt{\frac{300}{2}} = \sqrt{150}$ for hydrogen at 300 K. Ordering these results shows that the values increase from nitrogen at 300 K to oxygen at 400 K to hydrogen at 300 K, which determines the respective positions of points I, II, and III.

Since an open vessel maintains constant pressure and the container volume is fixed, the ideal gas law implies that the product of the molar amount and the absolute temperature remains constant, expressed as $n_1T_1 = n_2T_2$. Converting the starting temperature of 27 °C to 300 K and recognizing that losing two-fifths of the air leaves three-fifths of the initial moles, we represent the final mole count as $n_2 = \frac{3}{5}n_1$. Substituting these values into the equality gives $n_1(300) = (\frac{3}{5}n_1)T_2$. Solving for $T_2$ by canceling the mole variables and computing the final value results in $T_2 = 300 \times \frac{5}{3}$, which gives a final temperature of 500 K.

The compression factor $Z$ for a gas following the state equation $P = \frac{RT}{V-b}$ can be derived by rearranging the terms into $PV = RT + Pb$, which simplifies to $Z = \frac{PV}{RT} = 1 + \frac{Pb}{RT}$. This expression represents a linear relationship where the slope of the $Z$ versus $P$ plot is defined by the constant term $\frac{b}{RT}$. As the temperature $T$ is held constant, the steepness of the slope is directly proportional to the van der Waals constant $b$, which accounts for the excluded volume occupied by the gas molecules. Because larger atoms have a greater excluded volume, the value of $b$ increases with atomic size; consequently, Xenon, having the largest atomic size among the given noble gases, possesses the highest $b$ value and generates the steepest increase in the plot.

The critical temperature is determined by the relationship $T_c = \frac{8a}{27Rb}$, which indicates that a higher critical temperature is achieved with a larger Van der Waals constant $a$ and a smaller excluded volume constant $b$. These parameters reflect the strength of intermolecular attractive forces and the effective volume of the particles, respectively. Among the available options, Krypton features the highest value for $a$ at 5.1 atmdm^6 mol^-2 and the lowest value for $b$ at 1.0 x 10^-2 dm^3 mol^-1. This specific combination maximizes the ratio of $a$ to $b$, making it the gas with the highest critical temperature.

The ideal gas law, $PV = nRT$, relates the pressure, volume, and temperature to the total number of moles, which in this case is $n = 0.5 + x$. Substituting the given values yields $200 \times 10 = (0.5 + x) \times R \times 1000$, which simplifies to $2 = (0.5 + x)R$ after dividing by 1000. Dividing both sides by $R$ gives $\frac{2}{R} = 0.5 + x$, and subtracting 0.5 from both sides results in $x = \frac{2}{R} - 0.5$, which simplifies to $x = \frac{4 - R}{2R}$ using a common denominator.

The behavior of a real gas is determined by the Van der Waals constants, where $a$ represents the strength of intermolecular attractive forces and $b$ denotes the excluded volume of the particles. According to the principles of gas behavior, $\text{Compressibility} \propto a$ and $\text{Compressibility} \propto \frac{1}{b}$, meaning that a larger $a$ value signifies stronger intermolecular attractions and lower occupied volume, while a smaller $b$ value facilitates greater compressibility. When comparing Gas B ($a=155.21, b=0.04136$) to Gas D ($a=155.21, b=0.4382$), Gas B is more compressible because it has a lower $b$ constant despite sharing identical attractive forces. Furthermore, between Gas A ($a=642.32, b=0.05196$) and Gas C ($a=431.91, b=0.05196$), Gas A exhibits stronger intermolecular attractions due to its significantly higher $a$ value, which leads it to occupy less volume than Gas C.

The ideal gas equation $PV = nRT$ can be rearranged to express density $d$ as $d = \frac{PM}{RT}$, where $P$ is pressure, $M$ is molar mass, $R$ is the gas constant, and $T$ is temperature. At constant temperature and pressure, density is directly proportional to molar mass, leading to the ratio $\frac{d_{NH_3}}{d_{HCl}} = \frac{M_{NH_3}}{M_{HCl}}$. The molar mass of ammonia is calculated as $14 + 3(1) = 17 \text{ g/mol}$, while the molar mass of hydrogen chloride is $1 + 35.5 = 36.5 \text{ g/mol}$. Substituting these values into the ratio gives $\frac{17}{36.5} \approx 0.4657$, which rounds to 0.46.

