States Of Matter – JEE Main ChemistryPractice Questions & PYQs

According to Dalton's law of partial pressures, the ratio of the partial pressures of two gases in a mixture is equivalent to the ratio of their molar amounts, expressed as $\frac{P_A}{P_B} = \frac{n_A}{n_B}$. Assuming a 100 g sample, the number of moles for each gas is calculated using the molar masses of 28, 32, and 40 g/mol respectively: $n_{N_2} = \frac{70}{28} = 2.5$ mol, $n_{O_2} = \frac{27}{32} = 0.84375$ mol, and $n_{Ar} = \frac{3}{40} = 0.075$ mol. The ratio $\frac{P_{N_2}}{P_{O_2}}$ is given by $\frac{2.5}{0.84375} \approx 2.96$, and the ratio $\frac{P_{O_2}}{P_{Ar}}$ is given by $\frac{0.84375}{0.075} = 11.25$, which aligns with the values in the correct option.

The behavior of a gaseous mixture relies on Dalton's Law of Partial Pressures, which states that the pressure exerted by a specific component is the product of its mole fraction and the total pressure of the system. To determine the contribution of gas $X$, we calculate the number of moles for each gas by dividing the given mass by its respective molar mass.

For gas $X$, the molar amount is $\frac{0.6}{20} = 0.03$ mol, while for gas $Y$, it is $\frac{0.45}{45} = 0.01$ mol. The total amount of gas in the mixture is $0.03 + 0.01 = 0.04$ mol. Consequently, the mole fraction of $X$ is defined by the ratio $\frac{0.03}{0.04} = 0.75$. Applying the relationship $P_X = \chi_X \cdot P_T$, the partial pressure is calculated as $0.75 \times 740$, which results in 555 mm Hg.

The Andrews isotherm for $CO_2$ demonstrates that the substance exists exclusively as a gas from point (a) until it reaches point (b), confirming that the first statement is correct. Throughout the region between points (b) and (c), the system undergoes liquefaction, creating a phase equilibrium where liquid and gaseous $CO_2$ coexist, which validates the third statement. Since the compression process during this interval converts gaseous $CO_2$ into the liquid phase, the amount of liquid increases as the volume decreases, meaning the claims that the amount of liquid decreases or that it only appears at point (c) are both incorrect.

Kinetic molecular theory defines the root mean square speed of a gas as $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$ and the most probable speed as $v_{\text{mp}} = \sqrt{\frac{2RT}{M}}$. When the root mean square speed of gas $X$ at 600 K is equal to the most probable speed of gas $Y$ at 90 K, the relationship is given by the equality $\sqrt{\frac{3RT_X}{M_X}} = \sqrt{\frac{2RT_Y}{M_Y}}$. Substituting the given values $T_X = 600$ K, $M_X = 40$ gmol^-1, and $T_Y = 90$ K, and canceling the gas constant $R$, we arrive at the equation $\frac{3 \times 600}{40} = \frac{2 \times 90}{M_Y}$. Simplifying this result leads to $45 = \frac{180}{M_Y}$, which indicates that the molar mass of gas $Y$ is 4 gmol^-1.

The compressibility factor, denoted as $Z$, quantifies the deviation of real gases from ideal behavior and is fundamentally defined by the equation $PV = ZRT$. At point $A$ on the plot, the product of pressure and volume for a real gas is lower than that of an ideal gas, indicating that attractive intermolecular forces are dominant and the value of $Z$ is less than one. Under these specific conditions for one mole of gas, the mathematical expression derived from the van der Waals relationship simplifies to $Z = 1 - \frac{a}{RTV}$.

The van der Waals constant $a$ quantifies the strength of intermolecular forces of attraction between gas molecules, which inherently increases with higher molecular mass and larger molecular size. While simple noble gases like $Ar$ exhibit relatively weak attractions, the trend rises as we consider the larger $CH_4$ molecule and the significant dipole-dipole interactions arising from hydrogen bonding in $H_2O$. Ultimately, the large, complex structure of $C_6H_6$ results in the strongest attractive forces among these options, establishing the increasing sequence of $Ar < CH_4 < H_2O < C_6H_6$.

