States Of Matter – JEE Main ChemistryPractice Questions & PYQs

Determining the molar mass of an ideal gas trapped in a vessel requires first calculating the volume of the container and then applying the gas laws. Since the vessel has a fixed volume, we use the known density of the liquid to determine the space it occupies. Subtracting the mass of the empty glass vessel, which is 40.0 g, from the mass of the vessel filled with liquid, 135.0 g, tells us that the mass of the liquid is 95.0 g. Given the liquid density of 0.95 g/mL, the internal volume of the vessel is 95.0 g divided by 0.95 g/mL, which equals 100 mL, or 0.1 L.

With the volume established, we determine the mass of the ideal gas by calculating the difference between the vessel filled with gas, 40.5 g, and the empty vessel, 40.0 g, resulting in 0.5 g of gas. We then relate these parameters using the ideal gas equation $PV = nRT$, where the number of moles $n$ is represented as the mass $m$ divided by the molar mass $M$, yielding the expression $PV = \frac{m}{M}RT$. Substituting the given values into the equation $0.82 \times 0.1 = \frac{0.5}{M} \times 0.082 \times 250$ allows us to isolate $M$. Solving this equality, we find that $M = 125$ g/mol.

The behavior of ideal gases is governed by the ideal gas equation, which relates pressure, volume, temperature, and the amount of substance. By determining the total moles of hydrogen gas present and applying this relationship, we can calculate the total volume occupied by the gas under the specific conditions provided.

Calculate the number of moles of hydrogen gas by dividing the mass of 2.0 g by its molar mass of 2 g/mol, resulting in 1 mole. Using the ideal gas law:
$PV = nRT$
Rearranging this to solve for volume gives:
$V = \frac{nRT}{P}$
Substituting the 1 mole, the gas constant of 0.083, the temperature of 300 K, and the pressure of 1 bar into the equation results in:
$V = \frac{1 \times 0.083 \times 300}{1} = 24.9$
This total volume is 24.9 L. Dividing this by the 2.5 g of platinum powder yields 9.96 L per gram, which converts to 9960 mL per gram.

The ideal gas law establishes that for a fixed pressure and volume, the product of the amount of substance in moles and the absolute temperature remains constant. Because both gas A and the hydrogen sample share the same pressure and volume, we can equate the product of moles and temperature for both gases using the relation $n_A T_A = n_H T_H$.

For gas A, the number of moles is $n_A = \frac{3.0}{M}$, where $M$ is the molar mass in $\text{g mol}^{-1}$, and the temperature is $300\ \text{K}$. This gives the relationship $\frac{3.0}{M} \times 300 = \frac{PV}{R}$. For hydrogen, the number of moles is $n_H = \frac{0.2}{2.0} = 0.1\ \text{mol}$, and the temperature is $200\ \text{K}$, which implies $0.1 \times 200 = \frac{PV}{R}$. Equating these two expressions leads to $\frac{3.0}{M} \times 300 = 0.1 \times 200$. Solving this simplifies to $\frac{900}{M} = 20$, and dividing $900$ by $20$ results in a molar mass of $45\ \text{g mol}^{-1}$.

London dispersion forces arise from the interaction of instantaneous dipoles created by random fluctuations in electron distribution, which causes the potential energy between two particles to decay rapidly as they move apart. This physical behavior is governed by an inverse sixth-power law, expressed mathematically as $E \propto r^{-6}$. Comparing this standard relationship to the given form $r^x$, it is clear that the exponent $x$ is $-6$.

To determine the volume of oxygen adsorbed per gram of platinum, one applies the ideal gas equation $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is temperature. First, calculate the number of moles of oxygen by dividing the mass of oxygen by its molar mass, which gives $n = \frac{3.12}{32} = 0.0975$ mol. Using the ideal gas law to calculate the total volume $V$ at $P = 1$ atm and $T = 300$ K, with $R = 0.0821$ L atm K$^{-1}$ mol$^{-1}$, results in $V = \frac{nRT}{P} = \frac{0.0975 \times 0.0821 \times 300}{1} = 2.4014$ L. Finally, dividing this total volume by the 1.2 g of platinum used yields the volume per gram, which is $\frac{2.4014}{1.2} \approx 2$ L/g.