Real gases approximate ideal behavior under conditions where intermolecular forces are negligible and the finite volume of gas particles is insignificant, which typically occurs at low pressures, large volumes, or specifically at the Boyle temperature $T_B = \frac{a}{Rb}$. At very low temperatures, however, the decrease in molecular kinetic energy makes intermolecular attractive forces significant, leading to marked deviations from ideal behavior and often facilitating liquefaction. Therefore, the statement claiming that real gases exhibit ideal behavior at very low temperatures is incorrect because such conditions promote non-ideal characteristics rather than suppressing them.

The relationship between density $\rho$, pressure $P$, temperature $T$, and molar mass $M$ for an ideal gas is defined as:
$\rho = \frac{PM}{RT}$
Given that the density of the gaseous molecule $X$ is double that of $N_2$ at the same temperature, the relation $\rho_X = 2\rho_{N_2}$ becomes:
$\frac{P_X M_X}{RT} = 2 \left( \frac{P_{N_2} M_{N_2}}{RT} \right)$
Canceling the common terms $R$ and $T$ from both sides simplifies the expression to:
$P_X M_X = 2 P_{N_2} M_{N_2}$
Substituting the provided values $P_X = 2$ bar, $P_{N_2} = 4$ bar, and $M_{N_2} = 28$ g mol$^{-1}$ results in:
$2 M_X = 2(4 \times 28)$
Solving for $M_X$ gives:
$M_X = 4 \times 28 = 112 \text{ g mol}^{-1}$

The conservation of moles dictates that the total amount of gas in the system remains constant because the bulbs are connected and closed. Initially, each of the two bulbs contains a volume $V$ of gas at pressure $p_i$ and temperature $T_1$, giving a total molar amount of $n = \frac{p_iV}{RT_1} + \frac{p_iV}{RT_1} = \frac{2p_iV}{RT_1}$. After heating one bulb to temperature $T_2$, the pressure equalizes to a final value $p_f$ across the entire system, making the final molar content $n = \frac{p_fV}{RT_1} + \frac{p_fV}{RT_2}$. By setting the initial and final molar sums equal and canceling the common factors $V$ and $R$, we arrive at the equality $\frac{2p_i}{T_1} = p_f \left(\frac{T_1 + T_2}{T_1 T_2}\right)$. Rearranging this to isolate the final pressure results in $p_f = 2p_i \left(\frac{T_2}{T_1 + T_2}\right)$.

The behavior of real gases deviates from ideality as described by the Van der Waals equation for one mole, $(P + \frac{a}{V^2})(V - b) = RT$. Under conditions of very high pressure, the intermolecular attraction term $\frac{a}{V^2}$ becomes negligible compared to $P$, simplifying the expression to $P(V - b) = RT$. Expanding this leads to $PV - Pb = RT$, which can be rearranged as $PV = RT + Pb$. Dividing both sides by $RT$ expresses the compressibility factor as $Z = \frac{PV}{RT} = 1 + \frac{Pb}{RT}$.

The root mean square velocity of a gas is defined by the expression $v_{rms} = \sqrt{\frac{3RT}{M}}$, where $T$ represents the absolute temperature and $M$ is the molar mass. This relationship shows that the velocity is directly proportional to the square root of temperature and inversely proportional to the square root of molar mass. Initially, for nitrogen molecules with a molar mass of 28, the velocity is given by $u = \sqrt{\frac{3RT}{28}}$.

When the nitrogen molecules dissociate into nitrogen atoms, the molar mass of the particles is halved, changing from 28 to 14. Given that the temperature is simultaneously doubled to $2T$, we substitute these new values into the velocity formula to get $v_{new} = \sqrt{\frac{3R(2T)}{14}}$. Simplifying this expression results in $\sqrt{\frac{6RT}{14}}$, which can be further factored as $\sqrt{4 \times \frac{3RT}{28}}$. Extracting the square root of 4 shows that the new velocity is $2u$.