The compressibility factor, denoted by $Z$, characterizes the deviation of a real gas from ideal behavior and is defined by the relationship $Z = \frac{PV}{nRT}$. Since the amount of gas remains constant throughout the process, the product $nR$ is invariant, allowing us to equate the ratio $\frac{PV}{ZT}$ for both states as $\frac{P_1 V_1}{Z_1 T_1} = \frac{P_2 V_2}{Z_2 T_2}$. Rearranging this expression to isolate the unknown volume $V_2$ gives $V_2 = \frac{P_1 V_1 T_2 Z_2}{P_2 T_1 Z_1}$. By substituting the given values $P_1 = 100 \, \text{atm}$, $V_1 = 0.15 \, \text{dm}^3$, $T_1 = 500 \, \text{K}$, $Z_1 = 1.07$, $P_2 = 300 \, \text{atm}$, $T_2 = 300 \, \text{K}$, and $Z_2 = 1.4$, the equation becomes $V_2 = \frac{100 \times 0.15 \times 300 \times 1.4}{300 \times 500 \times 1.07}$. Simplifying this expression leads to $V_2 = \frac{0.03 \times 1.4}{1.07}$, which calculates to approximately $0.03925 \, \text{dm}^3$. Converting this into the required format results in $392 \times 10^{-4} \, \text{dm}^3$.

When dealing with a gas at a constant temperature, Boyle's Law dictates that pressure and volume are inversely related, meaning the product of pressure and volume remains constant as $P_1 V_1 = P_2 V_2$. Assigning an initial volume of 100 units simplifies the percentage change calculation; a 40% decrease leaves a final volume of 60 units. Substituting the initial pressure of 940.3 mm Hg into the governing relationship produces

$940.3 \times 100 = P_2 \times 60$

Evaluating the final pressure $P_2$ leads to approximately 1567.16 mm Hg, which rounds to an integer value of 1567 mm Hg.

The total amount of carbon dioxide in the drink is determined using the relation $n = M \times V$, where the molarity is the concentration of the dissolved gas. Converting the bottle volume to 0.3 liters and using the given concentration of 0.2 molar results in the following calculation for the number of moles:
$0.2 \times 0.3 = 0.06$

To find the corresponding volume of the gas at STP, you multiply the number of moles by the given molar volume of 22.7 liters per mole:
$0.06 \times 22.7 = 1.362$

Since this result is in liters, converting the value to milliliters by multiplying by 1000 yields 1362 milliliters.

When multiple gas bulbs are connected at a constant temperature, the total number of moles remains conserved, allowing us to equate the sum of the initial pressure-volume products to the final pressure and total volume product. Based on the conservation of moles, the system follows the relationship $P_f V_f = P_1 V_1 + P_2 V_2 + P_3 V_3$ where $V_f$ is the combined volume of all connected bulbs. Inserting the specific values provided for the three gases, the calculation becomes $P_f \times (2 + 3 + 4) = (2 \times 2) + (4 \times 3) + (3 \times 4)$ which simplifies to $P_f \times 9 = 4 + 12 + 12$. Solving for the final system pressure results \in $P_f = \frac{28}{9} = 3.11$, yielding a nearest integer value of 3.

The pressure of a moist gas is defined as the sum of the partial pressure of the dry gas and the vapor pressure of water. Since the vapor pressure of water depends only on the temperature, it remains constant when the volume of the container changes, meaning Boyle's Law applies exclusively to the dry gas component of the mixture.

Starting with the initial conditions at $27^{\circ}\mathrm{C}$, we subtract the vapor pressure of $0.4 \text{ atm}$ from the total moist gas pressure of $4 \text{ atm}$ to find the partial pressure of the dry gas, which is $3.6 \text{ atm}$. When the volume is doubled from $V_1$ to $2V_1$, the partial pressure of this dry gas decreases by half according to the relationship $P_1 V_1 = P_2 V_2$, resulting in a new dry gas pressure of $1.8 \text{ atm}$. Adding the constant water vapor pressure of $0.4 \text{ atm}$ back to this value yields a total moist gas pressure of $2.2 \text{ atm}$. Converting this into the required format of $\times 10^{-1} \text{atm}$ gives $22 \times 10^{-1} \text{atm}$, making the final numerical value 22.