The behavior of an ideal gas is fundamentally described by the ideal gas equation, $pV = nRT$. Because the temperature remains constant, the product of the number of moles, the gas constant, and the absolute temperature constitutes a fixed numerical value. As a result, the product $pV$ does not change as the pressure $p$ is varied, manifesting as a horizontal line on a graph representing $pV$ against $p$.

The ideal gas law $PV = nRT$ implies that for a fixed volume and constant ambient temperature, the pressure inside the cylinder is directly proportional to the mass of the gas. By determining the mass of the gas at both states, we can relate the pressure to the remaining quantity of LPG. The initial mass of the gas is $29.0 \text{ kg} - 14.8 \text{ kg} = 14.2 \text{ kg}$, which corresponds to an initial pressure of $3.47 \text{ atm}$. When the total weight of the cylinder decreases to $23.0 \text{ kg}$, the mass of the remaining gas is $23.0 \text{ kg} - 14.8 \text{ kg} = 8.2 \text{ kg}$.

Using the proportional relationship between pressure and mass, where $\frac{P_1}{m_1} = \frac{P_2}{m_2}$, we can set up the following equation:
$\frac{3.47}{14.2} = \frac{P_2}{8.2}$
Calculating the final pressure yields $P_2 = \frac{3.47 \times 8.2}{14.2}$, which equals approximately $2.003 \text{ atm}$. Rounding to the nearest integer, the final pressure inside the cylinder is $2 \text{ atm}$.

The behavior of chlorine gas under standard conditions can be accurately described by the ideal gas law $PV=nRT$, where the number of moles is determined by the relationship $n = \frac{PV}{RT}$. By substituting the pressure of 1 bar, the volume of 0.020 L, the temperature of 273 K, and the specified gas constant of 0.083, we calculate the molar quantity of the diatomic chlorine gas. Since each mole of chlorine molecules contains two chlorine atoms, the total number of atoms is found by multiplying this molar amount by two and then by Avogadro's number, 6.023 x 10^23, which yields approximately 1.06 x 10^21 chlorine atoms.

In any physically meaningful equation like the van der Waals relation $\left(p+\frac{a n^{2}}{V^{2}}\right)(V-n b)=n RT$, terms added together must share identical units according to the principle of dimensional homogeneity. Since the term $\frac{a n^2}{V^2}$ is added directly to the pressure $p$, it must possess the same units as pressure, which is atm. Rearranging the expression to isolate $a$ gives the formula $a = p \frac{V^2}{n^2}$, and by substituting volume in dm^3 and amount in mol, we determine the unit is atm dm^6 mol^-2.

The Ideal Gas Law, expressed as $PV = nRT$, defines the relationship between state variables and allows for the determination of the amount of chlorine present. Using the values of $P = 1\ \mathrm{atm}$, $V = 0.02\ \mathrm{L}$, $T = 273\ \mathrm{K}$, and $R = 0.083\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}$ \in this equation determines the number of moles $n$ of $\mathrm{Cl_2}$ gas.

With the molar amount established, the number of molecules is found by multiplying by Avogadro's constant, $6.023 \times 10^{23}$, which yields approximately $5.3 \times 10^{20}$ molecules. Because each $\mathrm{Cl_2}$ molecule is composed of two chlorine atoms, multiplying this amount by two results in $1.06 \times 10^{21}$ total atoms, which rounds to the integer $1$.

Because the volume of a car tire remains constant, the gas inside follows Gay-Lussac's Law, which states that pressure is directly proportional to its absolute temperature ($P \propto T$). Before applying this relationship, convert the initial temperature from 27 degrees Celsius to Kelvin by adding 273, which results in 300 K.

Using the relationship for a fixed volume, we can set up the following equation:
$\frac{P_1}{T_1} = \frac{P_2}{T_2}$
Substituting the known values of 35 psi, 300 K, and 40 psi into this expression yields:
$\frac{35}{300} = \frac{40}{T_2}$
Solving for $T_2$ gives approximately 342.86 K. Converting this absolute temperature back to Celsius by subtracting 273 results in approximately 69.86 degrees Celsius, which rounds to the nearest integer of 70 degrees Celsius.