The van der Waals equation for one mole of gas is $(P + \frac{a}{V^2})(V - b) = RT$. Expanding this expression yields the product $PV = RT + Pb - \frac{a}{V} + \frac{ab}{V^2}$. At low pressure, the gas occupies a large volume, which causes the terms $Pb$ and $\frac{ab}{V^2}$ to become negligible when compared to $RT$. By simplifying the remaining expression to $PV \approx RT - \frac{a}{V}$ and dividing both sides by $RT$, we arrive at the compressibility factor $Z = 1 - \frac{a}{RTV}$.

The kinetic molecular theory of gases defines the motion of particles using three distinct speed expressions: the most probable speed $C^* = \sqrt{\frac{2RT}{M}}$, the average speed $\bar{C} = \sqrt{\frac{8RT}{\pi m}}$, and the root mean square speed $C = \sqrt{\frac{3RT}{M}}$. By comparing these expressions under the assumption that the mass terms refer to the same molar mass, the relationship is defined by the ratio $\sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3}$. Converting these radicals into decimal form results in values of approximately 1.414 to 1.596 to 1.732. Normalizing this sequence by dividing each term by 1.414 simplifies the proportionality to 1 : 1.128 : 1.225.

The Van der Waals equation, $(P + \frac{a}{V^2})(V - b) = RT$, accounts for the non-ideal behavior of real gases by correcting for molecular attractions and the finite volume occupied by particles. At high pressure, the system pressure becomes so dominant that the attractive force term $\frac{a}{V^2}$ is negligible, simplifying the expression to $P(V - b) = RT$. Expanding this to $PV - Pb = RT$ and dividing by $RT$ yields the expression for the compressibility factor $Z = \frac{PV}{RT} = 1 + \frac{Pb}{RT}$. This result, where $Z > 1$, demonstrates that the finite volume of gas particles makes the gas less compressible than an ideal gas would be under these conditions.

The ease of liquefying a gas depends fundamentally on the nature of its intermolecular interactions and the physical size of its molecules, which are quantified by the van der Waals constants $a$ and $b$. The constant $a$ serves as a measure of the magnitude of the attractive forces between gas molecules; a higher value of $a$ indicates stronger attractions, which effectively pull molecules closer together to promote the transition into a liquid phase. On the other hand, the constant $b$ corresponds to the excluded volume or the effective size of the gas particles, where a lower value of $b$ signifies that the molecules occupy less space and can pack more efficiently. Since chlorine is more easily liquefied than ethane, it follows that chlorine must possess stronger intermolecular attractions and a smaller excluded volume than ethane, meaning $a$ for $\text{Cl}_2$ is greater than $a$ for $\text{C}_2\text{H}_6$ while $b$ for $\text{Cl}_2$ is less than $b$ for $\text{C}_2\text{H}_6$.

The ideal gas law, $PV = nRT$, effectively describes the behavior of water vapor occupying the flask's volume at equilibrium. Because the vapor fills the entire $1.0 \mathrm{dm}^3$ capacity of the container, the volume in SI units is $10^{-3} \mathrm{m}^3$. Rearranging the ideal gas law to solve for the number of moles gives $n = \frac{PV}{RT}$. Substituting the vapor pressure of $3170 \mathrm{Pa}$, the volume of $10^{-3} \mathrm{m}^3$, the universal gas constant of $8.314 \mathrm{J} \mathrm{K}^{-1} \mathrm{mol}^{-1}$, and the temperature of $300 \mathrm{K}$ yields the expression:

$n = \frac{3170 \times 10^{-3}}{8.314 \times 300}$

This calculation results in $1.27 \times 10^{-3} \mathrm{mol}$ of water present in the vapor phase.

According to Dalton's Law of Partial Pressures, the portion of the total pressure contributed by any gas in a mixture is equal to its mole fraction. Given equal masses $m$ of methane and oxygen, we calculate their respective molar amounts using their molar masses of $16 \, \text{g/mol}$ and $32 \, \text{g/mol}$, resulting in $\frac{m}{16} \, \text{mol}$ of methane and $\frac{m}{32} \, \text{mol}$ of oxygen. The total moles of gas in the container is the sum of these values: $\frac{m}{16} + \frac{m}{32} = \frac{3m}{32}$. Dividing the moles of oxygen by the total moles gives the mole fraction as
$\frac{m/32}{3m/32} = \frac{1}{3}$
Thus, oxygen exerts one-third of the total pressure.