To understand how pressure relates to density for an ideal gas, we begin with the ideal gas law, $PV = nRT$. By substituting density $d = \frac{m}{V}$ into the equation, we derive $P = d \frac{RT}{M}$, showing that for a constant molar mass and temperature, pressure is directly proportional to density. This results in a linear relationship where the graph passes through the origin with a slope equal to $\frac{RT}{M}$. Since this slope is directly dependent on temperature, higher temperatures yield steeper lines, which is why a graph with increasing temperature values exhibits a corresponding increase in slope.

Dalton's Law of Partial Pressures states that the pressure contributed by a specific gas in a mixture equals its mole fraction multiplied by the total system pressure, expressed as $P_{\text{gas}} = \chi_{\text{gas}} \times P_{\text{total}}$. Given a mixture that is 40% hydrogen by mass, we can evaluate a 100 g sample containing 40 g of hydrogen and 60 g of oxygen. Converting these masses to moles by dividing by the respective molar masses of 2 g/mol and 32 g/mol yields 20 mol of hydrogen and 1.875 mol of oxygen. The mole fraction of hydrogen, denoted as $\chi_{H_2}$, is found by dividing the moles of hydrogen by the sum of all moles in the mixture, resulting in the calculation $\chi_{H_2} = \frac{20}{20 + 1.875}$. This fraction equals approximately 0.914. Applying this to the total pressure of 2.2 bar provides a partial pressure of approximately 2.01 bar, which rounds to the nearest integer of 2 bar.

The behavior of an ideal gas is described by the ideal gas equation $PV = nRT$, which relates pressure, volume, temperature, and the amount of gas present. Since the number of moles equals the mass divided by the molar mass, denoted by $M$, we can set up the calculation to find this unknown value. Begin by converting the temperature of 27 degrees Celsius into Kelvin by adding 273, which results in 300 K.

Inserting the provided values into the equation yields the expression $1.5 \times 416 = \frac{100}{M} \times 0.083 \times 300$. Multiplying the pressure and volume on the left side results in 624, while simplifying the terms on the right side gives 2490 divided by $M$. Solving for $M$ by dividing 2490 by 624 results in approximately 3.99, which rounds to the nearest integer of 4.

In a mixture of non-reacting gases, Dalton's Law dictates that the total pressure is the sum of the individual partial pressures. Given that the total pressure in the cylinder is 25 bar and the partial pressure exerted by neon is 20 bar, the partial pressure of the oxygen must be 5 bar.

The partial pressure of any component gas is equal to its mole fraction multiplied by the total pressure. We first determine the moles of neon as $n_{\mathrm{Ne}} = 200/20 = 10$ mol, while the moles of oxygen are $n_{\mathrm{O}_2} = x/32$ mol. The definition of partial pressure leads to the following equation:
$5 = \left( \frac{x/32}{x/32 + 10} \right) \times 25$
Dividing both sides by 25 yields the equality $\frac{1}{5} = \frac{x/32}{x/32 + 10}$. Cross-multiplying results in the expression:
$\frac{x}{32} + 10 = \frac{5x}{32}$
Subtracting $x/32$ from both sides gives $10 = 4x/32$, which simplifies to $10 = x/8$. Multiplying both sides by 8 confirms that $x = 80$.

Since the nitrogen tank is rigid and sealed, the volume and amount of gas remain constant, which means the pressure of the gas is directly proportional to its absolute temperature. This relationship is governed by Gay-Lussac's Law, defined by the equation
$\frac{P_1}{T_1} = \frac{P_2}{T_2}$
where the pressure is measured in atmospheres and temperature must be expressed in Kelvin. Converting the Celsius temperatures to Kelvin by adding 273.15, we find the initial temperature $T_1 = 300$ K and the final temperature $T_2 = 318$ K. Substituting the known initial pressure $P_1 = 30$ atm into the relationship gives
$\frac{30}{300} = \frac{P_2}{318}$
Solving this calculation reveals $P_2 = 31.8$ atm, which rounds to an integer value of 32 atm.