The ideal gas law, $PV = nRT$, links the variables of an ideal gas, allowing us to compute the volume occupied by $4.75\,\mathrm{g}$ of acetylene. Converting the provided conditions to appropriate units yields a temperature of $323\,\mathrm{K}$ and a pressure of $\frac{740}{760}\,\mathrm{atm}$. The amount of substance in moles is calculated by dividing the mass by the molar mass of acetylene ($26\,\mathrm{g/mol}$), resulting in $n = \frac{4.75}{26}\,\mathrm{mol}$.

Rearranging the equation to $V = \frac{nRT}{P}$ allows for substitution of the known quantities: $n = \frac{4.75}{26}\,\mathrm{mol}$, $R = 0.0826\,\mathrm{L}\,\mathrm{atm}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$, $T = 323\,\mathrm{K}$, and $P = \frac{740}{760}\,\mathrm{atm}$. Evaluating the expression $V = \frac{\frac{4.75}{26} \times 0.0826 \times 323}{\frac{740}{760}}$ yields approximately 5.005 liters, which rounds to an integer value of 5 liters.

Since an LPG cylinder maintains a constant volume, the relationship between its internal gas pressure and temperature is governed by Gay-Lussac's Law, which dictates that pressure is directly proportional to absolute temperature. To calculate the bursting point, first convert the initial temperature of $27^{\circ}\mathrm{C}$ to $300\text{ K}$ and express the initial pressure of $300\text{ kPa}$ as $300 \times 10^3\text{ Pa}$ to match the units of the bursting pressure, $1.2 \times 10^6\text{ Pa}$. By applying the relationship $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ and substituting the known values into the equation, we get $\frac{300 \times 10^3}{300} = \frac{1.2 \times 10^6}{T_2}$. Simplifying this expression results \in a final temperature of $1200\text{ K}$, which corresponds to $927^{\circ}\mathrm{C}$ after converting from the Kelvin scale.

The state of an ideal gas is governed by the equation $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is temperature. To determine the mass of the gas used, we first convert the volume of 4000 cubic meters into liters, which results in $4.00 \times 10^6$ liters. Using the rearranged formula $n = \frac{PV}{RT}$, we substitute the known values to find the total moles: $n = \frac{1.0 \times 4.00 \times 10^6}{0.083 \times 300}$. Since the molar mass of methane is 16 grams per mole, the total mass is calculated by multiplying these moles by 16, resulting in $\frac{1.0 \times 4.00 \times 10^6 \times 16}{0.083 \times 300}$ grams. This calculation yields approximately $25.7 \times 10^5$ grams, which rounds to an integer value of 26 for $x$.

The compressibility factor $Z$ provides a measure of how significantly a gas deviates from ideal behavior and is defined by the ratio $\frac{p V_{m}}{R T}$. When a gas follows the equation of state $p(V_{m}-b)=R T$, we rearrange this expression to define the behavior of $Z$.

Expanding the equation gives $p V_{m} - p b = R T$, which simplifies to $p V_{m} = R T + p b$. Dividing by $R T$ gives the expression for the compressibility factor:
$Z = 1 + \frac{p b}{R T}$
Differentiating this with respect to $p$ while keeping $T$ constant yields:
$\left(\frac{\partial Z}{\partial p}\right)_{T} = \frac{b}{R T}$
By comparing this to the given value $\frac{x b}{R T}$, we identify that $x=1$.

The pressure exerted by a mixture of non-reacting ideal gases is calculated using the ideal gas law, $PV = nRT$, where the total pressure is directly proportional to the total number of moles of gas present in the container. Since the gases do not react, the total pressure is the sum of the partial pressures contributed by each gas species.

Calculating the mole counts reveals $0.4\ \mathrm{mol}$ of methane and $0.2\ \mathrm{mol}$ of carbon dioxide, resulting in a total of $0.6\ \mathrm{mol}$ of gas. Converting the temperature to absolute units, $27^{\circ}\ \mathrm{C}$ is equivalent to $300\ \mathrm{K}$. Applying the ideal gas law formula $P = \frac{nRT}{V}$ with a volume of $10\ \mathrm{L}$ and the gas constant $R = 8.314\ \mathrm{J}\ \mathrm{mol}^{-1}\ \mathrm{K}^{-1}$ yields $P = \frac{0.6 \times 8.314 \times 300}{10}$. This calculation results in approximately $149.652\ \mathrm{kPa}$, which rounds to an integer value of $150\ \mathrm{kPa}$.

Dalton's Law of Partial Pressures provides the framework for this calculation, stating that the pressure exerted by any single gas in a mixture is the product of its mole fraction and the total pressure of the container. Because the mixture contains 1 mole each of hydrogen, helium, and oxygen, there are 3 moles of gas in total, which makes the mole fraction of hydrogen $1/3$. With the partial pressure of hydrogen given as 2 atm, we can define the relationship as $2 = \left( \frac{1}{3} \right) \times P_T$ where $P_T$ represents the total pressure. Solving this equation demonstrates that the total pressure of the gases in the cylinder is 6 atm.

The Maxwell-Boltzmann distribution curve illustrates the spread of molecular velocities in a gas, with the horizontal axis representing increasing speed. Given that the magnitudes of these velocities follow the mathematical order $V_{\text{mp}} = \sqrt{\frac{2RT}{M}} < V_{\text{av}} = \sqrt{\frac{8RT}{\pi M}} < V_{\text{rms}} = \sqrt{\frac{3RT}{M}}$, the labels must be positioned sequentially along the x-axis from the lowest value to the highest. Consequently, point A located at the curve's peak corresponds to the most probable speed, point B represents the intermediate average speed, and point C on the far right signifies the highest root mean square speed.

Boyle's Law dictates that when the temperature and the amount of gas stay constant, the product of pressure and volume remains fixed, which is expressed as $P_1 V_1 = P_2 V_2$. For a spherical balloon, the volume is determined by the radius such that the ratio of volumes is equivalent to the cube of the ratio of the radii. By defining the initial pressure as $P_1 = 48 \times 10^{-3} \text{ bar}$ and utilizing radii $r_1 = 3 \text{ cm}$ and $r_2 = 12 \text{ cm}$, the final pressure $P_2$ is found using the relation $P_2 = P_1 (\frac{r_1}{r_2})^3$, which leads to the calculation:

$P_2 = (48 \times 10^{-3}) \times \left(\frac{3}{12}\right)^3$

Simplifying this expression results in $P_2 = (48 \times 10^{-3}) \times (\frac{1}{64})$, which calculates to $P_2 = 7.5 \times 10^{-4} \text{ bar}$. Converting this value into the final required unit scale gives $P_2 = 750 \times 10^{-6} \text{ bar}$.

The ideal gas law $PV = nRT$ is used to determine the number of moles of oxygen gas consumed by a person daily, where $P=1 \text{ atm}$, $V=492 \text{ L}$, $R=0.082 \text{ L atm mol}^{-1} \text{ K}^{-1}$, and $T=300 \text{ K}$. Substituting these values, the number of moles $n$ is calculated as $n = \frac{1 \times 492}{0.082 \times 300} = \frac{492}{24.6} = 20 \text{ mol}$ of $\mathrm{O}_2$.

The balanced chemical equation $\mathrm{NaClO}_3(s) + \mathrm{Fe}(s) \rightarrow \mathrm{O}_2(g) + \mathrm{NaCl}(s) + \mathrm{FeO}(s)$ indicates that the molar ratio between $\mathrm{NaClO}_3$ and $\mathrm{O}_2$ is 1:1. Consequently, 20 moles of $\mathrm{NaClO}_3$ are required to produce 20 moles of $\mathrm{O}_2$. Given that the molar mass of $\mathrm{NaClO}_3$ is $23 + 35.5 + (3 \times 16) = 106.5 \text{ g/mol}$, the required mass is $20 \text{ mol} \times 106.5 \text{ g/mol} = 2130 \text{ g}$.