To determine the molar mass of a gaseous compound, we start with the ideal gas law $PV=nRT$. By substituting the definition of moles as mass $w$ divided by molar mass $M$, the expression is rewritten as $PV=\frac{wRT}{M}$. Since density $d$ is defined as mass per unit volume, we rearrange the variables to isolate the molar mass as $M=\frac{dRT}{P}$.

Before performing the calculation, ensure that all variables align with the provided gas constant $R=0.082\,\text{L}\,\text{atm}\,\text{K}^{-1}\,\text{mol}^{-1}$. The temperature must be converted to the Kelvin scale by adding 273 to the Celsius value, resulting in $T=257+273=530\,\text{K}$. Similarly, convert the pressure from mm Hg to atmospheres by dividing by 760, giving $P=\frac{100}{760}\,\text{atm}$. Using the provided density value of 0.46 g L⁻¹ into the formula yields $M=\frac{0.46 \times 0.082 \times 530}{100/760}$. Solving this expression results in a molar mass of 152 g mol⁻¹.

When a real gas is subjected to high pressure, the volume occupied by the gas molecules becomes significant, making the excluded volume term dominant in the equation of state. Under these specific conditions, the compressibility factor is expressed by the relation:
$Z = 1 + \frac{Pb}{RT}$
Given that the compressibility factor $Z$ is 2 at a pressure $P = 99$ bar and temperature $T = 298$ K, we substitute these values directly into the equation. Setting $Z = 2$ leads to $2 = 1 + \frac{Pb}{RT}$, which simplifies to $1 = \frac{Pb}{RT}$.

Rearranging this equality to isolate the van der Waals constant $b$ gives the expression:
$b = \frac{RT}{P}$
Substituting the known values of $R = 0.083$ L bar K$^{-1}$ mol$^{-1}$, $T = 298$ K, and $P = 99$ bar, the calculation becomes $b = \frac{0.083 \times 298}{99}$. Solving this yields $b = 0.25$ L mol$^{-1}$. Since the final result is required in the form of $\times 10^{-2}$ L mol$^{-1}$, we represent $0.25$ as $25 \times 10^{-2}$, identifying the final value as 25.

The total number of moles $n$ is calculated using the ideal gas law $PV = nRT$. Using the given pressure, volume, and temperature, the molar amount is:
$n = \frac{PV}{RT} = \frac{6 \times 10^5 \times 0.0125}{8.3 \times 300} = 3$
The mixture contains hydrogen with a molar mass of 2 g/mol and helium with a molar mass of 4 g/mol. If $x$ denotes the mass of helium in grams, the mass of hydrogen is $10 - x$. Balancing the individual moles of each gas yields the equation:
$\frac{x}{4} + \frac{10 - x}{2} = 3$
Multiplying by 4 simplifies the equation to $x + 2(10 - x) = 12$, which expands to $x + 20 - 2x = 12$. Solving for $x$ confirms that the mass of helium in the mixture is 8 grams.

Since propane behaves as an ideal gas, the relationship between its pressure, volume, temperature, and the amount of gas present is governed by the ideal gas equation $PV = nRT$. First, determine the number of moles of propane by dividing the given mass by its molar mass, where $C_3H_8$ corresponds to a molar mass of $3(12) + 8(1) = 44\;\mathrm{g/mol}$, resulting in $n = 11\;\mathrm{g} / 44\;\mathrm{g/mol} = 0.25\;\mathrm{mol}$. To ensure consistency with the units of the gas constant, convert the volume to $2 \times 10^{-3}\;\mathrm{m}^3$ and the pressure to $2 \times 10^6\;\mathrm{Pa}$. Solving the equation for absolute temperature gives:

$T = \frac{PV}{nR} = \frac{2 \times 10^6 \times 2 \times 10^{-3}}{0.25 \times 8.3}$

This calculation results in approximately $1927.7\;\mathrm{K}$. Subtracting $273$ to convert this value to the Celsius scale yields approximately $1654.7^{\circ}\mathrm{C}$, which rounds to $1655^{\circ}\mathrm{C}$.